Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) Suppose \(a, b, c \in \mathbb{Z} .\) If \(a^{2}+b^{2}=c^{2},\) then \(a\) or \(b\) is even.

Short answer

Proof by contradiction successfully proves that in a Pythagorean triplet, at least one of the first two integers, 'a' or 'b', must be even.

Step-by-step solution

Assumption of the Contrary

Assume the contrary of the statement we wish to prove, meaning, let's assume that both 'a' and 'b' are odd numbers.

Mathematical Representation of Assumption

Define the odd numbers mathematically. An odd number can be written as \(2n + 1\), where n is any integer. Hence, 'a' and 'b' can be represented as \(2m + 1\) and \(2n + 1\) respectively, where 'm' and 'n' are integers.

Substitution into the Given Equation

Replace 'a' and 'b' in the given equation, \(a^{2}+b^{2}=c^{2}\), by the defined odd numbers, i.e., \((2m+1)^2 + (2n+1)^2 = c^2\). Simplify this to \(4m^2 + 4m + 1 + 4n^2 + 4n + 1 = c^2\), which simplifies to \(4(m^2 + m + n^2 + n) + 2 = c^2\).

Contradiction

The equation obtained in step 3 is a contradiction as the left side of the equation is even (since it's a multiple of 2), but the right side, \(c^2\), which is supposed to be equal to the left side, cannot be even unless 'c' itself is even. But if 'c' is even, this means that either 'a' or 'b' should also be even, contradicting our initial assumption that both 'a' and 'b' are odd.