Hi there, I am stuck at question 12, chapter 3.6 For all prime numbers a, b, and

Question

Hi there,

I am stuck at question 12, chapter 3.6

For all prime numbers a, b, and c, a² + b² <>c². <> denotes not equal to.

This statement has to be proved by contradiction.

So far, there's one approach I can think of right now but unclearhow this approach will lead to the answers.

To prove by contradiction, I negate the statement above asfollows:

There are prime numbers a,b, and c such that a² + b² =c².

b² = c² - a²
b² = (c - a)(c+ a)

In order for b to be a prime number, either (c - a) is 1 or (c + a)is one.

If neither (c - a) nor (c + a) equals to 1, then clearly b is notprime.

I am stuck here because I do not know how to prove that neither (c- a) nor (c + a) equals to 1.


I am not sure how to finish the rest of the question.

Explanation / Answer

instead of proving
c - a =1 or c + a = 1 use this approach if b² = (c - a)(c+ a) then c - a = b and c + a = b which is only possible if and only if a = 0 but "a" is a prime number (non zero) a = 0 is contrary to the given statement Hence, b² <> (c - a)(c+ a) b² <> c² - a²
or a² + b² <> c² proved.

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