Prime number problem Find natural numbers $x$,$y$ and prime $p$ so that $y^4+4=p^x$. 
In my opinion, i think we should consider cases y=0 and y=1. Hence (x;y;p)=(2;0;2), (1;1;5). Then we proof: if y>1 =>there's no solution.
 A: We have $y^4+4=(y^2+2y+2)(y^2-2y+2).$ We shall consider cases on $y.$
Case $(i).$ $y$ even. Since $y^2+2y+2\equiv y^2-2y+2\equiv0\pmod2$ and $y^4+4=p^x$ then $p=2;$ moreover, clearly, this implies that $x\geqslant2.$ Thus $y$ is even. Let $y=2k.$ Then $4(4k^4+1)=2^x$ so $4k^4+1=2^{x-2}.$ If $x>2$ then $2\mid1$ which is absurd. Hence $x=2$ so $4k^4=0$ which implies that $k=0.$ Hence $y=0,\;x=2$ and $p=2.$
Case $(ii).$ $y$ odd. Since $y^4+4=(y^2+2y+2)(y^2-2y+2)=p^x$ then $y^2+2y+2=p^a$ and $y^2-2y+2=p^b,$ where $a$ and $b$ are non-negative integers such that $a>b$ and $a+b=x.$ Clearly, since $y$ is a natural number, $b>0.$ Thus solving both quadratic equations for $y$ we have $(y+1)^2=p^a-1$ and $(y-1)^2=p^b-1.$ Thus $p^b(p^{a-b}-1)=4y.$ Since $p$ is an odd prime then $p^b=y$ and $p^{a-b}-1=4.$ Hence $p^{a-b}=5$ which implies that $p=5$ and $a=b+1.$ As $5^a-1=5^{b+1}-1=(5^b-1)\cdot5+4$ then $(y+1)^2=5(y-1)^2+4$ so $y\in\{1,2\}.$ A quick calculation shows that only $y=1$ works. Hence the only solution in this case is $y=1,$ $p=5$ and $x=1.$
