<s>[INST] Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$ 
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.

My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
I don't know how can I proceed.
 [/INST] First, observe that $8y^4$ is even, so $x^4+6x^2+1$ is even, so $x^4+6x^2$ is odd. Therefore, $x$ is odd. Rewrite $x$ as $2n+1$, where n is an integer and $n\geq0$. So the original equation can be rewritten as $(2n+1)^4+6(2n+1)^2+1=8y^4$. By simplifying it we see that $2n^4+4n^3+6n^2+4n+1=y^4$, which can be rewritten as $(n+1)^4+n^4=y^4$, where $n$ and $y$ are integers. It's obvious that when $n=0$ and $y=1$ the equation holds, but since $x>0$ this cannot be the case. So $n\geq1$. But by Fermat's Last Theorem, such $n$ and $y$ does not exist.
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