<s>[INST] Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime. 
Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$

Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equation, but it doesn't work. I have been taught Congruence 
 [/INST] Suppose $x,y\geq 0$ and let $g:= \gcd(x^2-2x+2,x^2+2x+2)$ 
Let prime $q\mid g$ then $q\mid 4x$. If $q\ne 2$ then $q\mid x$ but  then $$q\mid x^2+2x \implies q\mid 2 \implies q=2$$ 
So the only prime which can divide $ \gcd(x^2-2x+2,x^2+2x+2)$ is 2.


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*$\color{red}{Case\; 1}$: $2\mid g$ then $x$ is even, so $x=2z$ then we see $4\nmid g$ and thus $g=2$. But then $2\mid y$ so $16\mid 4(4z^4+1)$ which is impossible.

*$\color{red}{Case\; 2}$: $2\nmid g$, so $g=1$, then $$x^2-2x+2 = pa^4 \;\;\;\wedge \;\;\;x^2+2x+2=b^4$$ where $a,b$ are relatively prime or $$x^2-2x+2 = a^4 \;\;\;\wedge \;\;\;x^2+2x+2=pb^4$$

*$\color{red}{Case\; 2.1}$: $$(x+1)^2<x^2+2x+2 < (x+2)^2$$ 
so no solution.


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*$\color{red}{Case\; 2.2}$: If $x>1$ then $$(x-1)^2<x^2-2x+2 < x^2$$ 
so no solution. So $x=1$ ...


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