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

[Animuo]

Homework Statement

Prove by Contradiction: For all Prime Numbers a, b, and c, a^2 + b^2 =/= c^2

Homework 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.

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!

 

[HallsofIvy]

Homework Statement

Prove by Contradiction: For all Prime Numbers a, b, and c, a^2 + b^2 =/= c^2

Homework 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.

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)

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.

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!