When the square full right angle triangle divided by the repeated factor, \(2\). It seem an area of size \(1\) square unit does not count,
There are not considered squares there! And apparently, a lot of unit squares are being cut up to fit into a right angle triangle.
What childhood memories.
Prime are not congruent, by their definition! This is wrong.
When \(a=1\), for
\(b=2n\)
Area of triangle\(=\cfrac{1}{2}a*b=\cfrac{1}{2}.2n=n\)
But side \(c\) must also be rational.
\(\sqrt{1^2+(2n)^2}=\sqrt{4n^2+1}\)
which is the square of an odd number. \(c\) is odd.
\(\sqrt{4n^2+1}=2m+1\)
\(4n^2+1=(2m+1)^2\)
\(n=\cfrac{1}{4}\sqrt{(2m+1)^2-1}\)
So, maybe...
\(b=2n=\cfrac{1}{2}\sqrt{(2m+1)^2-1}\)
\(c=2m+1\)
\(a=1\)
This unfortunately does not happen. \(a=1\) only is wrong. Both is possible, \(a=odd\), \(b=even\) and \(c=odd\). Both sides \(a\) and \(b\) being even is not allow because they will generate a \(\sqrt{2}\) factor in \(c\) unless they form a perfect square. A pair of \(a\) and \(b\) both odd will give even \(c\) s. A pair of \(a\), \(b\), odd and even, will given odd \(c\) s.
But what about fractions? odd, even fraction?
In seem that this question is didactic, Kindergarten all over again. Aliens teachers...
Thank you.
