Saudi Arabia 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 $h:[2,\infty )\to \mathbb{R}$, $$h(x)={\left(\frac{3}{12}\right)}^x+{\left(\frac{4}{12}\right)}^x+{\left(\frac{5}{12}\right)}^x+{\left(\frac{8}{12}\right)}^x+{\left(\frac{10}{12}\right)}^x$$ 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.

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Alternative solution.

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 $$3^n+5^n\equiv (-1{)}^n+1(\mathrm{mod}4)\equiv 2(\mathrm{mod}4)$$ if $n$ is even, hence there are no even integers satisfying the property. If $n$ is odd, then we have $$3^n+5^n=(3+5)\left(3^{n-1}-3^{n-2}\cdot 5+\dots +5^{n-1}\right)=2^3\cdot a$$ where $a$ is an odd number. It follows $n\ge 3$, and we get the unique solution $n=3$.