Find the Squares

[JM Montolio A]

( 1* 5* 9)* 4 = 180 = 5* 36 = 4* 45
( 2* 6*10)* 4 = 480 = 12* 40 = 8* 60


g 5 ( 1* 5* 9)* 4 = 180 = 5* 36= 4* 45
g 7 ( 7*16*25)* 9 = 25200 = 112* 225= 63* 400
g 14 ( 7* 8* 9)* 1 = 504 = 9* 56= 7* 72
g 65 ( 5* 9*13)* 4 = 2340 = 45* 52= 20* 117
　
g 21 (-1* 3* 7)* 4 = 84 = 4* 21= 3* 28
g 34 (-1* 8*17)* 9 = 1224 = 9* 136= 8* 153
g 41 (-9*16*41)* 25 =147600 = 225* 656= 144* 1025


5* 6* 7* 1 = 210 (11.500)^2 -- ( 18.500)^2 -- ( 23.500)^2
1* 5* 9* 4 = 180 (15.500)^2 -- ( 20.500)^2 -- ( 24.500)^2
3* 12* 21* 9 =6804 (76.500)^2 -- (112.500)^2 -- (139.500)^2
　
15 ( 1* 3* 5)*2 = 30 =PQRS 3* 10 = 2* 15
5 ( 1* 5* 9)*4 = 180 =PQRS 5* 36 = 4* 45
5*13 ( 8*13*18)*5 = 9360 =PQRS 90* 104 = 40*234
　
Congruent forms: n(n+1)(2n+1), n^3-n, n^4-1, n(n+1)(n+2).


I think im right on assertions.
JM M

[science_man_88]

Originally Posted by JM Montolio A

n = pq = rs = r(p+q+r) ; "s means sum".


trivial proof.
n= 6 = 2*3= 1*(2+3+1). 6 is congruent.
1 x 6
2 x 3
-------
I named that form PQRS. Well, then:
* The product of 3 rationals , (d^2)-equidistanced, is PQRS.
for any d.
* Most, the product of 3 rational d-equidistanced by d, is also PQRS.


And i play to proof. congruent numbers using small Q numbers. I conjectured that any congruent number, multiplied by some square, is a PQRS number.

Examples are not proof.

[JM Montolio A]

Define PQRS. N = pq = rs =r(p+q+r); r< p< q< s.
Are PQRS: 24,54,60,84,96,120,150,180,210,...


Properties
q(p-r)=(r)(p+r).
2r=-(p+q) "+-" SQRT( (p^2) + (q^2) + 6pq).
q(p-r)=r(p+r).
if p=r+1, N=(r)(r+1)(2r+1).
if p=r+2, N=(r)(r+1)(r+2).
if p=r+k, kN=(r)(r+k)(2r+k)


PQRS are congruent numbers.
3 Squares N-equidistanced: (q-p) , (q+p), (r+s).
Triangle sides: (p+r), (q+r), (p+q).
Eliptic curve: "(x^3)-(N^2)*x = (y^2)"
Points: x=ps,qs; y=(p+r)x, (q+r)x.
--------------------------------------------------------------
If n = (a)(a+mm)(a+2*mm)*mm,
with "mm" as (m^2) and "aa" as (a^2).
Then 3 squares n-equidistance are: (d^2, e^2, f^2),
with:
d =(mm^2)-(a^2/2).
e =(m^2)((m^2)+a) +((a^2)/2)
f =(m^2)((m^2)+2a)+((a^2)/2)
--------------------------------------------


About the conjecture "there is some square for any congruent number g, that the product is a number PQRS":
g*(S^2) is PQRS
g 7 S 60 n 25200 =pq = 112* 225 =rs = 63* 400
g 39 S 10 n 3900 =pq = 13* 300 =rs = 12* 325
g 41 S 60 n 147600 =pq = 225* 656 =rs = 144* 1025


JM M

[CRGreathouse]

Originally Posted by JM Montolio A

Define PQRS. N = pq = rs =r(p+q+r); r< p< q< s.

https://oeis.org/A009112

[JM Montolio A]

Let pq=rs


Ten (q)(p-r)=(r)(p+r)


n=pq=rs


n[(p-r)^2]=...=r(p-r)p(p+r)


With r=1, n=product of 3 consecutives.
With r=(d^2), n=product of 3 numbers (d^2)-equidistanced.
For any r, n=product of 3 numbers r-equidistanced by r.


Thus.

The product of 3 numbers (d^2) equidistanced is congruent.
The product of 3 numbers (d ) equidistanced by d is congruent.


¿ Some mistake ?
JM M

[Dr Sardonicus]

Originally Posted by JM Montolio A

Let pq=rs


Ten (q)(p-r)=(r)(p+r)


n=pq=rs


n[(p-r)^2]=...=r(p-r)p(p+r)


With r=1, n=product of 3 consecutives.
With r=(d^2), n=product of 3 numbers (d^2)-equidistanced.
For any r, n=product of 3 numbers r-equidistanced by r.


Thus.

The product of 3 numbers (d^2) equidistanced is congruent.
The product of 3 numbers (d ) equidistanced by d is congruent.


¿ Some mistake ?

There appear to be counterexamples to your third assertion. Looking at OEIS A003273 Congruent numbers: positive integers n for which there exists a right triangle having area n and rational sides, the numbers

48 = 2 * (2 + 2) * (2 + 4);

66 = 1 * (1 + 5) * (1 + 10); and

105 = 3 * (3 + 2) * (3 + 4)

aren't on the list.

As to your first two assertions, the formulation

K = a * b * (a - b) * (a + b)

gives, with b = 1, a * 1 + (a-1) * (a+1) = (a - 1) * a * (a+1),

the product of three consecutive integers, as I pointed out before.

Taking b = d^2, we have

K = a * d^2* (a - d^2) * (a + d^2).

Since square factors can be dropped (similar triangles have areas differing by a factor which is the square of the ratio of similitude), the number

n = (a - d^2) * a * (a + d^2)

is congruent.

[JM Montolio A]

[QUOTE counterexamples to your third assertion. Looking at
OEIS A003273 the numbers
48 = 2 * (2 + 2) * (2 + 4);
66 = 1 * (1 + 5) * (1 + 10);
105 = 3 * (3 + 2) * (3 + 4)
aren't on the list.


REPLY.
My post is confuse. Best is
"the product of 3 number d-equidistanced and MULTIPLIED BY the d".
Is the product of 4 numbers.


At your comment,iIt must multiply by the d,
because this d is not square.
On the list: 48*2 (96), 66*5 (330), 105*2(210).

QUOTED Taking b = d^2, we have K = a * d^2* (a - d^2) * (a + d^2). Since square factors can be dropped (similar triangles have areas differing by a factor which is the square of the ratio of similitude), the number n = (a - d^2) * a * (a + d^2)is congruent.

REPLY. Yes. But not secure to be PQRS.

[JM Montolio A]

General form: n= d*f*c*l, with:


d: distance.
f: first number = c-d.
c: center number.
l: last number = c+d.


The values of p,q,r,s are: p=f*c, q=d*l, r=d*f, s=c*l
Exception: if q less p, change values. That is for d:1,2,3.


An PQRS implies congruent.


JM M