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n - odd => n=2k+1, k=0,1,2,...

n^2-1= (n+1)(n-1) = (2k+2)*2k= 4k*(k+1)

4k*(k+1)/8=1/2*k*(k+1) => k or k+1 is even => remainder (n^2-1)/8 is 0

by vorand

Fri Nov 12, 2010 11:49 am
Forum: Data Sufficiency
Topic: Remainder Prob
Replies: 74
Views: 40047