Question

Use a proof by contradiction to prove that if a, b, c E Z with a² + b2 = c, then a or b is even.

Answer 1

Answer:

See explanation below

Step-by-step explanation:

To prove by contradiction, we are going to assume that a and b are odd.

If a and b are odd, then there exists integers j, k > 0 such that

a = 2j + 1 and b= 2k + 1

We're going to rewrite the original expression substituting a and b by their odd expression.

$a^{2} +b^{2} = (2j+1)^{2} +(2k+1)^{2} =4j^{2} +4j+1+4k^{2} +4k+1=4j^{2} +4k^{2} +4j+4k+2\\=4(j^{2} +k^{2}+j+k) +2$

Now we have to cases, c is even or c is odd.

Case 1: If c is odd.

If c is odd, then c² is also odd, but we have that the expression above is even. Therefore, this is a contradiction.

Case 2: If c is even.

If c is even then it's multiple of 2, and c² is multiple of 4, but the expression above is not multiple of 4 (because it has the form 4g + 2). Therefore we have a contradiction.

Thus, a or b must be even.