Find the sum of all positive integers for which is a perfect square.
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Let (n^2)+96= x^2
=x^2-n^2=96
=(x+n)*(x-n)= 96
Listing all factors of 96=1,2,3,4,6,8,12,16,24,32,48,96
solving the above we find 4 solutions for x and n:
so adding all values of n =2+5+10+23= 40