What are the pairs ($p,n$) of non-negative integers where $p$ is a prime number, such that $$p^2+n-3=6^n+n^6$$
How can I solve this diophantine equation?
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Assume $p\neq 3$. Then $p^2\equiv 1\pmod 3$.
If $p=3$, then LHS ($n+6$) is smaller than RHS ($6^n+n^6$) $\forall n\ge 2$ and not equal for $n=0$, but equal when $n=1$. So $(p,n)=(3,1)$ is another solution.
If $n \equiv 0,1\pmod3$, we have $3$ divides $6^n+n^6 - n+3$, except for $n=0$. Further, for $n=1$, we have $6^n+n^6 - n+3 = 9 = 3^2$. Hence, $n \equiv 2\pmod3$. However, if $n \equiv 2 \pmod3$, we have $$6^n+n^6 - n+3 \equiv 2\pmod3$$ and no square is $2 \pmod 3$. Hence, the solutions are $(n,p) = (0,2)$ and $(n,p) = (1,3)$.