Kindergarten Again

 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.