Challenge Math Challenge - April 2021

[Not anonymous]

statement one:
Show that all Pythagorean triples $x^2 + y^2 = z^2$ can be found by
$(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N} , u > v$

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

 

[fresh_42]

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

You are right. I made a mistake (sloppy translation). It should have been ... of the form $d\cdot (u^2-v^2,2uv,u^2+v^2)$ ...

I apologize for that negligence, @Not anonymous.

 

[Not anonymous]

I apologize for that negligence, @Not anonymous.

No problem (problem ≠ math problem in this context ). Thanks to you as always for promptly reviewing my answers and patiently explaining what I may have missed or got wrong.