Integer solutions to $x^2=2y^4+1$.

Question

Find all integer solutions to $x^2=2y^4+1$.

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What I tried

The only solutions I got are $(\pm 1 ,0)$, I rewrote the question as : is $a_{n}$ a perfect square for $n>0$ were

$$a_0=0,\quad a_1=2, \quad a_{n+2}=6a_{n+1}-a_n.$$ I tried taking $\pmod{4}$ and $\pmod{12}$ but that led me nowhere.

Answer 1

There are no solutions other than those trivial ones. First we note that $x$ must be odd. Let $x = 2k+1$. This gives $2k^2+2k = y^4$. Thus $y = 2j$ and we have $k(k+1) = 8j^4$. Note that $\gcd(k,k+1) = 1$ so we have two cases.

Case 1: $k+1 = 8a^4, k = b^4$.

This case is impossible since this implies $b^4 \equiv -1 \pmod 4$ but $-1$ is a not a quadratic residue $\pmod 4$.

Case 2: $k+1 = b^4, k = 8a^4$.

This case is equivalent to solving the diophantine equation $8a^4+1 = b^4$. An elementary solution is given here. In the referenced paper, it is prove that the only non trivial solution to this equation is $(a,b) = (1,3)$ which gives $k = 8, k+1 = 9$ but $9 = b^4$ is impossible.