Are there any non-trival positive integer solutions to $n^{k}+k=2025!$?
For instance, I think that if there is a positive integer solution to this equation, $k$ cannot be equal to $2\pmod{4}$, as $4\mid2025!$, but $4\nmid n^2+2$ if $n$ is even.
If the last digit of $n$ is:
- $6$, then $k$ must be equal to $4\pmod{10}$, as the last of $2025!$ is zero.
- $4$, then $k$ should be equal to $2\pmod{4}$.
- $2, 8$, then $k\equiv3\pmod{4}$
Moreover, the result of the sum must have $505$ trailing zeroes at the end, and the last nonzero digit must be $2$.
Using Pari GP, I tried to check all values of $n, k\leq10^4$, but I was unsuccessful.
Best Answer
Because of $2^{19328}>2025!$ , we only have to check $2025!-k$ for $2\le k\le 19328$. Since none of those numbers if a perfect power , there cannot be a non-trivial solution. I checked this with brute force with PARI/GP , this is possibly not what is desired.