Solve a diophantine equation

Question

The questions is: Solve in the positive integers $$3^n=m^4+m+1.$$ I can prove that $n$ is odd, and $m\equiv 4(\text{mod }9)$, but i dont know why $n$ and $m$ are $1$.

Answer 1

Probably not of much use but some elementary observations about the numbers $m^4+m+1$ are as follows.

There is precisely one solution modulo $3^n$ of the equation $$m^4+m+1\equiv 0\pmod {3^n}$$ The solution for $3^{n+1}$ can be obtained from the solution for $3^n$ as follows:-

If $m^4+m+1\equiv 3^nd\pmod {3^{n+1}}$, where $d=1$ or $2$, then $$(m+3^nd)^4+(m+3^nd)+1\equiv 0\pmod {3^{n+1}}$$

The first few solutions are then

$m=1:$ $m^4+m+1\equiv 3\pmod {3^2}$

$m=1+3=4:$ $m^4+m+1\equiv 2\times 3^2\pmod {3^3}$

$m=4+2\times 3^2=22:$ $m^4+m+1\equiv 3^3\pmod {3^4}$

$m=22+3^3=49:$ $m^4+m+1\equiv 2\times 3^4\pmod {3^5}$

$m=49+2\times3^4=211:$ $m^4+m+1\equiv 3^7\pmod {3^8}$

$m=211+3^7=2398:$ $m^4+m+1\equiv 2\times 3^8\pmod {3^9}$