## Question 3

### Part 1

Let $d$ be an integer. The three requirements for $d$ are that

$$d|n+5 &(1) \\d|n^2+2 &(2) \\d|27&(3),$$

Where $n$ is any integer.

 Assume that $(1)$ and $(2)$ are true. The consequence of this is that there must be 2 integers, (let them be $a$ and $b$), such that

$$n+5&=ad &(3)\text{, derived from }(1)\\n^{2}+2&=bd &(4)\text{, derived from }(2).$$

Equation $(3)$ can be rearranged such that $n=ad-5$. Substituting this into $(4)$,

$$\begin{aligned}bd&=(ad+5)^{2}+2\\&=a^{2}d^{2}-10ad+25+2\\27&=bd-a^{2}d^{2}+10ad \\27&=d\left(b-a^{2}d+10a\right). &(5)\end{aligned}$$

Since $a$,$b$, and $d$ are all defined as integers, the right hand side of $(5)$ is an integer which is divisible by $d$. Therefore, 27 must also be divisible by $d$, satisfying $(3)$. As such, we can conclude that for all integers $n$, if $d$ is an integer such that $d|n+5$ and $d|n^2+2$, then $d|27$.