This is the outline of the proof found in the book "Proofs from THE BOOK" (P.17-22) by Aigner, Martin, Ziegler, Günter M. It is too long to put in a comment so I put it here. Please do not vote for it.
Lemma 1. For primes $p=4m+1$ the equation $s^2\equiv -1$ (mod $p$) has two solutions $s\in\{1,2,\dots,p-1\}$, for $p=2$ there is one such solution, while for primes of the form $4m+3$ there is no solution
Lemma 2. No number $n=4m+3$ is a sum of two squares.
Proposition. Every prime of the form $p=4m+1$ is a sum of two squares, that is, it can be written as $p=x^2+y^2$ for some natural numbers $x,y\in\mathbb{N}$
Theorem. A natural number $n$ can be represented as a sum of two squares if and only if every prime factor of the form $p=4m+3$ appears with an even exponent in the prime decomposition of $n$.
No it does not; consider the case where $m=1$, $r=5$, $s=2$. Then
$$2^m(r2^m+s)+1=2^1\times(5\times2^1+2)+1=25=5^2,$$
but $r$ is not a perfect square, so such an $n$ does not exist.
Moreover, it does not even imply that $s$ is even, as the values $m=3$, $r=s=5$ give
$$2^3\times(5\times2^3+5)+1=361=19^2,$$
which is a perfect square. There are many examples, the values $m=3$, $r=2$, $s=5$ give
$$2^3\times(2\times2^3+5)+1=169=13^2,$$
in some sense a 'smaller' example. And $m=3$, $r=1$, $s=7$ gives the smallest counterexample;
$$2^3\times(1\times2^3+7)+1=121=11^2.$$
Best Answer
For the purpose of this problem, since they are relatively small numbers, the squares must be $1, 4, 9, 16, 25, 36, 49, 64, 81,$ and $100$.
$81+25=106$. $100+9=109$. $81+36=117$.
So, 112, by process of elimination, cannot be written in this manner.