I have recently thought about a interesting question about Pythagorean Triples.
Consider such a right-angled trapezium formed by 3 right-angled triangle. Determine does it exist integral solutions for lengths of sides $AB, BC, CD, DE, EA, AC$ and $CE$.
Here's my ideas.
I know that Pythagorean Triple can be generated by substituting integer into $x^2-y^2, 2xy, x^2+y^2.$ So let $AB=m^2-n^2$, $BC=2mn$, $CD=2pq$, $DE=p^2-q^2$, $AC=m^2+n^2=u^2-v^2$, $CE=p^2+q^2=2uv$ and $ AE=u^2+v^2.$ To answer the question, I have to show that if $m^2+n^2=u^2-v^2$ and $p^2+q^2=2uv$ have integral solution(*). But I don't know how to show this.
Can anyone tell me if I'm right about (*)? If I'm right, how to show it? If I'm wrong, how to solve the question?
Sorry, I'm a poor question-tagger.