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correct formatting in q3 part 1

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### Part 1
Let d be an integer. The three requirements for d are that
Let $d$ be an integer. The three requirements for $d$ are that
d|n+5 &(1) \\d|n^2+2 &(2) \\d|27&(3),
$$d|n+5 &(1) \\d|n^2+2 &(2) \\d|27&(3),$$
Where n is any integer.
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
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).
$$n+5&=ad &(3)\text{, derived from }(1)\\n^{2}+2&=bd &(4)\text{, derived from }(2).$$
Equation (3) can be rearranged such that n=ad-5. Substituting this into (4),
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}
$$\begin{aligned}bd&=(ad+5)^{2}+2\\&=a^{2}d^{2}-10ad+25+2\\27&=bd-a^{2}d^{2}+10ad \\27&=d\left(b-a^{2}d+10a\right). &(5)\end{aligned}$$
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.
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 $d|27$.