Macedonian Mathematical Olympiad

Let us denote $a_n = (n^3+39n-2)n+17-21^n+5$.

If $n \ge 4$, then $8 \mid n!$. Moreover,
$$a_n \equiv 5^n + 5 \pmod{8}.$$
If $n$ is an even number, then $5^n \equiv 1 \pmod{8}$, so $a_n \equiv 6 \pmod{8}$. But, all full squares have remainder $0,1$ or $4$ when divided by $8$. So, if $n \ge 4$ and $n$ is even, then $a_n$ is not a full square.

Let $n \ge 7$. It is clear that $7 \mid n!$. Then $a_n \equiv 5 \pmod{7}$. On the other side, the remainders of the full squares when divided by $7$ are $0,1,2$ or $4$. So, $a_n$ is not a full square for $n \ge 7$. Having in mind the previous discussion, it remains to check for $n=1, n=2, n=3$ and $n=5$.

If $n=5$, $a_5 \equiv 2 \cdot 1^5 + 5 \equiv 2 \pmod{8}$.
Since the remainders of a full square when divided by $5$ are $0,1$ or $4$, $a_5$ is not a full square.

For $n=3$, we have $a_3 \equiv 3 \pmod{7}$, so $a_3$ is not a full square.

For $n=2$, we have $a_2 \equiv 1+5 \equiv 2 \pmod{4}$, so $a_2$ is not a full square.

For $n=1$, $a_1 = (1+39-2) \cdot 1 + 17 \cdot 2 + 5 = 400$.

This means that only for $n=1$, $a_n$ is a full square.