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 

Notice that $\text{gcd}(X^2+2X+2,X^2-2X+2)=\text{gcd}(X^2+2X+2,4X)=2$ or $1$. If $2|X$ then 2 must also divide $Y$ and thus: $4X^4+1=4Y^4$ a contradiction. 
Then $\text{gcd}(X^2+2X+2,X^2-2X+2)=1$ and since $(X^2+2X+2)(X^2-2X+2)=pY^4$ this implies:
1) $X^2+2X+2={Y_1}^4$ and $X^2-2X+2=p{Y_2}^4$ or 
2) $X^2+2X+2=p{Y_1}^4$ and $X^2-2X+2={Y_2}^4$. 
Let us consider the first case: write ${Y_1}^4-(X+1)^2=1$ and observe that $({Y_1}^2-X-1)({Y_1}^2+X+1)=1 \implies {Y_1}^2-X-1={Y_1}^2+X+1=1$. Can you take it from here for both cases?
