Subgroups of $\mathbb{Z}$?

First, can you show that 7 is in $I_a$?

Then, can you show that if there is even one number in $I_a$ that isn't a multiple of 7, then $I_a$ is all of $\bf Z$?

To ellaborate on my comment:
Hint: $I_a=\langle a\rangle+\langle 7-a^2\rangle=\langle \gcd(a,7-a^2)\rangle$ (where $\gcd$ is the greatest common divisor)
Denote $d=\gcd(a,7-a^2)$. $d$ divides $a$, and $d$ divides $7-a^2$ (clearly $d$ divides $a^2$). Can you proceed?

Hint $\rm\:(1)\:\ I = a\,\mathbb Z\! +\! b\, \mathbb Z\:$ is closed under subtraction so a subgroup of $\left<\mathbb Z, +\right>$ by the subgroup test.

$\rm(2)\:\ \{0\}\ne I\:$ has a least positive $\rm\:c.\:$ Being a group, $\rm\:I \supseteq c\:\mathbb Z.\:$ This is an equality (else there exists $\rm\:d\in I\:$ with $\rm\:nc < d < (n\!+\!1)c,\:$ so $\rm\ 0 < d\!-\!nc < c\:$ and $\rm\:d\!-\!nc \in I,\:$ contra leastness of $\rm\:c).\:$ Thus $\rm\:a,b\in I = c\,\mathbb Z\:\Rightarrow\:\color{#0a0}{c\:|\:a,b}.\:$ $\rm\:d\:|\:a,b\:\Rightarrow\:d\:|\:c = j\,a\!+\!k\,b\:\Rightarrow\: \color{#c00}{d\le c}.\:$ Thus $\rm\,\ c = gcd(a,b),\: $ being a $\color{#c00}{\rm greatest}\ \rm \color{#0a0}{common\ divisor}$ of $\rm\:a,b$.

But $\rm\:gcd(a,b\!-\!an) = gcd(a,b)\:$ since if $\rm\:c\:|\:a\:$ then $\rm\:c\:|\:b\!-\!an\iff c\:|\:b.\:$

So $\rm\:\ \ gcd(a,7-a^2) = gcd(a,7) = c,\:$ where $\rm\: c = 7\:$ if $\rm\:7\:|\:a,\:$ else $\rm\:c = 1$.