Doubt in given hint's interpretation for a question.

Question

The question is : Given $a,b$ as two odd integers, show $a^2+b^2$ is not a square number.

Book's hint states that any odd integer can be of the form: $n = 3q +r$, hence the $3$ cases based on value of $r$ are :
(i) if $r=0, n_1 = 3q \implies 3\mid n_1$
(ii) if $r=1, n_2 = 3q+1\implies 3\mid (2n_2+1)$
(iii) if $r=2, n_3 = 3q+2\implies 3\mid (n_3+1)$

I feel that the book's approach is to find all $6$ possible combinations, with the sum of individual squares as $m_i$ as :
(a) $m_1=n_1^2+n_1^2$
(b) $m_2=n_2^2+n_2^2$
(c) $m_3=n_3^2+n_3^2$
(d) $m_4=n_1^2+n_2^2$
(e) $m_5=n_1^2+n_3^2$
(f) $m_6=n_2^2+n_3^2$

I am doubtful about the interpretation made, hence want to confirm.

Answer 1

Yes, you can it confirm as so:

Any odd square is congruent to $1 (\mod 4)$.

Then, what will the sum of two odd squares be? And what conclusion can you arrive at?

Answer 2

HINT

Use congruence $\bmod 4$ and $(2k+1)^2=4k^2+4k+1$.