Pythagorean Triples: Is it Always True at Least One is Even?

[EV33]

Homework Statement

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

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

Homework Equations

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.

Thank you for your time.

 

[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
$$k = m^2+n^2+m+n$$

 

[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.