# 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?

• This is not quite true. Let $x$ be the even square $4$. – André Nicolas Jul 16 '15 at 5:20
• Whoops. I see the problem. Then I had to word the proposition the other way around. Sorry. – Cehhΐro Jul 16 '15 at 5:24
• I wrote too fast, there are counterexamples bigger than $4$. Can you check for the exact statement? – André Nicolas Jul 16 '15 at 5:44
• "For any odd square number $x$, there is an even square number $y$, such that $x+y$ is a square number." Which is the 'opposite' of what I tried to prove. – Cehhΐro Jul 16 '15 at 5:46
• That I can do, almost, doesn't work except trivially when $x=1$. – André Nicolas Jul 16 '15 at 5:48

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 $y=(2(w+1)(w))^2$.

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 $x>1$).

• Interesting approach. I like the simplicity, but would we able to prove the existence of the even square without finding one case in particular? (Which I believe is the only one, right?) – Cehhΐro Jul 16 '15 at 5:57
• It's not always the only solution. For example, this method gives $9^2+40^2=41^2$, but there is also $9^2+12^2=15^2$. – David Jul 16 '15 at 5:58
• So is there a way to prove that there are other ways besides exposing a specific solution? – Cehhΐro Jul 16 '15 at 6:00
• I really don't know. But, and I hope this doesn't sound rude, why would you want to? To prove an existence statement, nearly always the easiest method is to give a specific example. It's normally only if you are unable to find a specific example that you would try other methods. For instance: prove that there is a positive number $x<\frac\pi2$ such that $\sin x=e^{-x}$. – David Jul 16 '15 at 6:04
• That was by no means rude. I am learning Math a bit more formally (since I come from engineering). The notion of sufficient proof by exposing a case feels strange. – Cehhΐro Jul 16 '15 at 6:08