How can we prove that the square of any odd natural number is congruent to 1 mod 8?
Show that square of any odd natural number is congruent to 1 mod 8.
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discrete-mathematics
logic
computer-science
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0Hint: Every odd number can be expressed in the form $2k+1$ – 2017-02-16
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0$\mathbb Z_8^\times$ is isomorphic to the Klein 4 group. – 2017-02-16
2 Answers
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Squares of odd numbers modulo $8$: $$ 1^2 \equiv 1 \\ 3^2 = 9 \equiv 1 \\ 5^2 = 25 \equiv 1 \\ 7^2 = 49 \equiv 1. $$
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Any odd number can be written in the form $2n+1$, $n \in \mathbb{Z}$. $$ (2n+1)^2 = 4n(n+1)+1. $$ 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 $2 \mid n(n+1)$, so in fact the first term is divisible by $8$. Hence $$ (2n+1)^2 = 4n(n+1)+1 \equiv 1 \pmod 8. $$