Math, asked by rituraj4314, 1 year ago

For any odd integer a , show that a^2 is congruent to 1( mod8)​

Answers

Answered by aditiisas25
0

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 ???
aditiisas25: sorry my mistake...it will be odd no.
rituraj4314: but then it cannot be 2n+1
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