LVII OM - III - Problem 2

Determine all positive integers $ k $ for which the number $ 3^k+5^k $ is a power of an integer with an exponent greater than 1.
If $ k $ is an even number, then the numbers $ 3^k $ and $ 5^k $ are squares of odd numbers, giving a remainder of 1 when divided by 4. Hence, the number $ 3^k + 5^k $ gives a remainder of 2 when divided by 4, and thus is divisible by 2 but not by $ 2^2 $. Such a number cannot be a power of an integer with an exponent greater than 1.
If $ k $ is an odd number, then

The second factor on the right side of the above relationship contains an odd (equal to $ k $) number of odd summands.
Hence, the number $ 3^k + 5^k $ is divisible by 8 and not by 16. If this number is a power of an integer with an exponent greater than 1, then it must be a cube of an integer.
If $ k = 1 $, then the considered number is a cube of an integer: $ 3^1 + 5^1 = 2^3 $. Let us assume in the further part of the reasoning that $ k \geq 3 $. From the relationship

it follows that cubes of integers give remainders of 0, 1, 8 when divided by 9. For $ k \geq 3 $ we have $ 9 | 3^k $, so $ 3^k + 5^k \equiv 5^k (\mod 9) $. 
The remainders of the numbers $ 5, 5^2, 5^3, 5^4, 5^5, 5^6 $ when divided by 9 are 5, 7, 8, 4, 2, 1, respectively. Therefore, if $ 3^k + 5^k $ is a cube of an integer for $ k \geq 3 $, then $ 3 | k $. We have previously shown that $ k $ cannot be an even number.
Thus, the number $ k $ is of the form $ 6l + 3 $, where $ l $ is a non-negative integer.
From the relationship $ 3^3 \equiv 5^3 \equiv 6 (\mod 7) $ and $ 3^6 \equiv 5^6 \equiv 1 (\mod 7) $, it follows that

However, from direct verification, we obtain that the cube of an integer gives a remainder of 0, 1, or 6 when divided by 7:

Therefore, the considered number for $ k \geq 3 $ cannot be a cube of an integer, which concludes the solution of the problem.
Answer: $ k = 1 $.