Exercise 0.3.9 (If $n$ is odd, $n^2 \equiv 1 \pmod 8$)

Question

Prove that the square of any odd integer always leaves a remainder of $1$ when divided by $8$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $n = 2k +1$ be an odd integer. Then
$$n^2 = (2k+1)^2 = 4k^2 + 4k+ 1 = 8 \, \frac{k(k+1)}{2} +1.$$
Moreover $ \frac{k(k+1)}{2} = \binom{k}{2}$ is always an integer, so
$$n^2 \equiv 1 \pmod 8.$$
The square of any odd integer always leaves a remainder of $1$ when divided by $8$.
\end{proof}

Exercise 0.3.9 (If $n$ is odd, $n^2 \equiv 1 \pmod 8$)

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Exercise 0.3.9 (If $n$ is odd, $n^2 \equiv 1 \pmod 8$)

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