# Diophantine equations

Lets have the Diophantine equation $x^2-dy^2=f$ where $x=m^2-m+k$, $y=m-1$ and $d=m^2+2k$, where $m,k$ natural numbers. What are the necessary conditions to obtain infinite fundamental solutions of this equation?

For example, if$m=7$ and $k=6$ we have $48^2-61*6^2=108$ we can obtain infinite fundamental solutions of this equation: $867^2-61\times 111^2=108$, $72237^2-61\times 9249^2=108$, $1320393^2-61\times 169059^2=108$ and so on.

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I don't understand how the other solutions satisfy $x = m^2 - m + k$, etc. If there are no restrictions, the condition is just that $d$ is not a square and $f \ne 0$. –  Erick Wong Jun 23 '12 at 5:50
The phrase "fundamental solution" is being used in a nonstandard way in the question. –  Gerry Myerson Jun 23 '12 at 10:15
@Erick, The conditions apply to the first generating Diofantine equation. Like the equations 19^2-10*6^2=9 and 13^2-10*4^2=9, the values of x and y are different. Another way to think is, if d is given, formulate the Diofantine equation and find the fundamental solutions of the formulated Diofantine equation. –  Vassilis Parassidis Jun 23 '12 at 14:48
@Gerry, These Diofantine equations have applications to hyperelliptic equations with global solutions. As for the rest, see the reply to Erick. –  Vassilis Parassidis Jun 23 '12 at 14:50
Probably should be linked to math.stackexchange.com/a/147977/30402 , although I still don't see how this is a question. –  Erick Wong Jun 23 '12 at 17:59