For all prime numbers a, b, and c, a虏 + b虏 <> c虏. <> denotes not equal to. How to proof by contradiction?

This statement needs to be proofed by contradiction.For all prime numbers a, b, and c, a虏 + b虏 %26lt;%26gt; c虏. %26lt;%26gt; denotes not equal to. How to proof by contradiction?
John's proof is essentially correct, but he assumed a little more than he should have. He assumed that when you factored to





(c+a)(c-a)





neither of those factors are equal to 1. However, if c = 3 and a = 2, then the factor c-a is equal to 1. So, you would also have to show that this particular case is also impossible, probably with a direct argument.





It's also worth mentioning that this proof doesn't completely account for the case of a = c, all though, one extra step would take care of that. Since this would imply a-c = 0, we would have





b^2 = (a+c)(a-c) = 0





Of course, this is a contradiction, since it would mean b is both prime and zero.For all prime numbers a, b, and c, a虏 + b虏 %26lt;%26gt; c虏. %26lt;%26gt; denotes not equal to. How to proof by contradiction?
Suppose for contradiction that


c^2=a^2+b^2


then


b^2=c^2-a^2=(c+a)(c-a)





so we have found two integers not equal to 1 or b that multiply to give b. Contradicting the assumption that b is prime