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

(N/A) Let $x = 2m + 1$ and $y = 2n + 1$ be any two odd positive integers,where $m$ and $n$ are non-negative integers.
Then,$x^{2} + y^{2} = (2m + 1)^{2} + (2n + 1)^{2}$.
Using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,we get:
$x^{2} + y^{2} = (4m^{2} + 4m + 1) + (4n^{2} + 4n + 1)$.
$x^{2} + y^{2} = 4m^{2} + 4m + 4n^{2} + 4n + 2$.
$x^{2} + y^{2} = 4(m^{2} + m + n^{2} + n) + 2$.
Since $x^{2} + y^{2} = 2[2(m^{2} + m + n^{2} + n) + 1]$,it is clearly an even number.
However,when divided by $4$,it leaves a remainder of $2$. Therefore,$x^{2} + y^{2}$ is not divisible by $4$.