LVI OM - I - Task 2

Determine all natural numbers $ n>1 $ for which the value of the sum $ 2^{2}+3^{2}+\ldots+n^{2} $ is a power of a prime number with a natural exponent.
From the formula $ 1+2^{2}+3^{2}+ \ldots+n^{2}=\frac{1}{6}n(n+1)(2n+1) $, true for any natural number $ n $, we obtain



The task thus reduces to finding all natural numbers $ n>1 $, prime numbers $ p $, and positive integers $ k $ for which the equality



is satisfied.

Since the factor $ 2n^2 +5n+6 $ is greater than $ 6 $, it follows from equation (1) that it is divisible by $ p $. We will show that the factor $ n-1 $ is not divisible by $ p $.
Suppose $ p|n-1 $. Since the number



is also divisible by $ p $, $ p $ is a prime divisor of the number $ 13 $. Therefore, $ p = 13 $. Hence $ n - 1 \geq 13 $, which implies the inequality $ M \geq 36\cdot 13 $. From equation (1), we infer that $ M $ is a divisor of the number $ 6\cdot13^k $, and since $ M > 6\cdot13 $, it must be that $ 13^2 |M $. From relation (2), we then obtain that the number $ 9(n-1)+13 $ is divisible by $ 13^2 $, and thus



The number $ (n-1) $ is — by equation (1) — a divisor of the number $ 6\cdot13^k-1 $ and is simultaneously — by divisibility (3) — not divisible by 13. It follows that the number $ \frac{1}{13}(n-1) $ is equal to 1, 2, 3, or 6. However, in none of these four cases is divisibility (3) satisfied. We have obtained a contradiction, which proves that the factor $ n-1 $ is not divisible by $ p $.
From equation (1), we then infer that the factor $ n-1 $ must be one of the numbers: 1, 2, 3, or 6. Hence the number $ n $ is equal to 2, 3, 4, or 7. Direct verification shows that indeed each of these numbers satisfies the conditions of the problem:



Answer: The number $ n $ is equal to 2, 3, 4, or 7.