Question

show/prove that the triples (a,b,c) generated by a=2uv, b=u^2-v^2, c=u^2+v^2 where u and v are positive integers with u > v is a Pythagorean triple

Answer 1

Just show: \[
(2uv)^2+(u^2-v^2)^2=(u^2+v^2)^2
\]

Answer 2

does it have to be formal or can I just plug in values for 'u' and 'v'?

Answer 3

You need to show for all \(u,v\). One confirmation doesn't prove anything.

Answer 4

that's what I figured, but that's what i'm stuck on

Answer 5

Where did you get to?

Answer 6

(4u^2v^2) + (-2u^2v^2-v^4) = (u^4+2u^2v^2+v^4)

Answer 7

No, don't do both sides of the equation, work on only one side.

Answer 8

so: (4u^2v^2) + (u^4-2u^2v^2-v^4) = (u^2+v^2)^2 ?

Answer 9

Just show: \[
\begin{split}
(2uv)^2+(u^2-v^2)^2
&= 4u^2v^2 +u^4-2u^2v^2+v^4 \\
&= u^4 + 4u^2v^2 -2u^2v^2+v^4 \\
&= u^4 + 2u^2v^2 +v^4 \\
&= u^4 + u^2v^2+u^2v^2 +v^4 \\
&= u^2(u^2 + v^2)+v^2(u^2 +v^2) \\
&= (u^2 + v^2)(u^2 +v^2) \\
&= (u^2 + v^2)^2 \\

\end{split}
\]

Answer 10

thank you, I've literally been staring at this problem for hours i think my brain is exhausted