Prove that if \(x\) and \(y\) are both odd positive integers, then \(x^2 + y^2\) is even but not divisible by \(4\).
Even, not divisible by 4.
Step 1: Write odd numbers in standard form.
Since \(x\) and \(y\) are odd, there exist integers \(a\) and \(b\) such that
\(x = 2a + 1\)
\(y = 2b + 1\)
Step 2: Square each expression.
\(x^2 = (2a + 1)^2\)
\(\quad = 4a^2 + 4a + 1\)
\(\quad = 4\big(a^2 + a\big) + 1\)
Similarly, \(y^2 = (2b + 1)^2\)
\(\quad = 4b^2 + 4b + 1\)
\(\quad = 4\big(b^2 + b\big) + 1\)
Step 3: Add the squares.
\(x^2 + y^2 = \big[4(a^2 + a) + 1\big] + \big[4(b^2 + b) + 1\big]\)
\(\quad = 4\big[(a^2 + a) + (b^2 + b)\big] + 2\)
Step 4: Conclude parity and divisibility by 4.
The expression has the form \(4N + 2\) for some integer \(N\).
Therefore, \(x^2 + y^2\) is even (since it is \(2\) more than a multiple of \(4\))
but it is not divisible by \(4\) (a multiple of \(4\) cannot leave remainder \(2\)).
Optional check (modulo 4 view).
From the expansions above, \(x^2 \equiv 1 \pmod{4}\) and \(y^2 \equiv 1 \pmod{4}\).
Hence, \(x^2 + y^2 \equiv 1 + 1 = 2 \pmod{4}\), which matches \(4N + 2\).