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Method of solving $y^2+1=1250z$

In my answer to A curious coincidence for Wroblewski's solutions to $1^4+x_2^4+x_3^4+x_4^4+x_5^4 = y_1^4$ I used a solution to this Diophantine equation, for odd $y$,

$y=625b\pm182$ gives $(y^2+1)$ divisible by $1250$ for odd $b$

However, I only found this by experimenting on a spreadsheet.

My question

I’m looking for a simple, elementary, method of solving $y^2+1=1250z$, please.

Old Peter
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    http://www.wolframalpha.com/input/?i=diophantine++y%5E2%2B1%3D1250z – Eric Lee Jun 21 '17 at 19:10
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    You want $y^2\equiv -1\pmod{1250}$. This is equivalent to having both $y^2\equiv-1\pmod2$ and $y^2\equiv-1\pmod{625}$. The former congruence can be ignored ($y$ is odd). The latter can be solved by recalling the fact that $\Bbb{Z}{625}^*$ is cyclic of order $\phi(625)=500$. Therefore there are exactly two residue classes (those of order four) with square equal to $-1$. You can find them by Hensel lifting the solutions of $y^2\equiv-1\pmod 5$ to solutions modulo $5^2,5^3$ and finally $5^4$. Alternatively you can locate a generator $g$ of $\Bbb{Z}{625}^*$, and calculate $y=\pm g^{500/4}$. – Jyrki Lahtonen Jun 21 '17 at 19:24
  • @EricLee Thank you, but that only gives 2 solutions; I expect 4. I'll check again tomorrow. – Old Peter Jun 21 '17 at 19:25
  • @JyrkiLahtonen Sorry, far too advanced for me. – Old Peter Jun 21 '17 at 19:27
  • @EricLee My error, sorry. Your link does indeed give all solutions. – Old Peter Jun 22 '17 at 13:39
  • :D But I didn't do any work so thank wolfram alpha not me :D – Eric Lee Jun 23 '17 at 00:35

2 Answers2

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I found this parametric solution experimentally:

$ (x=157 , y =443)$, $(x=2293, y=1693)$

Suppose $ y=km^2+k_1$, we make following system of equations:

$km_1^2+k_1=1693$

$km_2^2+k_1=443$

$k(m_1^2-m_2^2)=1693-443=1250=2 ˣ 625$

Suppose $k=2$ and $m_1^2-m_2^2=625$

Compare this with:

$65^2-60^2=25^2$

We find $m_1=65$ and $m_2 = 60$ plugging one of thses in equation we get:

$2 ˣ 65^2 +k_1=1693$ ⇒ $k_1=-6757$

Therefore parametric formula for y is:

$y=2 m^2 -6757$

and for z is:

$z=\frac{(2m^2-6757)^2 +1}{1250}$

Here are some results for $60≤ m≤ 1000$ found by Python$ (x, y, m)$:

$(157, 443, 60 ), (2293, 1693, 65 ), (3044821, 61693, 185), (3426229,65443,190) ,(27511285, 185443, 310), (29396965, 191693, 315), (110524549, 371693, 435), (115789501, 380443, 440), (307959613, 620443, 560), (319228837, 631693, 565), (694441477, 931693, 685), (715089973, 945443, 690),(1363345141, 1305443, 810) ,(1397497909, 1321693, 815), (2426795605, 1741693, 935), (2479327645, 1760443,940)$

It seems this equation has infinitely many solutions.

sirous
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  • Old peter, please see my answer to your question. – sirous Nov 11 '17 at 09:34
  • Thank you so very much for attempting to answer my question. Unfortunately, it has already been fully answered in a comment; I’ll post that answer to make this clear to others. +1 for all the work! – Old Peter Nov 11 '17 at 15:58
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This question was fully answered by @EricLee on 21 June 2016

http://www.wolframalpha.com/input/?i=diophantine++y%5E2%2B1%3D1250z

Old Peter
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