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

Question

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.

Answer 1

This question was fully answered by @EricLee on 21 June 2016

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

Answer 2

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.