So, there are two attacks to proof it, both of them assume that a and b are odds:
1) If a and b are odds, c can not be rational, contradicting the assumption that all are rationals and primitives.
2) If a and b are odds, c can not be even which makes the area of the triangle, d, not integer...
But this could be also applicable in any triangle, namely, the area of any triangle can not be a positive integer unless one of its sides is an even number. But in the right triangle whose a, b are odd, c can not be even ( because c2 is not divisible by 4). This means in order for the area of...
If both a&b are odd, d can not be integer. So, d can not be integer except when one of a&b is even which completes the proof without need for Pythagoras theorem.
Ok, I tried that, if a, b, c are co-primes, then the area of the triangle is:
$$area=\frac{1}{2}ab=d$$
$$ab=2d$$
which means one of a or b must be even.
Homework Statement
Prove that if a right triangle has all sides rational and primitives (co-primes), then one of the smaller side must be even number.
Homework Equations
For a right triangle (a,b,c) with c is the hypotenuse.
$$a^2+b^2=c^2$$
The Attempt at a Solution
In order to create a...