4. Find the smallest natural $n>1$, for which the sum of no two natural powers is a perfect square of a natural number.

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# Answer.4.

Solution. For $n=2: 2^{1}+2^{1}=2^{2}$. For $n=3: 3^{3}+3^{2}=6^{2}$. Let $n=4$. Suppose there exist such $m, k$ that $4^{m}+4^{k}=a^{2}$.

Any power of 4 gives a remainder of 1 when divided by 3 (since $\left.4^{m}=(3+1)^{m}\right)$. Therefore, $4^{m}+4^{k}$ will give a remainder of 2 when divided by 3. But the square of an integer gives a remainder of 0 or 1 when divided by 3.

Comment. Correct solution - 20 points. Proven that $n \geq 4-12$ points. Proven that $n \geq 3-3$ points. Proven that $n \leq 5-3$ points. Solution not completed, but there is progress - 5 points. Solution started, but progress is minimal - 1 point.