# Prove by Contradiction: For all Prime Numbers a, b, and c

1. Aug 1, 2012

### Animuo

1. The problem statement, all variables and given/known data
Prove by Contradiction: For all Prime Numbers a, b, and c, a^2 + b^2 =/= c^2

2. Relevant equations
Prime number is a number whose only factors are one and itself.
Proof by contradiction means that you take a statement's negation as a starting point, and find a contradiction.

3. The attempt at a solution
The statement's negation is:
There exists prime numbers a, b, and c, such that a^2 + b^2 = c^2
Rearrange it:
a^2 = c^2 - b^2
a^2 = (c - b) (c + b)
a = √(c-b)(c+b)

I'm stuck here. To show that it's a contradiction I would have to show that it's factors are not equal to 1 or a, and I've been staring at this a little too long, my head just keeps going in circles.. some help would be appreciated!

2. Aug 1, 2012

### HallsofIvy

Staff Emeritus
Up to here, great. Since a is prime, the only factor of a^2 is a.
So you must have c- b= 1 and c+ b= a^2 or c- b= a and c+ b= a. Now if all of a, b, and c are odd that is impossible.

Last edited: Aug 1, 2012