Prove that,if $a$ and $b$ are odd positive integers,then $a^{2} + b^{2}$ is even but not divisible by $4$.

(N/A) Let $a$ and $b$ be two odd positive integers. Any odd positive integer can be expressed in the form $2n + 1$ or $2n - 1$ for some integer $n$.
Let $a = 2m + 1$ and $b = 2n + 1$ for some integers $m$ and $n$.
Then,$a^{2} = (2m + 1)^{2} = 4m^{2} + 4m + 1 = 4(m^{2} + m) + 1$.
Similarly,$b^{2} = (2n + 1)^{2} = 4n^{2} + 4n + 1 = 4(n^{2} + n) + 1$.
Now,$a^{2} + b^{2} = [4(m^{2} + m) + 1] + [4(n^{2} + n) + 1] = 4(m^{2} + m + n^{2} + n) + 2$.
Let $k = m^{2} + m + n^{2} + n$. Then $a^{2} + b^{2} = 4k + 2$.
Since $a^{2} + b^{2} = 2(2k + 1)$,it is clearly an even number.
However,when $4k + 2$ is divided by $4$,the remainder is $2$. Therefore,$a^{2} + b^{2}$ is not divisible by $4$.