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Old 2017-11-17, 11:18   #1
devarajkandadai
 
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Default Conjecture pertaining to modified Fermat's theorem

Let a = x + ysqrt(m) be a quadratic integer. As stated before modified
Fermat's theorem is valid for m = -1 and 5 as practically proved by Hardy (An introduction to the theory of numbers). Conjecture: it is valid for all integer values of m subject to conditions:
i) a and p are coprime
ii) m not equal to p.

Recall that modified Fermat's theorem is a^(p^2-1) = = 1 (mod p).

Here x,y and m belong to Z.

Last fiddled with by devarajkandadai on 2017-11-17 at 11:21 Reason: important poimt omited
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Old 2017-11-17, 15:09   #2
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Quote:
Originally Posted by devarajkandadai View Post
Let a = x + ysqrt(m) be a quadratic integer. As stated before modified
Fermat's theorem is valid for m = -1 and 5 as practically proved by Hardy (An introduction to the theory of numbers). Conjecture: it is valid for all integer values of m subject to conditions:
i) a and p are coprime
ii) m not equal to p.

Recall that modified Fermat's theorem is a^(p^2-1) = = 1 (mod p).

Here x,y and m belong to Z.
So to be explicit, your conjecture is:

Given any integer m which is not a square, any (rational) prime p, and any integers x and y, if
  • at least one of x and y is not divisible by p and
  • m is not equal to p
then (x + y*sqrt(m))^(p^2-1) = 1 + 0*sqrt(m) mod p.

Is this correct?
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Old 2017-11-17, 17:12   #3
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Quote:
Originally Posted by devarajkandadai View Post
Let a = x + ysqrt(m) be a quadratic integer. As stated before modified
Fermat's theorem is valid for m = -1 and 5 as practically proved by Hardy (An introduction to the theory of numbers). Conjecture: it is valid for all integer values of m subject to conditions:
i) a and p are coprime
ii) m not equal to p.

Recall that modified Fermat's theorem is a^(p^2-1) = = 1 (mod p).

Here x,y and m belong to Z.
I note that not all quadratic integers satisfy the condition that x, y be integers. The case m = 5, which you mention, affords examples; e.g. the roots of

x^2 - x - 1 = 0

which by quadratic formula are \frac{1\pm\sqrt{5}}{2}.

Second, you need to make it clear what you mean by "relatively prime to p."

But, assuming you mean "a" can be any algebraic integer in the maximal order R (ring of algebraic integers) in Q(sqrt(m)), such that the ideal aR + pR is all of R, or alternatively, a is invertible in the residue ring R/pR, then assuming p does not divide the field discriminant, your "conjecture" is trivial and has been known since the precambrian era. OK, maybe not that long, but Lordy, it's been known a long time. Under this additional assumption, the residue ring R/pR is either the field of p^2 elements, of which the invertible elements form a cyclic group of order p^2-1; or the direct product of two copies of the field of p elements. The group of invertible elements in this ring is the direct product of two cyclic groups of order p-1. In either case, the exponent of any invertible element divides p^2 - 1.

If p does divide the field discriminant, your conjecture is in trouble. For example, with m equal to -1, p = 2 (which is not equal to m as per your condition), and a = i, p^2 - 1 is 3, but i^3 is not congruent to 1 (mod 2). In fact, i^2 = -1 is congruent to 1 (mod 2).

Last fiddled with by Dr Sardonicus on 2017-11-17 at 17:14 Reason: Fixing typos and editing mistakes
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Old 2017-11-18, 11:08   #4
devarajkandadai
 
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Quote:
Originally Posted by Dr Sardonicus View Post
I note that not all quadratic integers satisfy the condition that x, y be integers. The case m = 5, which you mention, affords examples; e.g. the roots of

x^2 - x - 1 = 0

which by quadratic formula are \frac{1\pm\sqrt{5}}{2}.

Second, you need to make it clear what you mean by "relatively prime to p."

But, assuming you mean "a" can be any algebraic integer in the maximal order R (ring of algebraic integers) in Q(sqrt(m)), such that the ideal aR + pR is all of R, or alternatively, a is invertible in the residue ring R/pR, then assuming p does not divide the field discriminant, your "conjecture" is trivial and has been known since the precambrian era. OK, maybe not that long, but Lordy, it's been known a long time. Under this additional assumption, the residue ring R/pR is either the field of p^2 elements, of which the invertible elements form a cyclic group of order p^2-1; or the direct product of two copies of the field of p elements. The group of invertible elements in this ring is the direct product of two cyclic groups of order p-1. In either case, the exponent of any invertible element divides p^2 - 1.

If p does divide the field discriminant, your conjecture is in trouble. For example, with m equal to -1, p = 2 (which is not equal to m as per your condition), and a = i, p^2 - 1 is 3, but i^3 is not congruent to 1 (mod 2). In fact, i^2 = -1 is congruent to 1 (mod 2).
Am leaving town for only a week; will reply in detail to all queries on return
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Old 2017-11-27, 10:08   #5
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Originally Posted by devarajkandadai View Post
Am leaving town for only a week; will reply in detail to all queries on return
Back from a brief holiday. Why do we need to modify Fermat's theorem? Because it does not necessarily work when the base is an algebraic quadratic integer and hence we aim to arrive at one theorem that covers when the base is both a rational integer as well as cases when the base is an algebraic quadratic integer.
( to be continued).
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Old 2017-11-27, 20:45   #6
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Please consider that
(x+y)^p=x^p+y^p mod p if p is prime

for
(x+yI)^p = x^p+(yI)^p mod p
if p=1 mod 4
=> x^p+(y^p)I
if p=3 mod 4
=> x^p-(y^p)I

same calculation for A=sqrt (2) for example
(x+yA)^p = x^p +(yA)^p = x^p+(y^p)(A^p)

You get a criteria for an "extended Fermat"
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Old 2017-11-28, 12:26   #7
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Quote:
Originally Posted by devarajkandadai View Post
Back from a brief holiday. Why do we need to modify Fermat's theorem? Because it does not necessarily work when the base is an algebraic quadratic integer and hence we aim to arrive at one theorem that covers when the base is both a rational integer as well as cases when the base is an algebraic quadratic integer.
( to be continued).
Preliminary remarks: 1) Co-primality means co-primality with relevant Norm
2) If, with respect to a given base as given above Fermat's theorem is valid for a given p needless to say modified Fermat's theorem is valid; the converse is not necessarily true.
3) If for a given base and p modified Fermat's theorem is valid
in the relevant real field it will also be valid in the corresponding complex field (example: if it is valid in the field Mod(x^2 - 5) it will be valid in the field Mod(x^2 + 5)
Now let me state the conjecture in the case where the discriminant, m, is a prime number : Let a be an algebraic quadratic integer. Then a^(p^2-1)==1 (mod p) subject to the following conditions:
i) p is coprime with Norm of a ii) p not equal to m, the discriminant.
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Old 2017-11-29, 10:46   #8
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Quote:
Originally Posted by devarajkandadai View Post
Preliminary remarks: 1) Co-primality means co-primality with relevant Norm
2) If, with respect to a given base as given above Fermat's theorem is valid for a given p needless to say modified Fermat's theorem is valid; the converse is not necessarily true.
3) If for a given base and p modified Fermat's theorem is valid
in the relevant real field it will also be valid in the corresponding complex field (example: if it is valid in the field Mod(x^2 - 5) it will be valid in the field Mod(x^2 + 5)
Now let me state the conjecture in the case where the discriminant, m, is a prime number : Let a be an algebraic quadratic integer. Then a^(p^2-1)==1 (mod p) subject to the following conditions:
i) p is coprime with Norm of a ii) p not equal to m, the discriminant.
Note: unique factorisation not relevant.
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Old 2017-11-30, 05:16   #9
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Quote:
Originally Posted by devarajkandadai View Post
Preliminary remarks: 1) Co-primality means co-primality with relevant Norm
2) If, with respect to a given base as given above Fermat's theorem is valid for a given p needless to say modified Fermat's theorem is valid; the converse is not necessarily true.
3) If for a given base and p modified Fermat's theorem is valid
in the relevant real field it will also be valid in the corresponding complex field (example: if it is valid in the field Mod(x^2 - 5) it will be valid in the field Mod(x^2 + 5)
Now let me state the conjecture in the case where the discriminant, m, is a prime number : Let a be an algebraic quadratic integer. Then a^(p^2-1)==1 (mod p) subject to the following conditions:
i) p is coprime with Norm of a ii) p not equal to m, the discriminant.
Case ii) Let m, the discriminant be a squarefree composite number. Let m= p_1*p_2*....p_r . Now let a be an algebraic quadratic integer.

Then a^(p^2-1) = = 1 subject to the following conditions:
i) p is coprime with norm of a ii) p not equal to p_1, p_2,....p_r
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Old 2017-12-11, 05:04   #10
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Quote:
Originally Posted by devarajkandadai View Post
Case ii) Let m, the discriminant be a squarefree composite number. Let m= p_1*p_2*....p_r . Now let a be an algebraic quadratic integer.

Then a^(p^2-1) = = 1 subject to the following conditions:
i) p is coprime with norm of a ii) p not equal to p_1, p_2,....p_r
Counter examples pertaining to both cases invited.
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Old 2017-12-20, 05:19   #11
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Quote:
Originally Posted by devarajkandadai View Post
Counter examples pertaining to both cases invited.
Let us now consider the simplest case i.e. case (i) in which a = 0, b and c =1.Since m is
prime, raising sqrt(m) or sqrt(-m) to an even power and recalling that m is not equal to p reduces the case to nothing but Fermat's theorem. Hence partly proved. (to be continued).
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