# An extension to Pell's equation

During my number theory seminary, I found this interesting problem and I didn't know how to solve it. Given Pell's equation $$x^2-3y^2=1,$$ where $x,y \in \Bbb N,$ show that there are infinitely many solutions such that $x-1$ is a perfect square.

• An equivalent statement would be to prove there exists infinitely many integers where $x-1$ is a perfect square and $x+1$ is three times a perfect square. – S.C.B. Mar 29 '16 at 16:55
• This is because if $x-1=a^2$, $x+1=3b^2$, multiplying them together and rearranging gives us that $x^2-3(ab)^2=1$. So all you need to do is prove there exists infinite $a,b$ that $3b^2-a^2=2$. – S.C.B. Mar 29 '16 at 17:02
• @MXYMXY: More generally, since Legendre proved that $u^2-dv^2 = -2$ always has a solution for prime $d=8n+3$ and has infinitely many $u,v$, then this leads to infinitely many $x,y$, $$x^2-dy^2=(u^2+1)-d(uv)^2=1$$ where OP's was just the case $d=3$. – Tito Piezas III Mar 29 '16 at 19:28
• @TitoPiezasIII I don't know this result proved by Legendre. Can you tell me what is its name? – I. Stefan Mar 29 '16 at 20:01
• I don't remember if it has a specific name, but it is also mentioned in Dickson's History of the Theory of Numbers. Legendre also discovered the counterpart to your observation, namely, $u^2-dv^2=2$ always has a solution for prime $d=8n+7$ which leads to $$x^2-dy^2=(u^2-1)^2-d(uv)^2= 1$$ Essentially, what you saw was the nice tip of an iceberg. :) – Tito Piezas III Mar 29 '16 at 20:43

If $a^2-3b^2=1$, then $$((a+3b)^2+1)^2-3((a+3b)(a+b))^2=1 .$$
We also know that one solution where $a=7$ and $b=4$ exists. Thus there exist infinite such solutions.