2ⁿ + 65 = x² is an odd integer ,therefore x should be odd. As you know,any square of an odd integer ,should have 1 , 5, 9 as unit digit.

If n is odd, then the unit digit of 2ⁿ is 2 or 8 and this not accepted ,for when we add the unit digit of 65 with 2 or 8 it will give 7 or 3 which are not unit digits for any square of an integer.

As a conclusion : n is even and x is odd .

Suppose n = 2k and x =2p + 1 substituting in the formula :

2^2k + 65 = (2p + 1)² 2^2k - (2p + 1)² = - 65 (2^2k -2p - 1)(2^2k + 2p + 1)

= - 65 with - 65 = -1 * 65 or 13 * - 5 (2^2k -2p - 1)(2^2k + 2p + 1)         = -1 * 65                                                                                  Case1 : 2^2k -2p - 1 = - 1 ; 2^2k + 2p + 1 = 65  ; k = 5 and n = 10        Case 2 : 2^2k -2p - 1 = - 5 ; 2^2k + 2p + 1 = 13 resulting in k = 2 and n = 4 and these are the only solutions.

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