If $a,b,n \in \mathbb{Z}$, & $n \mid (a+b)$, prove $n$ does not necessarily divide $ (a^2+b^2)$

Question

In the exercise 2.4 of Wallace introductory book titled "Groups, Rings and Fields", it is stated that:

If $a,b,n \in \mathbb{Z}$, & $n \mid (a+b)$, prove $n$ does not necessarily divide $ (a^2+b^2)$.

I am unable to make any progress, as I feel it is only possibly true in complex domain, with $b$ as imaginary part, i.e. $(a+ib)(a-ib) = a^2 + b^2$.

A proof negating the divisibility of $a^2 + b^2$ by such $n$, for $a,b,n \in \mathbb{Z}$ is requested.

Answer 1

You need only a single counterexample.

$$a^2 + b^2 = (a-b)(a+b) + 2b^2$$

So, if $n$ divides $a+b$ but not $a^2+b^2$, then it must be that $n$ does not divide $2b^2$.

Any which way you look should produce counterexamples, e.g., $a = 3$, $b = 4$, so that $a+b = 7$ and $a^2 + b^2 = 25$. Pick $n = 7$. Clearly $n$ divides $a+b$ (they are equal) but it does not divide $a^2 + b^2$ since $7$ is not a factor of $25$.

Answer 2

The hint:

Use $$a^2+b^2=(a+b)^2-2ab$$

and try for example $a=3$, $b=5$ and $n=8$.