n is an odd positive integer. We take n=2k+1, where k"\\geq 0"
"n^{2}-1=(n+1)(n-1)=" (2k+2)2k=4(k+1)k
Note that k(k+1) is always an even number for any value of k.
Therefore k(k+1)=2 m
This imply 8"|n^{2}-1"
Hence "n^{2}\\cong 1(mod 8)"
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