For any odd integer a , show that a^2 is congruent to 1( mod8)
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Any odd number can be written in the form 2n+1, n∈Z.
(2n+1)2=4n(n+1)+1.
The second term is equivalent to 1 modulo 8 (obviously). The first term is divisible by 4, so it is either 0 or 4 modulo 8. But one of n, n+1 is always even, so 2∣n(n+1), so in fact the first term is divisible by 8. Hence
(2n+1)2=4n(n+1)+1≡1 (mod8)
Hope it helps!
rituraj4314:
it is odd integer u wrote number ???
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