<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] Certainly, $x$ must be odd.  Note that $(x^2+1)^2 = x^4+2x^2+1 = 2y^2+2x^2$ and $(x^2-1)^2 = x^4-2x^2+1 = 2y^2-2x^2.$  Multiply these two equations together to get
$$(x^4-1)^2 = 4(y^4-x^4).$$
Since $x^4\equiv 1 \pmod{4}$, we can write
$$\left(\frac{x^4-1}{2}\right)^2 = y^4-x^4.$$
This implies a solution to the better-known Diophantine equation
$$X^4-Y^4 =Z^2.$$ 
This equation can be proved to have only trivial solutions by infinite descent, so we must have $x^4-1 = 0$.  Hence $x=\pm 1$ which forces $y=\pm 1$.
For the descent (switching to lower case):
We have $z^2 + (y^2)^2 = (x^2)^2.$ By the usual construction of solutions to the Pythagorean equation, either $y^2=2mn$ or $y^2=m^2-n^2.$ If $y^2 = 2mn,$ then $m = u^2,$ and 
$n = 2v^2.$ Then $z^2 = m^2 - n^2 = u^4 - v^4,$ and $u^4 = m^2 < m^2 + n^2 <
x^2 < x^4,$ so we have a smaller positive solution. If $y^2 = m^2 - n^2,$ 
$z = 2mn,$
and $x^2 = m^2 + n^2,$ then $x^2y^2 = m^4 - n^4$ which is a smaller solution, because $m^2 < x^2.$
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