Saudi Arabian Mathematical Competitions

We have $3^{3}+4^{3}+5^{3}+8^{3}+10^{3}=12^{3}$, that is $n=3$ satisfies the property. We shall prove that this is the unique solution. Consider the function $h:[2, \infty) \rightarrow \mathbb{R}$,
$$h(x)=\left(\frac{3}{12}\right)^{x}+\left(\frac{4}{12}\right)^{x}+\left(\frac{5}{12}\right)^{x}+\left(\frac{8}{12}\right)^{x}+\left(\frac{10}{12}\right)^{x}$$
which is a strictly decreasing function with $h(3)=1$. It follows that for $n>3$, we have $h(n)<h(3)=1$. That is
$$\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}<12$$
On the other hand, it is clear that
$$10<\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}$$
hence if $\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}$ is an integer, then it must be 11. We get $3^{n}+4^{n}+5^{n}+8^{n}+10^{n}=11^{n}$, relation which is not possible since the left hand side is an even number.

Assume that
$$\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}=k$$
for some positive integer $k$, that is $3^{n}+4^{n}+5^{n}+8^{n}+10^{n}=k^{n}$. Since the left hand side is an even number, it follows that $2 \mid k^{n}$, hence $2^{n} \mid k^{n}$. We obtain $2^{n} \mid 3^{n}+5^{n}$.
On the other hand, we have
$$3^{n}+5^{n} \equiv (-1)^{n}+1 \quad (\bmod 4) \equiv 2 \quad (\bmod 4)$$
if $n$ is even, hence there are no even integers satisfying the property.
If $n$ is odd, then we have
$$3^{n}+5^{n}=(3+5)\left(3^{n-1}-3^{n-2} \cdot 5+\ldots+5^{n-1}\right)=2^{3} \cdot a$$
where $a$ is an odd number. It follows $n \geq 3$, and we get the unique solution $n=3$.