Find two Pythagorean right triples A,B,C and D,E,F such that A=D and B=F, where C and F are the length of the two hypotenuses,
The required triples do not exist.
Simplifying the given expressions as indicated by the problem: A^2+E^2=B^2, A^2+B^2=C^2.
Canonically, the parts of a pythagorean triple (PT) can be analysed as:
(u^2-v^2)^2+ (2uv)^2= (u^2+v^2)^2
Let A=u^2-v^2, E=2uv (or vice versa); then B=u^2+v^2, for some u,v.
Then (u^2-v^2)^2+(u^2+v^2)^2=C^2, so C^2=2(u^4+v^4)
But (u^4+v^4)=((u^2)^2+(v^2)^2), so there would exist another PT, with hypotenuse((u^2)^2+(v^2)^2), exactly half C^2, which is impossible, since no square can be twice the size of another.
Edited on February 28, 2024, 11:37 pm
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Posted by broll
on 2024-02-28 23:26:54 |
some thoughts
