- #1

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## Homework Statement

For [tex]a,b, \in Z[/tex], prove that if ab is odd, then [tex]a^2 + b^2[/tex] is even

**question 1**

ab = 2k + 1 for some k is integer (1)

It doesn't seem logical to rearrange some a and b on (1)

So I need the following

product of two odd numbers is always odd,product of two even numbers is always even, product of odd and even integers is ???

## The Attempt at a Solution

**Case 1**

Let ab be odd, i.e. for [tex]k, l \in \mathbb{Z}[/tex]

a = 2k + 1

b = 2l + 1

ab = (2k + 1)(2l + 1) = [tex]4kl + 2k + 2l + 1 = 2(2lk + k + l) + 1[/tex]

(2lk + k + l) is an integer, so ab is odd. Then

[tex](2k + 1)^2 + (2l + 1)^2 = 4k^2 + 2k + 1 + 4l^2 + 2l + 1 = 4k^2 + 2k + 2 + 4l^2 + 2l = 2(2k^2 + k + 2l^2 + l + 1)[/tex]

[tex](2k^2 + k + 2l^2 + l + 1) \mathbb{Z}[/tex], so [tex]a^2 + b^2[/tex] is even (check)

**Case 2**

Let a be even and b be odd (WLOG --> means I shouldn't need to do a not odd and b is odd) and that [tex]s, t \in \mathbb{Z}[/tex]

a = 2s

b = 2t + 1

Then

ab = (2s)(2t + 1) = 4st + 2s = 2(st + s)

[tex]st + s \in \mathbb{Z}[/tex], so ab is unfortunately even

Something went wrong here, because I will only have one case.