# Primitives triangle with smallest side an even number

1. Nov 20, 2015

1. The problem statement, all variables and given/known data
Prove that if a right triangle has all sides rational and primitives (co-primes), then one of the smaller side must be even number.

2. Relevant equations
For a right triangle (a,b,c) with c is the hypotenuse.
$$a^2+b^2=c^2$$

3. 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.

2. Nov 20, 2015

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.

3. Nov 20, 2015

### RUber

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.

4. Nov 20, 2015

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.

5. Nov 22, 2015

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.

6. Nov 22, 2015

### haruspex

There is no reason why d should be an integer, so no contradiction.

7. Nov 23, 2015

### RUber

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.

8. Nov 23, 2015

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.

9. Nov 23, 2015

### haruspex

If a and b are odd, the area is not an integer. The reference to c being even is irrelevant, and it certainly does not by itself imply d is not an integer.
You have no basis for that statement. Consider a=1, b=1. The area is not an integer, but you can construct a right triangle.