<s>[INST] Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$. Find all pairs of intergers $(x,y)$ such that $x^4 + 1 = 2y^2$.
I'm thinking of Gaussian integers, since the LHS can be factored in $\mathbb{C}$. But I don't know how to continue here.
 [/INST] We consider $x, y$ positive (if $x$ is solution then $-x$ is solution, same with $y$). From $x^4+1=2y^2$ we deduce that $2y^2=t^2+1$, where $t=x^2$ must be odd. Let $t=2k+1$, so $y^2=k^2+(k+1)^2$. We can now use the fact that $y=(m+n)^2$ and $k=(m-n)^2$ and $(k+1)=2mn$ or $k=2mn$ and $(k+1)=(m-n)^2$, where $m, n\in \mathbb Z$. 
Case 1. $k=(m-n)^2$ and $(k+1)=2mn$.
Then $k+2(k+1)=(m+n)^2=y$, so $y=3k+2$ so $$9k^2+12k+4=2k^2+2k+1$$ and we find k.
Case 2. $k=2mn$ and $(k+1)=(m-n)^2$
Then $2k+k+1=(m+n)^2=y$, so $y=3k+1$ so $$9k^2+6k+1=2k^2+2k+1$$ and we find k.
EDIT: It is NOT right, sorry for the confusion
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