Below is a complete solution to question 7. It is based on several ideas in the contestants solutions.

Proof: Please consider 2ⁿ + 65 = m², where m and n are positive integers. Since 2ⁿ + 65 is clearly odd, m must be odd. if n where odd, then .                                                  It follows: . But the quadratic residues of an odd number modulo 10 are { 1, 5, 9 } – a contradiction. Therefore, n must be even, and n = 2k for some positive integer k. We have: 2ⁿ + 65 = 2^2k + 65 = m² 65 = m² - 2^2k = (m – 2^k)(m + 2^k) 1(65) = 5(13) = (m – 2^k)(m + 2^k) ; m + 2^k is obviously positive, and  m – 2^k must be positive as well; otherwise, their product would be negative. Moreover, m – 2^k < m + 2^k. We observe 65 has precisely two factorizations 1(65) and 5(13), in positive integers, with the multiplier less than the multiplicand. Consequentially, (*) m – 2^k = 1 and m + 2^k = 65 or (**) m – 2^k = 5 and m + 2^k = 13 In both cases, we add the equations together to deduce m = 33 (case *) or m = 9 (case **). Hence, m = 33, n = 10 and m = 9, n = 4 are the only positive integer solutions for the equation: 2ⁿ + 65 = m².

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