Is it possible to prove that if $x$ and $y$ are co-prime, then $(x-y)$ and $\sqrt {xy}$ are also co-prime?

Question

I was trying to prove that pythagorean triplets exists in Natural Number domain.

Here's simplified argument that I did:

consider two natural numbers $x$ and $y$ such that $x > y \ge 1$.

1. $(x + y)^2 = (x - y)^2 + (2 . \sqrt {xy})^2$
2. The above equation is of form $a^2+ b^2 = c^2$
3. so, $a = (x - y), b = (2 . \sqrt {xy})$ and $c = (x + y)$
4. $a$ and $c$ are Natural numbers. For $b$ to be a natural number, $xy$ must be a perfect square
5. proof that for some $x$ and $y$, $xy$ is perfect square: considering $xy = p_1^{2i} . p_2^{2j} . p_3^{2k} . \cdots . p_n^{2z}$, where $p_1,p_2, \cdots, p_n$ are prime factors of $xy$ and $i, j, k, ... , l$ are whole numbers. So it is evident that there can be atleast one combination of $x$ and $y$ which can result in $xy$ being a perfect square
6. STATEMENT (CONCLUSION): there exists natural numbers $a, b$ and $c$ such that $a^2+b^2 = c^2$, where:

1. $a = x – y, b = 2\sqrt {xy},$ and $c = x + y$
2. $x$ and $y$ are natural numbers
3. $xy$ is a perfect square.

To test whether the above statement is true, I just took

1. $xy = 81$ and $x = 27$ and $y = 3$. So, $a = 24, b = 18, c = 30$. This is a true triplet.
2. $xy = 4$ and $x = 4$ and $y = 1$. So, $a = 3, b = 4$ and $c = 5$. This is a true triplet (and also primitive).

Now, using above statement, I want to define Primitive pythagorean triplet. So, a and b should be co-prime. So I have to introduce atleast one more condition along with previous statement to define Primitive Pythagorean triplet.

I intuitively begin to think that if natural numbers x and y are co-prime, then $(x-y)$ and $\sqrt {xy}$ are also co-prime, but I dont know whether this is true and if its true, then how to prove.

Is there any theorems that I could use to prove this?

Answer 1

From what you said, $xy$ must be an integer. As $x, y$ are co-prime, this implies that both $x$ and $y$ are perfect squares.

Consider a prime $p$ dividing $\sqrt{xy}=\sqrt{x}\sqrt{y}$. Both of these factors must be relatively prime integers, so for $p$ to divide their product, $p$ must divide exactly one of them. Without loss of generality assume $p|\sqrt{y}$. Thus, $p|y$. As $x$ and $y$ are relatively prime, $p\nmid x$. Thus, $p\nmid x-y$. This shows that $\sqrt{xy}$ and $x-y$ have no common factors other than one, completing the proof.