The Number Of Integers ' ' For Which is a Perfect Square Are
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For n^2 +96 to be a perfect square
n^2 +96= m^2 where m is any positive integer m^2 - n^2 =96 (m+n)(m-n)=96 (m+n)(m+n -2n)=96 (m+n)^2 - 2n(m+n) =96 Now on RHS we have a even no. So LHS should be even too For this subtraction of two even no.s is only and always even So in a way we proved LHS to be even Thus initially m+n and m-n are even On factorizing 96 the solutions are 2×48 ,4 × 24 , 6 × 16 , 8 × 12 Thus 4