Proofs

Let the odd integer be $n = 2k + 1$ for some integer $k$. [1]

Then $n^{2} = 4k^{2} + 4k + 1 = 4k(k+1) + 1$. [1]

$k$ and $k+1$ are consecutive integers, so $k(k+1)$ is even. Write $k(k+1) = 2m$. [1]

$n^{2} = 8m + 1$, so the remainder on division by $8$ is $1$. [1]

Total: 4 marks

Let $P(n)$ be the statement that $6$ divides $7^{n} – 1$.

Base step: $7^{1} – 1 = 6$, which is divisible by $6$, so $P(1)$ is true. [1]

Inductive hypothesis: assume $P(k)$ is true for some positive integer $k$, so $7^{k} – 1 = 6m$ for an integer $m$, and $7^{k} = 6m + 1$. [1]

Inductive step: $7^{k+1} – 1 = 7 cdot 7^{k} – 1 = 7(6m + 1) – 1 = 42m + 6 = 6(7m + 1)$. [3]

$7m + 1$ is an integer, so $P(k+1)$ is true. By induction, $P(n)$ holds for every positive integer $n$. [1]

Total: 6 marks