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

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?

• This is hard to follow. if $\sqrt {xy}$ is an integer, then $x,y$ must both be perfect squares. Did you mean to require that?
– lulu
Commented Oct 16, 2023 at 11:52
• And what is the core question? The complete parameterization of Pythagorean triples is well known, there's no mystery left there.
– lulu
Commented Oct 16, 2023 at 11:53
• Please provide an example of two relatively prime natural numbers $x,y$ which are not squares but which are such that $\sqrt {xy}$ is an integer.
– lulu
Commented Oct 16, 2023 at 11:55
• Anyway, if we just assume that $x,y$ are perfect squares then the prime divisors of $\sqrt {xy}$ are just the prime divisors of $x,y$ and it is obvious that none of those can divide $x-y$.
– lulu
Commented Oct 16, 2023 at 11:56
• Ok, do you see how my last comment resolves your problem? And, as an exercise, you should prove the claim I made initially: if $\gcd(m,n)=1$ and $mn$ is a perfect square, then $m,n$ are each perfect squares. here is a proof of that claim, if you get stuck.
– lulu
Commented Oct 16, 2023 at 12:03

## 1 Answer

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.