# Primitives triangle with smallest side an even number

## 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 contradiction, I assume both a and b are odd, so.
$$a=2n_1 +1$$
and.
$$b=2n_2+1$$
applying Pythagorean theorem,
$$a^2+b^2=4(n_1^2+n_2^2-n_1-n_2)+2$$.
This only gives me that c must be even but it does not tell me whether it is still rational and co-primes to a and b or not.

## Answers and Replies

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.

RUber
Homework Helper
There is nothing in your second proof forcing d to be an integer. So I am not sure that is the best method.
Using your expansion from the Pythagorean theorem,
## a^2 + b^2 = c^2 = 4( n_1^2 + n_2^2 -n_1 - n_2 ) + 2 ##
That gives that
##c = \sqrt{4( n_1^2 + n_2^2 -n_1 - n_2 ) + 2} = 2 \sqrt{ n_1^2 + n_2^2 -n_1 - n_2 +1/ 2} = 2\sqrt{K - 1/2} ##
for some ##K \in \mathbb{Z}##.
It should be pretty clear to show that c is not an integer.

There is nothing in your second proof forcing d to be an integer. So I am not sure that is the best method.
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.

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.
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 the right triangle to be a positive integer, a or b must be even.

haruspex
Homework Helper
Gold Member
2020 Award
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.
There is no reason why d should be an integer, so no contradiction.

RUber
Homework Helper
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.
What is d? d is the area. Nothing in your original problem said that the area had to be an integer...just that the lengths of all the sides were rational.
But in the right triangle whose a, b are odd, c can not be even ( because c^2 is not divisible by 4).
I think your proof was supposed to show that if a and b are both odd, then c cannot be a rational number. Thus contradicting your original assumption and proving that if a, b, and c are all rational and primitives, either a or b must be even. You had the right attack using the Pythagorean thm.

What is d? d is the area. Nothing in your original problem said that the area had to be an integer...just that the lengths of all the sides were rational.

I think your proof was supposed to show that if a and b are both odd, then c cannot be a rational number. Thus contradicting your original assumption and proving that if a, b, and c are all rational and primitives, either a or b must be even. You had the right attack using the Pythagorean thm.
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 which is a must for any right triangle. This means a or b must be even.

haruspex