Doubt whether proof is correct I'm not very good at number theory, so I would rather ask. The proposition states that for every even square $x$ there is an odd square $y$ such that $x+y$ is a square number. The proposition comes from Liber Quadratorum.
I tried to proceed by contradiction. (First I think it refers to squares of integers, so $x,y \in \mathbb{N}$.) We proceed by assuming both number are even. Then we define numbers such that:
$$
x=(2n)^2 \\
y=(2m)^2
$$
Now we need to have a number $c \in \mathbb{N}$ where
$$
\frac{x+y}{c}=c.
$$
Substituting,
$$
\frac{4n^2+4m^2}{c}=\frac{4n^2}{c}+\frac{4m^2}{c}=c.
$$
We define constants $k_1,k_2$ as
$$
4n^2=k_1c^2 \\
4m^2=k_2c^2
$$
Then
$$
\frac{k_1c^2}{c}+\frac{k_2c^2}{c}=c \\
k_1+k_2=1
$$
This means that at least one of the constants $k_1,k_2$ is less than $1$. An the other one its reciprocal, but if that happens, then in $x+y=c^2$ either $x$ or $y$ would be larger than $c^2$, which isn't possible, since all the numbers are positive. Therefore at least one of the numbers must be odd. (We assumed that both were even, so if it isn't even it is odd?)
I'm not sure if I followed a correct logic. I feel that assuming that $x+y$ is a perfect square is a weak point. Is this good enough?
 A: For the revised question as given by the OP in comments: if $p^2$ is an odd square then
$$p^2+\Bigl(\frac{p^2-1}{2}\Bigr)^2=\Bigl(\frac{p^2+1}{2}\Bigr)^2\ .$$
That is:

for any odd square $x$ there is an even square $y$ such that $x+y$ is a square

(and we can take $y\ne0$ as long as $x>1$).
A: It is not true that for every even square $x$ there is an odd square $y$ such that $x+y$ is a square. For example, let $x=4$ or $x=36$. Indeed it cannot be done  if $x$ is $4$ times an odd square. For general theory, one can use the fact that if $s$ and $t$ are relatively prime of opposite parity, then $(|s^2-t^2|, 2st, s^2+t^2)$ are a primitive Pythagorean triple, and all primitive triples  are of this form.
We prove the assertion asked for in a comment, that if $x$ is an odd square greater than $1$, there is a positive even square such that $x+y$ is a square. 
Let $x=(2w+1)^2$, where $w\gt 0$. Note that $2w+1=(w+1)^2-w^2$ and
$$((w+1)^2-w^2)^2 +(2(w+1)(w))^2=((w+1)^2+w^2)^2.$$
So we can take $y=(2(w+1)(w))^2$.
