20200626, 22:34  #1 
Jun 2020
26_{8} Posts 
New {function, criterion, theorem, conjecture} for discriminant of congruence
New function, new criterion, new theorem and new conjecture for discriminant of congruence
Zhou CongYao College of Information Science and Engineering, Hunan University Yu Wei Department of statistics & financial engineering,College of mathematics and statistics,ningbo university TangXiaoNing Beijing Haitian Start Technology Service Co., Ltd. ABstruct In this paper,a new function of congruence discrimination is proposed,namely,the upper bound estimation funtion of the number of solutions of the congruence. If the value of this function is 1,the posttive integer A is noncongruent,this is the new N1 criterion.Using this guideline, to history all 3 congruence negative theorms,a very simple elementary proof is given;and proved the author's newly discovered three modular 8type congruence negative theorems; At the same time,eight new congruences negation theorems(not model 8 types) are proved,by the way,we derive 1511 infinite sequences of noncongruent If A is a congruence number,the value of this function is integer power of 2,that is,there are at least two types of solutions for congruence,one of them must be Fermat type,and each type has an infinite number of solutions. Using this upper bound function, some new theorems for the number of congruence solutions are derived: the exact value of the solution is 2, 4, 8, 16, 32; On this basis, a new conjecture is put forward: If the type of the congruence solution is used as an element, then these elements form a group,and the order of this group is integer power of 2 ,for any congruence, the conversion relationship between the types is fixed. Keyword：congruence number，Elliptic Curves，Number Theory,Group Last fiddled with by Zcyyu on 20200626 at 23:10 Reason: 9. Conclusion 1) This paper proves all 3 congruence negative theorems in history using the N1 criterion and the simplest e 
20200627, 00:01  #2 
6809 > 6502
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Aug 2003
101×103 Posts
10001010111010_{2} Posts 
Can you provide a link to the paper or a pdf? English preferred.

20200628, 02:24  #3 
Jun 2020
22_{10} Posts 
New function, new criterion, new theorem and new conjecture for discriminant of congruence
New function, new criterion, new theorem and new conjecture for discriminant of congruence
Zhou CongYao College of Information Science and Engineering, Hunan University Yu Wei Department of statistics & financial engineering,College of mathematics and statistics,ningbo university TangXiaoNing Beijing Haitian Start Technology Service Co., Ltd. ABstruct In this paper,a new function of congruence discrimination is proposed,namely,the upper bound estimation funtion of the number of solutions of the congruence. If the value of this function is 1,the posttive integer A is noncongruent,this is the new N1 criterion.Using this guideline, to history all 3 congruence negative theorms,a very simple elementary proof is given;and proved the author's newly discovered three modular 8type congruence negative theorems; At the same time,eight new congruences negation theorems(not model 8 types) are proved,by the way,we derive 1511 infinite sequences of noncongruent If A is a congruence number,the value of this function is integer power of 2,that is,there are at least two types of solutions for congruence,one of them must be Fermat type,and each type has an infinite number of solutions. Using this upper bound function, some new theorems for the number of congruence solutions are derived: the exact value of the solution is 2, 4, 8, 16, 32; On this basis, a new conjecture is put forward: If the type of the congruence solution is used as an element, then these elements form a group,and the order of this group is integer power of 2 ,for any congruence, the conversion relationship between the types is fixed. Keyword：congruence number，Elliptic Curves，Number Theory,Group 0 History background Since the 10th century AD, the Persian mathematician AlKaraji has raised the problem of congruence, and it has been more than a thousand years [1] 1855,Genocchi proved the primes p≡3(mod 8) are all not congruence [2]； 1874,Genocchi proved if primes p≡5(mod 8),then 2p are all not congruence [2]； 1915,Bastien proved if primes p≡9(mod16),then 2p are all not congruence [2]。 1952,Heegner & Birch, if primes p≡3(mod4). then 2p are all congruence [2] 1975,Stephens the primes p≡5,7(mod 8) are all congruence [2] Alter,Kurtz & Kubota came up with conjecture: primes p≡5,6,7(mod 8) are all congruence(Not certified yet) [2] Identify the following issues with congruence: 1．Seeking a discriminant rule to determine whether a n∈N is a congruence (not completely resolved yet) 2．If the n∈N is a congruence , how to find the corresponding right triangle. (Not resolved at all) 3．Structure of the solution of congruence no one has ever studied, this paper finds and proves that the number of types of congruence solutions is the power of 2.new conjecture is proposed for the types of congruence solutions. The author of this paper constructs a new function for the first time, which is called the upper bound estimation function. 1. When the value of this function is 1, the corresponding positive integer A must be a noncongruence number. (Developed the author's N1 criteria) 2. When the positive integer A is a congruence, the value of this function corresponding to A is 2^k,where k≥1 3. When the solution of congruence A is not easy to find,using the intermediate result of this function can provide the search direction, For example, A = 822, we have not find the corresponding (s, t) solution, but this function provides 822. there is a solution of type (1,1, 2,411), which is helpful for the solution. 1 Existing judgment theorem or conjecture: 1.1 Fermat, Keqin Feng, Ye Tian and other existence theorems. Theorem A) Fermat's discriminant theorem: A is the congruence ,if and only if the system of equations: x^2+Ay^2=z^2 x^2Ay^2=w^2 There is a positive integer solution with y≠0 Theorem B). Existence theorem (Keqin Feng 1996, Delang LiYe Tian 2000, Chunlai Zhao2001) We have an infinite number of noncongruences with an arbitrary number of odd prime factors. In addition, there are an infinite number of congruences of a specified number of odd prime factors. [3] Theorem C). (2009, Xungui Guan) The necessary and sufficient condition for n to be the congruence is: there are positive integers a, b,v such that: nv^2=6a^2 b^2a^4b^4 or 4ab(a^2b^2) (Note: There are only two forms [4] 1.2 Practicality theorem: Theorem 1) Positive integer n=2p,where p≡3 (mod 4), n is the congruence .(Heegner,Birch) Theorem 2) Positive integer p≡5,7 (mod 8), p is the congruence .(Stephens[2],p276) Theorem 3) Set p_0≡5,7(mod 8) (Respectively p_0≡3(mod 4) ), p_i≡1(mod 8), i=1 …k, p_0,p_i Is odd prime and satisfies: (p_i 〖/p〗_0)=1), i=1, … ,k, and (p_i 〖/p〗_j)=1, 1≤ j ≤i1 then: n=p_0 p_1…p_k (Respectively n=2p_0 p_1…p_k） is the congruence (Ye Tian,2012)[5] Theorem 4) Let the expansion coefficient of the BSD modular function be a_n, b_n ,where n Is a square free positive integer, when a_n≠0, then n is not congruence, when b_n≠0, Then 2n is not a congruence r..[Coteswiles,1979] Theorem 5) Prime number p≡3 (mod 8), then p is not a congruence . (1855,Genocchi Theorem 6). Positive integer p≡5 (mod 8), then 2p is not a congruence.(1874,Genocchi) Theorem 7) Prime number p≡9 (mod 16),then 2p i s not a congruence .(1915,Bastieni) Theorem8)Integer n=pq,here p,q≡3 (mod 8), n Is’t a congruence (2019, ConyaoZhou,Wei Yu,XiaoningTang)[6] Theorem 9)Integer n=pq here p,q≡3(mod 8),2pq Is’t a congruence (2019,ConyaoZhou,Wei Yu,XiaoningTang) [7] Theorem10) Prime number p,q≡5 (mod 8),then 2pq is’t congruence (2019 ConyaoZhou,Wei Yu,XiaoningTang) [7] Theorem11) For the Integer A, as long as it has no solution at the N1 positions of the factor combination, then A is not a congruence (2019 ConyaoZhou,Wei Yu,XiaoningTang) [8] For more than 1,000 years, only these 11 theorems have practical value,Three of the judgments were yes and eight were negative. 1.3 Three conjectures: BSD Conjectural modular function: Let g_n=T∏_(n=1)^∞▒〖(1T^8n )(1T^16n ), 〗 θ_k=1+2∑_(n=1)^∞▒T^(2kn^2 ) k=1,2 g_n θ_1=∑_(n=1)^∞▒〖a_n T〗^n , g_n θ_2=∑_(n=1)^∞▒〖b_n T〗^n 1)BSD conjecture:Let n be a positive integer without a square factor, when a_n≠0,then n is not congruence, b_n≠0,then 2n is not congruence, when a_n=0, then n is congruence, b_n=0, then 2n is congruence, 2）AKK conjecture:(Alter，Kurtz & Kubota):p≡5,6,7(mod 8) is all congruence, 3）Tunnell conjecture:(Actually, it is an equivalent conjecture, but its calculation of 〖 a 〗_n,b_n is much easier than BSD ) Assuming that the BSD conjecture is true, If n is an odd square free number, define: f_1={(x,y,z)∈Z2x^2+y^2+8z^2=n}, f_2=2{(x,y,z)∈Z2x^2+y^2+32z^2=n }, a_n=f_1〖2f〗_2 If n is an even square free number, define: f_1={(x,y,z)∈Z4x^2+y^2+8z^2=n/2},f_2=2{(x,y,z)∈4x^2+y^2+32z^2=n/2 } a_n=f_1〖2f〗_2 Then n is Is the congruence quivalent to a_n=0 (Not proven yet). Since the case of analytical rank 0 has been proved, therefore a_n≠0 can be launched n is’n congruence. Using this method, you can effectively verify a noncongruence number. The problem is that when n is relatively large, the calculation f_1，f_2 very slow , In particular, calculating the coefficients a_n,b_n not only is it slow, but the value of the intermediate process is huge, Despite this, the BSD conjecture is still the most wonderful at present, but unfortunately it cannot be proved. 2. Preliminary theorem Preliminary theorem 1: Let f(x)=a_n x^n+a_(n1) x^(n1)+…+a_0 Is an algebraic equation with integral coefficients, m=p_1 p_2…p_k, T(f,m) express the number of solutions of f(x)≡0 (mod m) ,where p_i is prime. Then this congruence equation and congruence equations:f(x) ≡0 (mod p_i) i=1,…k, is equivalent, and T(f,m)=T(f, p_1) T(f, p_2)…T(f, p_k) [9] Theorem 1 of p182 That is to say, as long as one equation in the congruent equation system has no solution, the congruent equation system has no solution as a whole. 2.2 Preliminary theorem 2: Set ax^2+by^2+cz^2=0, where abc has no square factor, a、b、c not all are positive or negative, If and only if S^2≡ab (mod c), S^2≡bc (mod a and S^2≡ca (mod b) when all have a solution, quadratic equation ax^2+by^2+cz^2=0 only have integer solutions.[9] Theorem 1 of p282.. Preliminary theorem 3:(Congyao Zhou,Wei Yu,Xiaoning Tang,2019): If A is a congruence number, there are at least two types of solutions (s,t),(S,T), The latter is of type Fermat. Where S=〖(s^2+t^2)〗^2, T=4Am^2=4st(s^2t^2) 【Prove】Any positive integer s,t,(s,t)=1 s>t, and s,t An odd and an even s,t is not of type: 〖(a^2+b^2)〗^2 and 4ab(a^2b^2) Assume (s,t) is Is a solution of congruence A, by definition then Am^2=st(s^2t^2). Let S=〖(s^2+t^2)〗^2, T=4Am^2=4st(s^2t^2), For (S,T): S is a perfect square,T=A〖(2m)〗^2 then T = A multiply with a perfect square S+T=〖(s^2+t^2)〗^2,+ 4st(s^2t^2)=(s^2t^2+2st)^2, is a perfect square. ST=〖(s^2+t^2)〗^24st(s^2t^2)=(s^2t^22st)^2, is a perfect square. so another solution (S,T) corresponding is also the congruence A, and (ST,T,S,S+T)=(1,A,1,1）, Which is of type Fermat, Conversely, if a positive integer s,t, (s,t)=1 s>t s,t An odd number and an even number s,t is the type of 〖(a^2+b^2)〗^2 and 4ab(a^2b^2) can be obtained s_0=a,〖 t〗_0=b, s_0,t_(0 ) and s,t are all correspond to congruence A, (s_0,t_(0 ))≠(s,t), That is also at least 2 types of solutions. Theorem is proved. note: In fact, always get a small set of solutions first (s,t), Then use the formula to get the second type of solution (S, T) For example, the solutions of the 6 smallest two types are (2,1) and (25,24), the types are (1,1,2,3), (1,6,1,1) the solutions of the smallest two types of 1254 are (11,8) and (34225,20064), the types are (3,2,11,19), (1,1254,1,1) 3. Symbols and definitions Definition 1: Type of solution Let the explicit solution of primitive congruence A be: A×m^2=(st)ts(s+t),there m=xyzw, Here x,y,z,w is respectively maximum square factor of (st),t,s,(s + t),thisis x^2(st), y^2t, z^2s, w^2 (s+t) Let a=(st)/ x^2 ，b=t/y^2, c=s/z^2, d=(s+t)/w^2 ,then A=abcd is primitive congruence. st(s+t)(st) is the area of a right triangle, m^2 Is the square factor in the area of the right triangle. We call s, t the solution of the congruence A; The quaternion (a, b, c, d) is the type of solution; At the same time, we call (a, b, c, d) = (1, A, 1, 1) a Fermat type solution. Algorithm from solution to type (s,t)→(st,t,s,s+t) respective, remove the maximum square factor, get (a,b,c,d), On the contrary, there is no effective way. E.g : Minimum solution of 5 is (s,t)=(5,4), (s,t) → (st,t,s,s+t)=(1,4,5,9), → (a,b,c,d)=(1,1,5,1). Minimum solution of 1254 is (s,t)=(19,8), →(st,t,s,s+t)=(11,8,19,27), → (a,b,c,d)=(11,2,19,3). When it does not cause confusion, sometimes a symbol, such as english letters or numbers or greek letters, is used as a short name for the type. Definition 2: Number of factor combinations N=Y(A), Here A is a positive integer without a square factor, and the element of the factor combination is denoted as C1, C2, ..., Cn, ... ,the lowercase symbol ci = its value ( 1 or 0 ) The number of factor combinations of A is as: Y (1) =1, Y(p)=4 where p is an odd prime number. For composite number A, multiply each odd prime factor by 4, if it contains 2, then multiply by 2. ∵ 6=2*3, ∴Y(6)=8； ∵ 15=3*5, ∴ Y(15)=16； ∵1254=2*3*11*19, ∴Y(1254)=128 ∵992221230= 2*3*5*7*11*13*19*37*47, ∴ Y(992221230)=131072 The algorithm from A to Ci is as follows: Factor A and arrange the prime factors from small to large; let the number of factors be r, Here Two intermediate arrays A1, A2 are used, and the final result is stored in the array Ci. If A is even, the initial state of A1 is two elements (1,2,1,1),(1,1,2,1); Otherwise A1 is four elements (p_1,1,1,1), (〖1,p〗_1,1,1), (1,1,p_1,1), (1,1,1,p_1); Let the number of elements in A1 be s 3) A1→A2. If factor number of A r=1, then goto 5) 4) if factor number of A r>1,, then for k=2 to r do: Change each element in array A1 from 1 to 4, write to array A2: Clear A2, A1(i,j)* p_k Order write into array A2(* , j). there i=1 to s, j=1 to 4 5)) Array A2 write to array Ci E.g 1: A=5 then A1=A2={(5111),(1511)(1151)(1115)}={C1,C2,C3.C4} E.g 2: A=6 then A1={(1211),(1121)}, A2={(3211),(1611)(1231)(1213), (3121),(1321),(1161),(1123)}={C1..C8} E.g 3: A=15 then A1={(3111),(1311),(1131),(1113)}, A2= {(15,1,1,1),(3,5,1,1),(3,1,5,1)(3,1,1,5), (5,3,1,1),(1,15,1,1),(1,3,5,1),(1,3,1,5) (5,1,3,1),(1,5,3,1),(1,1,15,1),(1,1,3,5), (5,1,1,3),(1,5,1,3),(1,1,5,3),(1,1,1,15)}={C1..C16} The correspondence between the factor combination Ci and A is fixed, We have an algorithm to automatically generate a complete set of factor combinations (see Appendix 10.1) Definition 3: Four equations Let a solution of primitive congruence A be A×M^2=(st)ts(s+t) Where:(s,t)=1; s>t, s,t is integer, an odd and an even , st,s+t must be odd, and st,t,s,s+t are mutually prime. For st,t,s,s+t see its parity, There are only two combinations: case A: odd odd even odd; case B: odd even odd.odd The corresponding right triangle side length is 2st, s^2t^2,s^2+t^2, area=(st)ts(s+t), The basic primitive congruence number A, the area of the right triangle, and the basic relationship of the right side are: :Am^2=st(s^2t^2 ), we call (s,t) an explicit solution of congruence, For the congruence number A, if (s, t) is obtained, then the solution is successful. There are four factors for the area of the triangle: st, ②t, ③s, ④s+t Corresponding to: ax^2, by^2,cz^2, dw^2 By defined ①②③④, Obtain the following four equations: (For any combination of factors (a, b, c, d) of A, if there is a solution, it is said (a, b, c, d) is the type of solution), the four equations are named Q1Q4: ax^2+by^2=cz^2 (Q1） by^2+cz^2=dw^2 (Q2) ax^2+dw^2=2*cz^2 (Q3） dw^2ax^2=2*by^2 (Q4) Definition 4: 12 discriminant equations There are three ways to use each equation, mod a or mod b or mod c,Corresponds to the second subscript j=1,2,3; There are 12 kinds of judgment methods. For example: from Q1) ax^2+by^2= cz^2 get S^2≡bc (mod a),notes q11, Available in total q11,q12,q13; q21,q22,q23; q31,q32,q33; q41,q42,q43. referred to as qij (i=1,2,3,4;j=1,2,3) As long as one of the 12 formulas is not true, this factor combination abcd has no solution, at this time, define qij=0 The converse is not possible. even if all the 12 formulas are true, it does not mean that this combination must be solvable, it is just suspicious. now define qij = 1 Definition 5: M8 set (Type of solution Modulo 8) For (s,t)=1,s>t,s,t an odd number and an even number, for (st,t,s,s+t)：caseA :ooeo case B: oeoo, where st=ax^2, t= by^2, s=cz^2, s+t=dw^2 And odd^2≡1(mod 8) E.g a=27, a’=3, x=3, 3^2≡1(mod 8) Also t=18, b’=2,y=3, y^2≡1(mod 8); t=12,b’=3,y=2, y^2≡4(mod 8); t=48, b’=3,y=4 y^2≡0(mod 8) In short b is even y^2≡{0,1,4}(mod 8) . Take the modulus 8 of the factor combination (abcd), we get (a’ b’ c’ d’), There are 512 subdivisions: , a’={1,3,5,7},b’={1,2,3,5,6,7},c’={1,2,3,5,6,7},d’={1,3,5,7}, But b’,c’ can not be even at the same time. set of M8 have 512 items (Among them, 416 items in the negative set are marked as black; 96 items in the suspicious set are marked as red), M8={( a’ b’ c’ d’)}={ 1111 1113 1115 1117 1121 1123 1125 1127 1131 1133 1135 1137 1151 1153 1155 1157 Subtotal 1161 1163 1165 1167 1171 1173 1175 1177 (24) 1211 1213 1215 1217 1231 1233 1235 1237 1251 1253 1255 1257 1271 1273 1275 1277 (16) 1311 1313 1315 1317 1321 1323 1325 1327 1331 1333 1335 1337 1351 1353 1355 1357 1361 1363 1365 1367 1371 1373 1375 1377 (24) 1511 1513 1515 1517 1521 1523 1525 1527 1531 1533 1535 1537 1551 1553 1555 1557 1561 1563 1565 1567 1571 1573 1575 1577 (24) 1611 1613 1615 1617 1631 1633 1635 1637 1651 1653 1655 1657 1671 1673 1675 1677 (16)1 1711 1713 1715 1717 1721 1723 1725 1727 1731 1733 1735 1737 1751 1753 1755 1757 1761 1763 1765 1767 1771 1773 1775 1777 (24) 3111 3113 3115 3117 3121 3123 3125 3127 3131 3133 3135 3137 3151 3153 3155 3157 3161 3163 3165 3167 3171 3173 3175 3177 (24) 3211 3213 3215 3217 3231 3233 3235 3237 3251 3253 3255 3257 3271 3273 3275 3277 (16)2 3311 3313 3315 3317 3321 3323 3325 3327 3331 3333 3335 3337 3351 3353 3355 3357 (24) 3361 3363 3365 3367 3371 3373 3375 3377 3511 3513 3515 3517 3521 3523 3525 3527 3531 3533 3535 3537 3551 3553 3555 3557 (24) 3561 3563 3565 3567 3571 3573 3575 3577 3611 3613 3615 3617 3631 3633 3635 3637 3651 3653 3655 3657 3671 3673 3675 3677 (16) 3711 3713 3715 3717 3721 3723 3725 3727 3731 3733 3735 3737 3751 3753 3755 3757 3761 3763 3765 3767 3771 3773 3775 3777 (24)3 5111 5113 5115 5117 5121 5123 5125 5127 5131 5133 5135 5137 5151 5153 5155 5157 5161 5163 5165 5167 5171 5173 5175 5177 (24) 5211 5213 5215 5217 5231 5233 5235 5237 5251 5253 5255 5257 5271 5273 5275 5277 (16) 5311 5313 5315 5317 5321 5323 5325 5327 5331 5333 5335 5337 5351 5353 5355 5357 5361 5363 5365 5367 5371 5373 5375 5377 (24) 5511 5513 5515 5517 5521 5523 5525 5527 5531 5533 5535 5537 5551 5553 5555 5557 5561 5563 5565 5567 5571 5573 5575 5577 (24)4 5611 5613 5615 5617 5631 5633 5635 5637 5651 5653 5655 5657 5671 5673 5675 5677 (16) 5711 5713 5715 5717 5721 5723 5725 5727 5731 5733 5735 5737 5751 5753 5755 5757 5761 5763 5765 5767 5771 5773 5775 5777 (24) 7111 7113 7115 7117 7121 7123 7125 7127 7131 7133 7135 7137 7151 7153 7155 7157 7161 7163 7165 7167 7171 7173 7175 7177 (24)5 7211 7213 7215 7217 7231 7233 7235 7237 7251 7253 7255 7257 7271 7273 7275 7277 (16) 7311 7313 7315 7317 7321 7323 7325 7327 7331 7333 7335 7337 7351 7353 7355 7357 7361 7363 7365 7367 7371 7373 7375 7377 (24) 7511 7513 7515 7517 7521 7523 7525 7527 7531 7533 7535 7537 7551 7553 7555 7557 7561 7563 7565 7567 7571 7573 7575 7577 (24) 7611 7613 7615 7617 7631 7633 7635 7637 7651 7653 7655 7657 7671 7673 7675 7677 (16) 7711 7713 7715 7717 7721 7723 7725 7727 7731 7733 7735 7737 7751 7753 7755 7757 7761 7763 7765 7767 7771 7773 7775 7777 } (24)6 M8 Negative set ={1113,,1115,1117,1121……} 416 items (The black combination in the table above M8,its value m8= 0) E.g1 (Negative set) :For a’b’c’d’=1113, correspond Q1: ’x^2+y^2≡z^2 Q2: ’y^2+z^2≡3w^2 CaseA look Q1,left≡1+1,right≡1,Module 8 left≠right(caseA Cannot be established, abbreviated as A ×) caseB:look Q1,1+{014}≡1,may be, as y≡0,z≡1,substitute Q2 left≡0+1,right≡3(abbreviated as B√×) when a’b’c’d’=1113,for caseA,B, none of them are true,so the ternary 2 degree congruence equations of this factor combination have no solution, and 1113 belongs to the negative set Similarly, it can be proved that other 415 combinations are negative sets Here define M8=0; E.g2 (Suspicious Set) For a’b’c’d’=1123, correspond Q1: ’x^2+y^2≡2z^2 Q2: y^2+2z^2≡3w^2 caseA：1+1≡2{014}may be,as y=1,z=1 , Substitute Q2 left=y^2+2z^2≡3, right=3w^2≡3 Can't Certainly no solution (A√√) caseB：left≡1 +{04}≡{1,5}, right≡2, left≠ right (abbreviated as B×), In case A and B are all not established, we can guarantee that 1123 has no solution; as long as one of cases A and B cannot be determined, the whole cannot be determined. For the suspicious set, define M8 = 1 for this combination. There are 96 items in the suspicious set. (The red combination in the above table, define M8 = 1 at this time) Similarly, it can be proved that other 95 combinations are suspicious sets Here define its value m8=1; Let the factor combination be (a,b,c,d),some of them are definitely unsolvable, such as 1113; some are suspicious, that is, the solution of this combination cannot be determined, such as 1111 E.g: p=41, a’b’c’d’=1111, ∵41is congruence, ∴ The ternary quadratic system of course has solutions; Also p=17,a’b’c’d’=1111,∵17 is’t congruence, ∴The congruent equation system must be unsolved. Thus, M8=1111 cannot be determined by the M8 method, but it is only suspicious. Definition 6: The function of the upper bound estimate of the type of solution, abbreviation Z(A) The factor combination Y (A) has been defined. A as a positive integer without a square factor. We have an algorithm 1, which automatically generates all factor combinations. When searching to find the number of types of congruence number = Z (A),the type number is exact; define the value as T(A) = type number of congruence solution Definition 7: New function Let A is a positive integer without a square factor, If the qij and M8 of this combination are all 1, then this combination contributes ci=1, this is Z(A)=∑ci (lowercase c) E.g: A=17,Y(17)=4, C1= (abcd)=(17.1.1.1) Q1; 〖17x〗^2+y^2 〖z〗^2=0, S^2≡1(mod 17),q11=1; S^2≡17(mod 1) q12=1; S^2≡17(mod 1) q13=1 ∴Q1 has sulotion Q2; y^2+z^2 〖w〗^2=0 S^2≡1(mod 1), 21=1; S^2≡1(mod 1),q22=1; S^2≡1(mod 1) , q23=1 ∴Q2 has sulotion Q3; 〖17x〗^2+w^2 〖2z〗^2=0 S^2≡2(mod 17)q31=1; S^2≡34(mod 1),q32=1; S^2≡17(mod 2),q33=1 ∴Q3 has sulotion Q4; w^217x^2 〖2y〗^2=0 S^2≡34(mod 1)q41=1; S^2≡2(mod 17)q42=1; S^2≡17(mod 2)a43=1 ∴Q4 has sulotion M8=(abcd)(mod 8)=(1111) belong suspicious set of M8; qij, M8 of this combination be all 1, C1 contributes 1 to Z (A), C2,C3,C4 C1 contributes 1 to Z (A), so Z(41)=4 Definition 8．N1 guidelines If the value of the new function Z(A)=1, Then A is a noncongruence number, where N Is the number of factor combinations (see definition 2), [Prove] By Preparation Theorem 3,if A is congruence, Then there are at least two different types of solutions that is, at the N positions of factor combinations, at least, there are two places that appear to be solvable, In other words, at most at the N2 factor combination, the corresponding congruence equation may be found to have no solution And we have determined that the corresponding congruence equation has no solution at the N1 combination of factors, so A can only be noncongruence. If the value of the new function Z(A)>1, there is no clear conclusion. Example, Z(17)=4, And 17 is not a congruence number;Z(41)=4, but 41 is the congruence. Definition 9 . The exact value of the number of types of congruence solutions T (A) When the sequential search method and I⊕J method, or other methods, The number of types of solutions found is equal to the upper bound estimate Y(A), Define this value to be T(A), E.g T(6)=2, That is, there are only two types of solutions of the 6: (1123) and (1611). 4. The Elementary Proof of the Negative Theorem of Congruence in History 4.1 History Theorem 1: If prime number p≡3 (mod 8),then All of them not to be noncongruence (Genocchi,1855) 【prove】Prove with our method , first of all the prime factor combination N = 4 C1: p.1.1.1 look Q3, px^2 + w^22z^2=0, by preparation Theorem 1,S^2≡2(mod p),and(2/p)=1, ∴c1=q31=0 C2: 1.p.1.1 The Fermat type belongs to the suspicious set and cannot be proved. ∴c2=1 C3: 1.1.p.1 look Q3, x^2 + w^22pz^2=0,by preparation Theorem 1 S^2≡1(mod 2p) , And among them (1/p)=1 so S^2≡1(mod 2p) no solution,then Q3 no solution, ∴c3=q33=0 C4: 1.1.1.p look Q3, x^2 + pw^22z^2=0, by preparation Theorem 1,S^2≡2(mod p), ∴ c4=q32=0 By N1 guidelines, History theorem 1 was established. New method, so concise , The proof process is abbreviated as {c1,,..c4}={0,1,0,0}, this is Z(A)=1 4.2 History Theorem 2: If prime number p≡5 (mod 8), then 2p all not to be congruence (Genocchi,1874) 【prove】For 2p, its number of factor combinations is N=8 C1: p.2.1.1 look Q1,px^2 + 2y^2z^2=0, get S^2≡2(mod p),(2/p)=1, ∴1=q11=0 C2: 1.2p.1.1 The Fermat type belongs to the suspicious set and cannot be proved. ∴c2=1 C3: 1.2.p.1 look Q2, 2x^2 + py^2w^2=0 get S^2≡2(mod p), ∴c3=q22=0 C4: 1.2.1.p look Q2, 2y^2 +z^2pw^2=0, get S^2≡2(mod p),(2/p)= 1, ∴c4=q23=0 C5: p.1.2,1 look Q4,w^2px^22y^2=0, get S^2≡2(mod p), ∴ c5=q42=0 C6: 1.p.2.1ook Q1,x^2+py^22z^2=0, get S^2≡2(mod p), ∴ c6=q12=0 C7:1.1.2p.1 Only the modulo 8 method can be used:1121 Belongs to the negative set, ∴c8=m8=0 C8: 1.1.2.p look Q2, y^2 +2z^2pw^2=0, get S^2≡2(mod p), c8=q23=0 The inspection sequence is {0,1,0,0,0,0,0,0},this is Z(A)=1, by N1 guidelines, History theorem 2 was established. 4.3 History Theorem 3: If prime p≡9 (mod 16), then 2p all are not congruence (1915,Bastieni) (Our elementary proof) C1: p,2,1,1 all qij=1 and M8 belongs to the suspicious set, these methods are invalid! Look Q2) z^2+2y^2=w^2,[4] Exercise 9 for p102,the general solution of (z,y,w)=1 is z= 〖u〗^22v^2 , y=2uv, w=u^2+2v^2, There (u,v)=1,u,v one odd one even. substitute Q1,get px^2=z^22y^2=u^4+4v^48u^2 v^2 that do not meet two solutions of the theorem of [7], so factors C1 no solution , this c1=0, C2: 1,2p,1,1 The Fermat type belongs to the suspicious set and cannot be proved. notes c2=1 C3: 1,2,p,1 all qij=1 and M8=1211 belongs to the suspicious set Look Q4) w^2x^2=4y^2, this x^2+(2〖y)〗^2=w^2, the general solution x=s^2t^2, 2y=2st, w=s^2+t^2 substitute Q1,get z^2=w^22y^2=s^4+t^4, that do not meet two solutions of the theorem of [7], so c3=0, C4: 1,2,1,p all qij=1 and M8=1211 belongs to the suspicious set Look Q1) x^2+2y^2=z^2,see exercise 9 of p102[4], the general solution of (z,y,w)=1 is x= u^22v^2 , y=2uv, z=u^2+2v^2, There (u,v)=1,u,v one odd one even, substitute Q1,get pw^2=2y^2+z^2=u^4+4v^4+12u^2 v^2 that do not meet two solutions of the theorem of [7], so c4=0 C5: p,1,2,1 similar C4, px^2=2z^2y^2=〖12u^2 v^2u〗^44v^4, do not meet two solutions of the theorem of [7], c5=0, C6: 1,p,2,1 M8=1121 belongs to the negative set, c6=m8=0, C7: 1,1,2p,1 M8=1121 belongs to the negative set, c7=m8=0, C8: 1,1,2,p M8=1121 belongs to the negative set, c8=m8=0 The proof process is abbreviated as {C1,…,C8}={0,1,0,0,0,0,0,0}. this is Z(A)=1, By N1 guidelines, History theorem 3 was established. 5. Elementary proof of the negative theorem of the author's 3 congruences 5.1. Theorem B33 published in the China Science and Technology Online Community [5] If p≡3 (mod 8),q≡3 (mod 8),then n=pq is not congruence number. For n=pq, number of factor combinations N=16, use the new N1 criterion, look factor combination C6 C6=(1,pq,1,1) is Fermat type, there is no need to prove,also can't prove, other 15 ci=0, this is Z(A)=1, so n=pq is not congruence [5] 5.2 Theorem C233 (First published in China Science and Technology Online) If p≡3 (mod 8),q≡3 (mod 8),then n=2pq.all are not congruence number. For n=2pq , number of factor combinations N=32. Use the new N1 criterion , factor combination C6=(1,2pq,1,1) is Fermat type, there is no need to prove,also can't prove, other 31 ci=0, this is Z(A)=1, so n=2pq is not congruence [6] 5.3 Theorem C255 (First published in China Science and Technology Online) If p≡5 (mod 8),q≡5 (mod 8),then n=2pq is not congruence number.. For n=2pq, number of factor combinations N=32. Use the new N1 criterion, factor combination C6=(1,2pq,1,1) is Fermat type,no need to prove,also can't prove, other 31 ci=0, this is Z(A)=1, so n=2pq is not congruence [6] 6. Congruence negative theorems(nonmodulo 8, 1511 noncongruence sequences) Theorem B13,B57,B31,B75,C215,C237,C251 and C273 in references[8] , all using N1 criteria and elementary methods.here, p≡u (mod 8a),u Depending on the value of a E.g: IF a=17, then u=3,11,27,75,91,99,107,131 (mod136), 8 infinite sequences, equivalent to 8 small theorems. one of them: If p≡3(mod 136),then 17p all are not congruence (Similar to History Theorem 4.3) a total of 1511 explicit noncongruence infinite sequences were found 7. Theorems on the number of types of congruence solutions Theorem7.1:Let prime number p≡5(mod 8),for such p, exists and only exists 2 types of solutions, the type of solution is (1,1, p, 1), (1, p, 1,1), the latter is a Fermat type solution F = (1A11) . There are an infinite number of solutions under each type. [Prove] We already know that the prime numbers p≡5(mod 8) are congruence, according to the preparation theorem, know the number of types of solutions ≥2. In addition, the number of factor combinations N=Y(5)=4.∴ C1: p.1.1.1 LookQ3, px^2 + w^22z^2=0,from q31 get S^2≡2(mod p),C1 no solution, note c1=q31=0 C2: 1.p.1.1 Fermat type: all qij = 1, meanwhile (a’b’c’d’)=(1511) belongs to suspicious set,m8=1,.note c2=1 C3: 1.1.p.1 all qij = 1, (1151) belongs to suspicious set, note c3=1 C4: 1.1.1.p look Q3, x^2 + pw^22z^2=0, from q42get S^2≡2(mod p) Q3,this C4 no solution, note c4=q42=0 Above C1C4, c2=c3 = 1 (possible solution); c1=c4 = 0 (definitely without solution) This is, the four combinations, two may have solutions, so the number of solutions ≤ 2,this is Z(p)=2; By the preparation theorem 3, the number of types of solutions of the congruence number ≥2; Therefore, the number of types of the solution is exactly = 2; the types are: a = (1, 1, p, 1), (1, p, 1, 1), the latter type (1A11) is the Fermat class, which is abbreviated as F. Take 5 as an example, a1=(5,4) type=1151 a2= (397335659233067745125,506744562502300651240) type=1151 F1=(1681,720) type=1511 F2=(11183412793921,11170580662080)type=1511 … (Note: With the above types referred to as symbols ai, Fi, the first symbol indicates the type of solution, and the second symbol indicates a solution of different size in the same class) ... Theorem7.2: Let prime number p≡7(mod 8),for such p, exists and only exists 2 types of solutions, the type of solution is (p,1,1, 1), (1, p,1,1), the latter is a Fermat type solution F = (1A11) . There are an infinite number of solutions under each type. [Prove] We already know that the prime numbers p≡5(mod 8) are congruence, according to the preparation theorem, know the number of types of solutions ≥2. In addition, the number of factor combinations N=Y(5)=4.∴ C1:p.1.1.1 all qij = 1, and M8=(7111) belongs to suspicious set,note c1=q31=1 C2:1.p.1.1 Fermat type: all qij = 1, meanwhile (a’b’c’d’)=(1711) belongs to suspicious set,M8=1,.note c2=1 C3:1.1.p.1 look Q1, x^2+y^2pz^2=0,from q13,get S^2≡1(mod p),no solution, notes c3=q13=0 C4:1.1.1.p look Q2, y^2+z^2pz^2=0, from q43get S^2≡1(mod p),no solution,notes c4=q43=0 Above C1C4, c1=c2=1(possible solution); c3=c4=0 (There must be no solution) , This is, Z (p) = 2, that is, the number of types of solutions = Z(A)=2; respectively , a=(p,1,1,1),(1,p,1,1) The latter (1A11) is Fermat type F, taking 7 as an example: a1=(16,9)type=7111 a2=(463864517776, 38766853449)type=7111 F1=(113569,100800)type=1711 F2=(531697086017373333121,125342742889328428800) ype=1711 Theorem7.3:Let prime number p≡3(mod 4),then 2p for such p, exists and only exists 2 types of solutions, There are infinitely many solutions under each type. 7.3A) Prime numbers in the form of 4m + 3, m is an even number, such as 3,11,19… The type of solution is (1,1,2, p), (1,2p, 1,1), the latter is Fermat type F = (1A11) [Prove]From the preparation theorem 3,we already know that if prime number p≡3(mod 4),then 2p is congruence, so the number of types of solutions is ≥ 2. In addition, the 2p factor combination is N=Y(2p)=8,(a,b,c,d) There are eight factor combinations: C1:p.2.1.1 look Q1,px^2 + 2y^2z^2=0, for q11get S^2≡2(mod p) no solution,notes c1=q11=0 C2:1.2p.1.1 is Format type,all qij=1 and (1611) ,belongs to the suspicious set, notes c2=1 C3:1.2.p.1 look Q2, 2y^2 + py^2z^2=0,from q22,get S^2≡2(mod p) no solution,notes c3=q11=0 C4:1.2.1.p look Q3, x^2 + pw^22z^2=0, 从q32 get S^2≡2(mod p) no solution,notes c4=q32=0 C5:p.1.2.1 look Q1,px^2 + y^22z^2=0,从q11 get S^2≡2(mod p) no solution,notes c5=q11=0 C6:1.p.2.1 look Q1,x^2 + py^22z^2=0,从q12 get S^2≡2(mod p) no solution,notes c6=q12=0 C7:1.1.2p.1 lookQ1,x^2+y^22pz^2=0,从q13 get S^2≡1(mod 2p),其中(1/p)=1, no solution,notes c7=q13=0 C8:1.1.2.p all qij=1 and (1123) belongs to the suspicious set, notes c8=1 Above C1C8, c2=c8=1(possible solution); other ci =0 (without solution).Z(A)=2 That is, the number of types of solutions = 2; respectively: a = (1,1,2, p), F=(1,2p, 1,1)=(1A11)， The latter is Fermat type, taking 6 as an example: a1=(2,1)type1123 a2=(2738, 529)type1123……, F1=(25,24)type1611 F2=(1442401,117600) type1611 …… 7.3B) Prime numbers in the form of 4m + 3, m is an odd number, such as,如7,23,31…, The type of solution is (p, 1,2,1), (1,2p, 1,1), the latter is Fermat type F = (1A11) [Prove] The method is similar to 7.3A). Among the eight factor combinations, only C1 and C5 may have a solution, this is Z (p) = 2. Taking 14 as an example a1=(8,1) type 7121 a2=(160348232,141681409)type 7121 ……, F1=(4225,2016) type 1A11 F2=(480262009244161,469707106377600) type 1A11…… Explanation of Theorem 7: There are an infinite number of congruences of type number = 2. Theorem 7.4) Some examples of the type T (A) = 4 of congruence solutions 7.4A) The type of the solution of 41 is 4, the few examples of Y (A) = Z (A) = T (A) [Prove] Sequential search found 6 solutions, 4 types a, b, c, F, as follows: s=25 t=16 a1 type=111p s=441 t=400 b1 type=p111 s=1025 t=64 c1 type=11p1 s=296225 t=135424 c2 type=11p1 s=776161 t=590400 F1 type=1p11 s=1610361 t=547600 b2 type=p111 for p=41,N=Z(41)=4,, So 41 has exactly 4 types of solutions; that is T (41) = 4, The 6 × 6 symmetric matrix is as follows: Let vector V=(a1,b1,c1,c2,F1,b2) by group I⊕J with elliptic curve group to get matrix M=V^TV a1 b1 c1 c2 F1 b2 The result of row i and column j in the table is obtained by I⊕J a1 F1 c1 b1 b3 a1 c3 b1 F2 a1 a1 b3 F1 The main diagonal is summarized as α⊕α=F c1 F3 F1 c3 a1 Column F is summarized as F⊕α=α’ c2 F4 c1 a1 F1 F5 b1 (Note: Only the type is considered here, not the size of the value) b2 F6 7.4B) Congruence number 999809706= 2*3*7*31*101*7603 exists and only exists 4 types of solutions, [Prove] Factor combinations =2048.Using the algorithm in Appendix 1, you can immediately get the complete set of 2048 factor combinations (C1—C2048) We have found two different solutions, so it should have at least four different solutions a, b, F, c Through sequential search, and elliptic curve group plus I⊕J, five solutions, four classes: a s=126250 t=109443 b s=1130291 t=380150 F1 s=779349550735638897001 t=218935532452227465000 F2 s=2022288111013292268686761 t=1947385466590353310722600 c s=28275200 t=5854667 Calculated using our new function Z(999809706)=4, as follows: （abcd） qij=1 and m8=1 01: 21 15206 3131 5631∈suspicious set 02: 1 A 1 1 1211∈suspicious set 03: 93 707 2 7603 5323∈suspicious set 04: 7 3 202 235693 7325∈suspicious set All other 2044 combinations qij=0, so T(999809706)=4 We have calculated and proved that there are 19101 congruences, which have exactly 4 different types of solutions. Theorem 7.5) Some examples of the type T (A) = 8 of congruence solutions Theorem 7.5A) The congruence number 1254 exists and only exists 8 types of solutions (Y(1254)=128,Z(1254)=8,T(1254)=8) [Prove] Through sequential search we found 13 solutions of 1254 as follows(Actually found 25, limited space, omit some) No. type=(a b c d) abbr. solution =(s,t) 01) 3 2 11 19 a1 11 8 02) 11 2 19 3 b1 19 8 03) 19 3 22 1 c1 22 3 04) 3 11 38 1 d1 38 11 05) 19 2 3 11 e1 147 128 06) 1 1 2 627 f1 338 289 07) 11 19 6 1 g1 486 475 08) 19 2 3 11 e2 507 32 09) 1 1 2 627 f2 578 49 10) 11 19 6 1 g2 1350 19 11) 3 11 38 1 d2 1862 1859 12) 19 3 22 1 c2 11638 11163 13) 1 1254 1 1 F1 34225 20064 By group addition using elliptic curves I⊕J, no new types found, 8 types in total. The 13×13 symmetric matrix formed by I⊕J is as follows: a1 b1 c1 d1 e1 f1 g1 e2 f2 g2 d2 c2 F1 Note a1 F1 e2 d2 c2 b2 g2 f2 b3 g1 f1 c1 d1 a7 Main diagonal α⊕α=F The other summary F⊕α=α’ b1 F2 f1 g1 a3 c1 d1 a1 c3 d3 g3 f3 b2 c1 F3 a2 g1 b4 e1 g2 b5 e2 a4 F4 c2 d1 F4 f1 e1 b2 f2 e2 b7 F1 a1 d2 α⊕α’=F e1 F5 d4 c4 F2 d2 c2 f4 g2 e2 f1 F6 a2 d2 F2 a6 e2 b2 f4 If only the first symbol is considered do not consider the second symbol (value size then α⊕α=F, α⊕F=α g1 F7 c2 A5 F2 b5 e3 g4 e2 F8 d6 c5 f5 g1 e1 f2 F9 a2 e1 b3 f5 g2 FA b4 e4 g5 104 locations in the upper triangle,only 8 types ag,F d2 FB a7 d7 c2 FC c1 F1 a7 b2 c2 d2 e2 f4 g4 e1 f5 g5 d7 c1 FD Summarized as F⊕α=α’ N(1254)=128, Call the Appendix 1 algorithm to immediately obtain N(1254)=128, and call the Appendix 2 algorithm to obtain Z (1254) = 8 type all qij=1 and M8=1∈Suspicious set, (a’b’c’d’) ∈Suspicious set a 3 2 11 19 3233 F 1 1254 1 1 1611 e 19 .2 .3. 11 3233 b 11. 2 .19. 3 3233 d 3. 11. 38. 1 3361 c 19. 3 .22 . 1 3361 g 11. 19. 6 . 1 3361 f 1. 1. 2. 627 1123 The number of types found is equal to the upper bound Z (1254), so the type of its solution is exactly = 8, that is, T (1254) = 8 定理7.5B) The congruence number 992503974 exists and only exists 8 types of solutions [Prove] ∵ 992503974 = 2*3*7*11*41*151*347, ∴Y(992503974)=8192 Using the sequential search method we have obtained 3 different solutions a,b,c, type=(a,b,c,d) Abbreviation solution =(s, t) 1057 1 2 469491 a 260642, 208849 41 576357 6 7 b 760416, 576367 3 302 41 26719 c 1623641, 1609358 The i⊕J algorithm is used for the first time, and a new d,e,f,F, abbreviated as 03+F+03 a,b,c,d,e,f,F, using the i⊕J algorithm again, only a new class g is added, and a total of 8 types of solutions are obtained. This solution process is abbreviated as 03+F+03；+01 (a, b, c, d, e, f, F has a new type g, marked in red bold in the matrix below) a1 b1 c1 d1 e1 f1 F1 F2 F3 Main diagonal α⊕α=F ∵∴ a1 F1 e1 f2 g2 b1 c5 a1 a2 a3 (a1+F1=a1…Abstract α⊕F=α’) b1 F2 d2 c4 a1 g2 b2 b1 b3 (b1+F1=b2…Abstract α⊕F=α’) c1 F3 b3 g1 a3 c2 c3 c6 (c1+F3=c6…Abstract α⊕F=α’) d1 F6 f1 e3 d3 d1 d4 (d1+F1=d3…Abstract α⊕F=α’) e1 F7 d1 e4 e5 e2 (e1+F2=e5…Abstract α⊕F=α’) f1 F8 f1 f3 f4 (f1+F3=f4…Abstract α⊕F=α’) F1 F9 F7 F4 F1+F1=F9,F1+F2=F7,F1+F3=F4 F2 FB F5 F2+F2=FB,F2+F3=FA F3 FA A congruence can be verified infinitely α⊕α’= F type type(a,b,c,d)  Minimum solution (s, t) a 1057 1 2 469491 260642 208849 b 41 576367 6 7 760416 576367 c 3 302 41 26719 1623641 1609358 d 1 1057 2 469491 948823922 88281697 e 453 2 287 3817 7649815931078447 3613908587282450 f 6191 3817 42 1 9120552 645073 g 1057 938982 1 1 90011878276681 39160961692248 F 1 A 1 1 12443883723543998011225 5294624069956300680216 By the algorithm in Appendix 2,get Z(992503974)=8, Manual calculation of 8192 factor combinations, each combination of up to 12 qij, plus an M8 judgment,Not only hard but almost impossible, We have discovered and proved that the type of solutions for 3076 different congruences is 8 Theorem 7.6) Examples of T (A) = 16 with Congruent Solution 7.6A) , Congruence number 29274 exists and only exists 16 types of solutions [Prove] By sequential search we got 41 solutions, we only used 16 of them as follows: (including F type) type solution (s,t) type solution (s,t) 01) 7 17 6 41 24 17 02) 3 7 34 41 34, 7 03) 7 34 41 3 41 34 04) 119 2 1 123 121, 2 05) 123 119 2 1 242 119 06) 51 287 2 1 338, 287 07) 7 41 6 17 384 41 08) 287 1 102 1 408, 121 09) 7 82 17 3 425 82 10) 1 42 1 697 529, 168 11) 1 42 697 1 697 672 12) 697 42 1 1 2209, 1512 13) 287 2 1 51 2209 1922 14) 119 1 246 1 8856, 3025 15) 3 7 82 17 9922 4375 16) 1 A 1 1 748225,468384 16 types have been found, and group addition of elliptic curves can’t generate new types, so there is no need to continue I⊕J, operation ∵29274= 2*3*7*17*41 ∴ N=Y(29274)=512 Calculate Z (29724) = 16, found 16 types, ∴ 29274 T(29274)=16 7.6B) Congruence number 46274 exists and only exists 16 types of solutions [Prove] 8 solutions of 46274 have been found, (Fermat type F is not included, must continue to calculate) Type(a,b,c,d) Abbreviation solution (s,t) 01) 1 34 1 1361 a 1225, 136 02) 1361 2 1 17 b 1369, 8 03) 1 2 1361 17 c 12249, 2048 04) 1 2 1 23137 d 12769, 10368 05) 1 2 23137 1 e 23137, 14112 06) 1361 2 17 1 f 240737, 228488 07) 17 2 1 1361 g 978121, 504008 08) 23137 2 1 1 h 1168561, 34848 By addition of groups I⊕J On the original 8 types, there are 8 new types :i,j,k,l,m,n,o,F, There are 16 types in total. And Z(46274)=16, also 46274=2*17*1361 ,N=Y(46274)=32,, The complete set of factor combinations is as follows: a b c d qij, M8’ Abbr. C1： 23137 2 1 1 qij is all 1 1211 h C2： 17 2722 1 1 qij is all 1 1211 C3: 17 2 1361 1 qij is all 1 1211 C4: 17 2 1 1361 qij is all 1 1211 g C5: 1361 34 1 1 qij is all 1 1211 C6: 1 46274 1 1 qij is all 1 1211 C7: 1 34 1361 1 qij is all 1 1211 C8: 1 34 1 1361 qij is all 1 1211 a C9: 1361 2 17 1 qij is all 1 1211 f C10: 1 2722 17 1 qij is all 1 1211 C11: 1 2 23137 1 qij is all 1 1211 e C12: 1 2 17 1361 qij is all 1 1211 C13 1361 2 1 17 qij is all 1 1211 b C14 1 2722 1 17 qij is all 1 1211 C15 1 2 1361 17 qij is all 1 1211 c C16 1 2 1 23137 qij is all 1 1211 d (1211∈suspicious sets and m8 =1) C17 23137 1 2 1 qij is all 1 but M8=1121∈negative set,c17=0 C18 17 1361 2 1 qij is all 1 but M8=1121∈negative set,c18=0 C19 17 1 2722 1 qij is all 1 but M8=1121∈negative set,c19=0 C20 17 1 2 1361 qij is all 1 but M8=1121∈negative set,c20=0 C21 1361 17 2 1 qij is all 1 but M8=1121∈negative set,c21=0 C22 1 23137 2 1 qij is all 1 but M8=1121∈negative set,c22=0 C23 1 17 2722 1 qij is all 1 but M8=1121∈negative set,c23=0 C24 1 17 2 1361 qij is all 1 but M8=1121∈negative set,c24=0 C25 1361 1 34 1 qij is all 1 but M8=1121∈negative set,c25=0 C26 1 1361 34 1 qij is all 1 but M8=1121∈negative set ,c26=0 C27 1 1 46274 1 qij is all 1 but M8=1121∈negative set,c27=0 C28 1 1 34 1361 qij is all 1 but M8=1121∈negative set,c28=0 C29 1361 1 2 17 qij is all 1 but M8=1121∈negative set,c29=0 C30 1 1361 2 17 qij is all 1 but M8=1121∈negative set,c30=0 C31 1 1 2722 17 qij is all 1 but M8=1121∈negative set,c31=0 C32 1 1 2 23137 qij is all 1 but M8=1121∈negative set,c32=0 Upper bound estimation function Z(46274)=16, ∴ T(46274)=16 The discovery process is abbreviated as 08+F+07 (Note: The sequential search algorithm gets 08 types ah; once I⊕J gets F and (io), a total of 16 types) We have discovered and proved that the number of solutions of 280 congruences is 16 theorem 7.7) Congruence number A=253375590, The type of solution constitutes a 32order group [Prove] The proof is divided into the following steps: 01: The sequential search method found 9 types of solutions: Note: As long as the Fermat solution is not included, or although the Fermat type is included, but the type number is not 2^k,you can use the elliptic curve group addition algorithm continue to solve Type=(a’b’c’d’) Abbr. sulotion=( s,t) 01) 601 10 611 69 1 611 10 02) 3005 94 69 13 2 3381 376 03) 4485 1202 47 1 3 5687 1202 04) 4485 2 47 601 4 16967 12482 05) 5 94 41469 13 5 41469 24064 06) 1 10 367211 69 6 367211 364810 07) 367211 115 6 1 7 408726 41515 08) 1 421590 601 1 8 1688209 1686360 09) 1 10 611 41469 9 1027091 1004890 02: For the first time, I⊕J algorithm of elliptic curve group add , 16 types of solutions were found: (The smallest solution of the class)) Type Abbr. sulotion=(s,t) The sequential method can't get such a large value 10) 1 6010 611 69 a 14892495059, 96935290 The red S in the table is＜F1 11) 1 421590 1 601 b 305235841 284994840 12) 3 5 2 8445853 c 6051008522026112 4555132942345445 13) 3 5 1202 14053 d 149328445832 104145641645 14) 3 3005 2 14053 e 80797472 16013645 15) 5 94 69 7813 f 34880949 27005824 16) 5 56494 69 13 g 7350089829 2126208184 17) 115 13 1202 141 h 8000843025992 2814274927477 18) 235 3 598 601 i 6540186810118 1448036856003 19) 235 1803 598 1 j 34732438 14281563 20) 601 421590 1 1 k 81393939815881 2130506751360 21) 611 115 6 601 L 816005406534 645133923115 22) 4485 2 28247 1 m 2074487927 1409698802 23) 69115 13 2 141 n 19294472 10931557 24) 2695485 2 47 1 o 3250943 555458 25) 1 A 1 1 F 139443243241 9121521240 The smallest class F 03: The second time I used elliptic curve group plus I⊕J algorithm, and found 7 types of solutions 26) 115 13 2 84741 U 8536513665152 2567422343317 27) 115 7813 2 141 V 1270034842001408 645393512861773 28) 235 3 359398 1 w 486110593462 189601354827 29) 611 115 3606 1 X 726973206 536047315 30) 611 69115 6 1 Y 23897989926 3194840875 31) 1803 5 2 14053 Z 1251737003016992 101659237686125 32) 141235 3 598 1 T 87085202140838278 40706048314280163 A total of 32 types of solutions including F have been found, and then the group plus I⊕J algorithm cannot be used to add new types. 04: Calculate N=Z(A)=32,the number of types found is equal to the upper bound estimate, so T(A)=32 05: Observe the 32 × 32 relationship matrix 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k L m n o F U V W X Y Z T Note： 1 F 3 2 5 4 b e 9 8 k 6 X L 7 m o i h n a d f j g 1 W T U c Z Y V In the matrix: 2 3 F 1 6 b 4 j f m o 5 h U n 8 k c X 7 g W 9 e a 2 d Z L i T V Y The format of element 3 2 1 F b 6 5 n m f g 4 i W j 9 a X c e o U 8 7 k 3 L Y d h V T Z All should be α_i, 4 5 6 b F 1 2 h o g f 3 j T i a 9 7 e c m V k X 8 4 Y L Z n U W d among them 5 4 b 6 1 F 3 i g o m 2 n V h k 8 e 7 X f T a c 9 5 Z d Y j W U L 6 b 4 5 2 3 F c a k 8 1 7 Y X o m j n h 9 Z g i f 6 T W V e d L U 7 e j n h I c F L d Z X 6 9 1 W T 4 5 2 Y 8 U 3 V 7 m o f b k a g 8 9 f m o g a L F 1 6 k Z e d 2 5 V T W b 7 3 U 4 8 n h j Y X c I 9 8 m f g o k d 1 F b a Y 7 l 3 4 T V U 6 e 2 W 5 9 j i n Z c X h The previous a k o g f m 8 Z 6 b F 9 L X Y 4 3 W U V 1 c 5 T 2 a i j h d e 7 n step 0103 b 6 5 4 3 2 1 X k a 9 F e Z c g f n j i 8 Y o h m b V U T 7 L d W The first symbol c X h I j n 7 6 Z Y L e F k b V U 2 3 4 d a T 5 W c g f o 1 9 8 m represents the type, d L U W T V Y 9 e 7 X Z k F 8 n h g o m c 1 j f i d 2 5 3 a 6 b 4 second digit express e 7 n j I h X 1 d L Y c b 8 F U V 5 4 3 Z 9 W 2 T e f g m 6 a k o numerical size f m 8 9 a k o W 2 3 4 g V n U F b Z Y L 5 j 1 d 6 f e c 7 T i h X The first 32 values g o k a 9 8 m T 5 4 3 f U h V b F d l Y 2 i 6 Z 1 g c e X W j n 7 only the minimum value h i c X 7 e j 4 V T W n 2 g 5 Z d F 1 6 U o Y b L h k 8 a 3 m f 9 of each type i h X c e 7 n 5 T V U j 3 o 4 Y L 1 F b W g Z 6 d i a 9 k 2 f m 8 j n 7 e c X h 2 W U V i 4 m 3 L Y 6 b F T f d 1 Z j 9 a 8 5 g o k k a g o m f 9 Y b 6 1 8 d c Z 5 2 U W T F X 4 V 3 k h n i L 7 e j l d W U V T Z 8 7 e c Y a 1 9 j i o g f X F n m h L 3 4 2 k b 6 5 m f 9 8 k a g U 3 2 5 o T j W 1 6 Y Z d 4 n F L b m 7 X e V h i c n j e 7 X c i 3 U W T h 5 f 2 d Z b 6 1 V m L F Y n 8 k 9 4 o g a o g a k 8 9 f V 4 5 2 m W i T 6 1 L d Z 3 h b Y F o X 7 c U n j e F 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k L m n o F U V W X Y Z T U W d L Y Z T m n j i V g 2 f e c k a 9 h 3 7 8 X U F b 1 o 4 5 6 V T Z Y L d W o h i j U f 5 g c e 8 9 a n 4 X k 7 V b F 6 m 3 2 1 W U L d Z Y V f j n h T o 3 m 7 X a k 8 I 2 e 9 c W 1 6 F g 5 4 b X c I h n j e b Y Z d 7 1 a 6 T W 3 2 5 L k V 4 U X o m g F 8 9 f Y Z T V U W d k X c e L 9 6 a i j m f g 7 b h o n Y 4 3 5 8 F 1 2 Z Y V T W U L a c X 7 d 8 b k h n f m o e 6 i g j Z 5 2 4 9 1 F 3 T V Y d L U g I h n n W m 4 o X 7 9 8 k j 5 c a e T 6 1 b f 2 3 F The following laws can be summarized: 1) The addition of solutions of the same kind must be of type F, α+α=F (main diagonal) 2) Any type plus F, the type remains unchanged. (Row 25 or column 25) 3) Any row or column is an arrangement of 32 types, but it is not a permutation group. 4) F is equivalent to the 0 element of the group 5) Anything of the same type is inverse to each other 6) Satisfying the combination law (a⊕b)⊕c=a⊕(b⊕c); Satisfy the exchange law a⊕b= b⊕a 06: The most special ones are: If the three points of the elliptic curve P⊕Q=R, then Q⊕R≠P R⊕P≠Q,but the types of solutions corresponding to these three points αβγ, but if α⊕β=γ,then β⊕γ=α γ⊕α=β,we call this ternary cycle, This is a deeper relationship than the point group relationship we found. Here α and so on represent one of the solution types 19, ao, U, V, W, X, Y, Z, T, The above 32 groups have more than 180 ternary cycles, such as:123,145,16b,17e,189,1ak, 1cX, 1dL, 1fm, 1go, 1hi, 1jn, 1F1, 1UW, 1VT, 1YZ…： It can be seen from the matrix 1+U=W, We can use the ternary cycle to verify the relationship of any ternary cycle of this matrix : 1+U=(2+3)+(2+d)=3+2+2+d=3+(2+2)+d=3+F+d=3+d=W, ∴1UWForming a ternary cycle. Fill in the type at ①②③ 1,U,W, Rotate counterclockwise, there is ①⊕②=③,②⊕③=①, ③⊕①=②; Or rotate clockwise, there is ①⊕③=②,③⊕②=①,①⊕③=② Fill in type a, a, F, rotate counterclockwise a⊕a=F,a⊕F=a,F⊕a=a 07: Closed and complete (For some examples are insufficient due to sequential search. although a closed set is formed, it is not a complete set) The above matrix is fully affirmed 253375990 its type of solution is a group of order 32. Note ：Due to the limitation of calculation time or space, the sequential search method and the elliptic curve group addition algorithm ⊕, for most congruences, can only reach incomplete partial solution types, Misjudgment may occur， Take this example, if it is restricted by time and space, the following will happen: 1):If only the first solution is found J1：s=611, t= 10 its type(abcd)=(601.10.611.69), With elliptic curve group addition, only one new type F=(1A11) can be added, The solution is s=139443243241,t=9121521240 The solution process is abbreviated as 01 + F, although Z (A) = 32, it is also useless, and can only be misinterpreted as a 2order group. 2):If only the first two solutions are founds J1,J2, look at the two rows and two columns of the above matrix： J1⊕J1=F,J2⊕j2=F,J1⊕J2=J3Misjudged as 4th order group, the solution process is abbreviated as 02+F+01 3): If only the first 3 solutions are found J1,J2,J3, Look at the first three rows and columns of the above matrix: J1⊕J1=F,J2⊕j2=F,J3⊕j3=F, J1⊕J2=J3,J1⊕J3=J2, J3⊕J1=J2, Would be misjudged as a 4order group,03+F 4): If only the first 4 solutions are found J1,J2,J3,J4, Look at the first 4 rows and 4 columns of the above matrix: J1⊕J1=F,J2⊕j2=F,J3⊕j3=F,J1⊕J2=J3,J3⊕J1=J2, J2⊕J3=J1…, The situation has improved, the solution process is abbreviated as 04+F+03， It means that the sequential search found 4 types 1, 2, 3, 4 and ⊕ operation found F, 5, 6, and b. Since the I⊕J operation can no longer add new classes, it can only be misjudged as an 8order group. 5): If only the first 5 solutions are found J1,J2,J3,J4,J5 Look at the first 5 columns of the 5 rows of the above matrix J_i⊕J_i=F,J1⊕J2=J3,J1⊕J3=J2, J3⊕J1=J2… The situation has not improved, the solution process is abbreviated as 05+F+02, It means that the search found 5 types 15, ⊕ operation found F, 6, b, but it can only be misjudged as an 8order group 6) If only the first 6 solutions are found J1J6 Look at the first 6 columns of the 6 rows of the matrix, The solution process is abbreviated as 06 + F + 01, and the type is misjudged as an 8 order group 7): If only the first 7 solutions are found J1J7 Look at the first 7 columns of the 7 rows of the matrix, The solution process is abbreviated as 07+F+07;+01, type misjudged as 16. 8): If only the first 8 solutions are found J1J8, Look at the first 8 columns of the 8 rows of the matrix,, The solution process is abbreviated as 08+F+09;+14, type misjudged as 32. 9): If only the first 8 or more solutions are found, they can all be correctly identified as order 32, but the discovery process x + F + y; z is different 08: This 32order group has more than 160 4order subgroups, and more 8order subgroups. (For example, delete c from the abcF of the ternary cycle and add one of the 28 types, which is the 8order group) We have discovered and proved that the number of types of solutions for 5 different congruence numbers is the 32order group, announced as follows: A Y(A) z(A) T(A Solution process 253375590 8192 32 32 09+F+15;+07 277872990 8192 32 32 13+F+17;+01 565004406 2048 32 32 06+F=12;+13 840075621 16384 32 32 14+F+16;+01 904789886 8192 32 32 06+F=12;+13 Theorem 7.8) Any type of solution α satisfies the relationship α⊕α=F [Prove] Observe the relationship on the main diagonal of the matrix：α⊕α=F Let a solution of A be s,t,then Am^2=st(st)(s+t), Corresponds to a point on an elliptic curve P=(x,y) Its coordinates x=((s^2+t^2)/2m)^2 y=(s^2+t^2 )(s^2t^2+2st)(s^2t^22st)/(8m^3 ) (8.1) Use the tangent method of the elliptic curve: Slope L_2=(3x^2A^2)/2y = (3((s^2+t^2)/2m)^4A^2)/(2 (s^2+t^2 )(s^2t^2+2st)(s^2t^22st)/(8m^3 )) = (3〖〖(s〗^2+t^2)〗^4〖16A〗^2 m^4)/(4m(s^2+t^2 )((s^2+t^2 ) )  〖〖(s〗^2t^2)〗^24s^2 t^2 ) Let 2P = (X, Y) solution correspond to S, T according to the author's own lemma AM^2=ST(ST)(ST) S=〖〖(s〗^2+t^2)〗^2 T=4Am^2 L_2=(3〖〖(s〗^2+t^2)〗^4〖16A〗^2)/2M where M=2m(s^2+t^2 )((s^2+t^2 ) )  〖〖(s〗^2t^2)〗^24s^2 t^2  X=L_2^22x=〖((3〖〖(s〗^2+t^2)〗^4〖16A〗^2 m^4)/2M)〗^2(2〖〖(s〗^2+t^2)〗^2)/(4m^2 ) = 〖〖(3S〗^2T^2)〗^2/(4M^2 )  2S/(4m^2 ) =〖(3S^2〖16A〗^2 m^4)〗^2/〖(2M)〗^2 (2SM^2/m^2)/(4M^2 )=(9S^46S^2 T^28S^4+8S^2 T^2)/(4M^2 )=〖〖(S〗^2+T^2)〗^2/(4M^2 ) , The X of the 2P point is the same as the x structure,and it is substituted into the curve equation Y^2= X(X^2A^2) get Y. Replace the lowercase s,t,m in the y formula of (8.1) with uppercase, In this way, the structure of X and Y is exactly the same as that of x and y in (8.1). For ST,T,S,S+T Its structure is 1A11, That is, Fermattype solution, proved complete. Note:∵AM^2=ST(S^2T^2)=S(4Am^2)( S^2T^2) Both sides Multiply on both sides 2S/(Am^2) get 2SM^2/m^2=8 S^2( S^2T^2) substitute into the above formula theorem 7.9:Types of congruence solutions=2,4,8,16,32, Proof of the calculation of millions of congruences [Prove] Introduced tys_200w_tj in Appendix 10.4 The calculation and significance of the three fields bz, IJ and up in the table , here is a table to illustrate our proof. Note red numbers represents the number of congruence 
20200628, 10:38  #4 
Jun 2020
2·11 Posts 
New function, new criterion, new theorem and new conjecture for discriminant of congruence number
New function, new criterion, new theorem and new conjecture for discriminant of congruence
Zhou CongYao College of Information Science and Engineering, Hunan University Yu Wei Department of statistics & financial engineering,College of mathematics and statistics,ningbo university TangXiaoNing Beijing Haitian Start Technology Service Co., Ltd. ABstruct In this paper,a new function of congruence discrimination is proposed,namely,the upper bound estimation funtion of the number of solutions of the congruence. If the value of this function is 1,the posttive integer A is noncongruent,this is the new N1 criterion.Using this guideline, to history all 3 congruence negative theorms,a very simple elementary proof is given;and proved the author's newly discovered three modular 8type congruence negative theorems; At the same time,eight new congruences negation theorems(not model 8 types) are proved,by the way,we derive 1511 infinite sequences of noncongruent If A is a congruence number,the value of this function is integer power of 2,that is,there are at least two types of solutions for congruence,one of them must be Fermat type,and each type has an infinite number of solutions. Using this upper bound function, some new theorems for the number of congruence solutions are derived: the exact value of the solution is 2, 4, 8, 16, 32; On this basis, a new conjecture is put forward: If the type of the congruence solution is used as an element, then these elements form a group,and the order of this group is integer power of 2 ,for any congruence, the conversion relationship between the types is fixed. Keyword：congruence number，Elliptic Curves，Number Theory,Group Mod note: This post has been moved here from a new thread. Please don't make new threads with similar content to existing threads. 
20200628, 14:30  #5  
Nov 2003
2^{6}×113 Posts 
Quote:
"the congruence". One must ask: What congruence?? Does he refer to what is known as "the congruent number problem"? 

20200629, 07:38  #6  
Jun 2020
2×11 Posts 
Quote:
the conjecture that all congruence solutions all to be a group and the order of the group is a power of 2, it may be one of the most important discoveries in the field of millennial congruence. Last fiddled with by Zcyyu on 20200629 at 08:04 

20200629, 11:02  #7 
Feb 2017
Nowhere
111011101101_{2} Posts 
An introduction to the subject of "congruent numbers" may be found here: THE CONGRUENT NUMBER PROBLEM by Keith Conrad.

20200630, 06:52  #8  
Jun 2020
2×11 Posts 
New function, new criterion, new theorem and new conjecture for discriminant of congruence
Quote:
1) Elementary proof of three theorems in history 2) Proof of our three modular 8 theorems 3) Our proof of 8 modeless 8 theorems and 1511 noncongruent rsequences 4) 8 theorems proving of the types of solutions 5) Three new conjectures 6) One new funtion 7) One new criterion Last fiddled with by Zcyyu on 20200630 at 07:03 

20200630, 20:36  #9 
Jun 2020
26_{8} Posts 
New function, new criterion, new theorem and new conjecture for discriminant of congruent number

20200630, 20:44  #10 
If I May
"Chris Halsall"
Sep 2002
Barbados
10010001101110_{2} Posts 

20200701, 08:16  #11 
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
10202_{10} Posts 

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