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Solution 7
by Soarer Let 2n + 65 = x2 , since x2 (mod 5) as 5 does not divide 2n +65, 2n (mod 5) Which forces n must be even. Let n=2k then, 65=(x-2k )(x+2k )=1*65=5*13 Thus 2k+1 =(x+2k )-(x-2k )=65-1 or 13-5 Therefore n=10 or 4.
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Solution 7
by Soarer Let 2n + 65 = x2 , since x2 (mod 5) as 5 does not divide 2n +65, 2n (mod 5) Which forces n must be even. Let n=2k then, 65=(x-2k )(x+2k )=1*65=5*13 Thus 2k+1 =(x+2k )-(x-2k )=65-1 or 13-5 Therefore n=10 or 4.
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