# Homework Help: Pythagorean triples

1. Aug 11, 2010

### EV33

1. The problem statement, all variables and given/known data

Let a,b,c be integers such that a2+b2=c2.

Is it always true that at least one of {a,b} is even?

2. Relevant equations

3. The attempt at a solution
I say yes, and I am going to try and prove it with a proof by contrdiction.

Suppose a,b,c be integers such that a2+b2=c2 and that a and b are odd. By definition of odd a=2m+1, and b=2n+1, for some m,n in Z. By substitution we get c2= (2m+1)2+(2n+1)2. By simple arithmatic we get c2=2(2m2+2n2+2m+2n+1).
c=$$\sqrt{2}$$$$\sqrt{2m^2+2n^2+2m+2n+1}$$.
Because the second square root has an odd number in it means that we can not pull out a 1/$$\sqrt{2}$$ to cancel out the $$\sqrt{2}$$. This means that we will have an irrational answer, rather than an integer for all m,n in Z. Thus we have reached a contradiction.//

I was wondering if my logic is correct on this.

2. Aug 11, 2010

### lanedance

look reasonable to me, I think it is enough to note
$$c^2=2(2m^2+2n^2+2m+2n+1) = c^2=2(2k+1)$$

where k is the integer given by
$$k = m^2+n^2+m+n$$

3. Aug 11, 2010

### Dick

I think it would be clearer if you skip the square root arguments and just talk about remainders mod 4 which is what I think lanedance is getting at.