Diophantine equation:$x^5+x^4+1=p^y$

Question

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??

Answer 1

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 $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.

Answer 2

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 $(2, 2, 7)$.