Estonian Mathematical Olympiad

Denote $f(n)=2n^3+4n^2-3n+12$. The following table shows the remainders of $n^2,n^3,n^5$ and $f(n)$ upon division by 11:

$n$	0	1	2	3	4	5	6	7	8	9	10
$n^2$	0	1	4	9	5	3	3	5	9	4	1
$n^3$	0	1	8	5	9	4	7	2	6	3	10
$n^5$	0	1	10	1	1	1	10	10	10	1	10
$f(n)$	1	4	5	5	5	6	9	4	3	7	6

As one can see from the table, the only remainders upon division by 11 that the fifth power of an arbitrary integer $n$ can give are 0, 1 and 10. On the other hand, integers of the form $f(n)$ give only remainders 1, 3, 4, 5, 6, 7, and 9 upon division by 11, whereby the remainder is 1 only if $n$ is divisible by 11. Consequently, $f(p)$ can be the fifth power of an integer only if $p$ is divisible by 11. As $p$ is prime, the only possibility is $p=11$. And indeed, $f(11)=2\cdot 11^3+4\cdot 11^2-3\cdot 11+12=3125=5^5$.