<s>[INST] Diophantine equation:$x^5+x^4+1=p^y$ 
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.

My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tried factorising the equation i.e. $x^5+x^4+1=(x^2+x+1)(x^3-x+1)=p^y$
Now I am stuck. I tried considering the gcd of the common factors but it does not help.
Any ideas??
 [/INST] Continuing your approach note that from the equality
$$
(x^3-x+1)(x^2+x+1)=p^y
$$
we obtain that $x^3-x+1=p^n$ and $x^2+x+1=p^m$ for some nonnegative integers $m$ and $n$.
For $x=1$ and $x=2$ we have solutions $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
Now note that for $x\geq 3$ we have
$$
x^3-x+1=(x^3-1)-(x-2)=(x-1)(x^2+x+1)-(x-2)>x^2+x+1,
$$
so $p^n>p^m$ or $n>m$. Hence, $p^m\mid p^n$, so $x^2+x+1\mid x^3-x+1$. Since
$$
x^3-x+1=(x-1)(x^2+x+1)-(x-2)
$$
we have $x^2+x+1\mid x-2$. However, for $x\geq 3$ we have $0<x-2<x^2+x+1$, so there are no solutions in case $x\geq 3$.
Therefore, all solutions to this equation are $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
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