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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?

Bill Dubuque
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Cinverse
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    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 Oct 16 '23 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 Oct 16 '23 at 11:53
  • @lulu The first statement is a general definition of pythagorean triplet. in that, only the xy should be a perfect square, no need of x and y to be perfect squares, individually. – Cinverse Oct 16 '23 at 11:55
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    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 Oct 16 '23 at 11:55
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    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 Oct 16 '23 at 11:56
  • @lulu I was just trying to prove (on my own) its existence, define its general definition and define for primitive one. It's to learn how to prove and define things in maths and learn along with it. Infact got some of the ideas by reading Euclid's formula. – Cinverse Oct 16 '23 at 12:00
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    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 Oct 16 '23 at 12:03
  • @lulu yeah i get it that if x and y are co-prime, then the only way in which xy is perfect square is when even x and y are perfect sqaures. And, this is the new condition that is needed to define Primitive Pythagorean triplet. Thanks. – Cinverse Oct 16 '23 at 12:10

1 Answers1

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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.

Shreya Mundhada
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