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??
For anybody like me who is unable to view the solution at Diophantus Era begins! that Math Lover's question comment linked to because they're not a member of Brilliant and doesn't want to join, here is a solution.
First, apart from $x = 1$, which leads to the solution of $(1, 1, 3)$ you've already found, then both of the factors on the left are greater than $1$ and, thus, must be positive powers of $p$. This gives
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
$$x^3 - x + 1 \equiv 0 \pmod{p} \tag{2}\label{eq2A}$$
Next, \eqref{eq2A} minus \eqref{eq1A} gives
$$x^3 - x^2 - 2x \equiv 0 \pmod{p} \implies x(x + 1)(x - 2) \equiv 0 \pmod{p} \tag{3}\label{eq3A}$$
This means $x \equiv 0 \pmod{p}$, $x \equiv -1 \pmod{p}$ or $x \equiv 2 \pmod{p}$. The first $2$ cases give in \eqref{eq1A} that $1 \equiv 0 \pmod{p} \implies p = 1$, which is not allowed. With the third case, \eqref{eq1A} gives $7 \equiv 0 \pmod{p} \implies p = 7$ is the only possibility.
Checking $x = 2$ itself shows it works to give on the left side $49 = 7^2$, so $(2, 2, 7)$ is another solution. Next, consider $x \gt 2$, so
$$x = 7z + 2 \tag{4}\label{eq4A}$$
for some integer $z \ge 1$. Substituting this into the $x^2 + x + 1$ factor, and letting it be equal to $7^m$ for some integer $m \ge 1$, gives
$$\begin{equation}\begin{aligned} 7^m & = (7z + 2)^2 + (7z + 2) + 1 \\ & = 49z^2 + 28z + 4 + 7z + 3 \\ & = 49z^2 + 35z + 7 \\ & = 7(7z^2 + 5z + 1) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Thus, since $z \ge 1 \implies 7z^2 + 5z + 1 \gt 1$, then $7 \mid 7z^2 + 5z + 1$, so
$$5z + 1 \equiv 0 \pmod{7} \tag{6}\label{eq6A}$$
Next, letting $x^3 - x + 1 = 7^n$ for some integer $n \ge 1$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned} 7^n & = (7z + 2)^3 - (7z + 2) + 1 \\ & = (7)^3z^3 + 3(7^2)(2)z^2 + 3(7)(4)z + 8 - 7z - 1 \\ & = (7)^3z^3 + 6(7^2)z^2 + (11)(7)z + 7 \\ & = 7((7)^2z^3 + 6(7)z^2 + (11)z + 1) \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Similar to before, this results in
$$11z + 1 \equiv 0 \pmod{7} \implies 4z + 1 \equiv 0 \pmod{7} \tag{8}\label{eq8A}$$
Next, \eqref{eq6A} minus \eqref{eq8A} gives
$$z \equiv 0 \pmod{7} \tag{8}\label{eq9A}$$
However, this contradicts both \eqref{eq6A} and \eqref{eq8A} since it results in $1 \equiv 0 \pmod{7}$, showing there is no such positive integer $z$.
In summary, the only $2$ triplets $(x, y, p)$ that satisfy the equation are $(1, 1, 3)$ and $(2, 2, 7)$.
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 $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
