To prove that if $a^2 + b^2$ is odd, then $a$ and $b$ must both be odd, let us analyze the problem step by step.

---

Proof:

1. Odd and Even Numbers

• An integer $x$ is even if $x = 2k$ for some integer $k$.
• An integer $x$ is odd if $x = 2k + 1$ for some integer $k$.

2. Squaring Odd and Even Numbers

• If $x$ is even ($x = 2k$), then $x^2 = 4k^2$, which is also even.
• If $x$ is odd ($x = 2k + 1$), then: $x^2 = (2k + 1)^2 = 4k^2 + 4k + 1,$ which is odd (since $4k^2 + 4k$ is divisible by 2, leaving a remainder of 1).

Thus:

• The square of an even number is even.
• The square of an odd number is odd.

---

3. Analyzing $a^2 + b^2$

• Suppose $a^2 + b^2$ is odd.
• The sum of two numbers is odd only if one is odd and the other is even. Specifically:
  • Even + Even = Even.
  • Odd + Odd = Even.
  • Odd + Even = Odd.

4. Assume $a$ or $b$ is even

• Let $a$ be even. Then $a^2$ is even.

• For $a^2 + b^2$ to be odd, $b^2$ must be odd.

• If $b^2$ is odd, $b$ must also be odd (as shown above).

  But this results in $a^2 + b^2 = \text{Even + Odd} = \text{Odd}$, which does not hold because $a^2 + b^2$ is given to be odd.

---

5. Conclusion:

• Both $a$ and $b$ cannot be even because their squares would sum to an even number.
• If $a^2 + b^2$ is odd, then $a$ and $b$ must both be odd so that $a^2 + b^2 = \text{Odd + Odd} = \text{Odd}$.

Thus, $a^2 + b^2$ being odd implies that $a$ and $b$ are odd.

---

Would you like further clarification or examples to illustrate this proof?

---

Related Questions:

1. Can you prove the contrapositive: If $a$ or $b$ is even, then $a^2 + b^2$ is even?
2. What happens if $a^2 + b^2$ is even? Can one or both numbers still be odd?
3. How does this property relate to Pythagorean triples (e.g., $a^2 + b^2 = c^2$)?
4. Can this reasoning extend to more than two variables (e.g., $a^2 + b^2 + c^2$)?
5. How do these results apply to modular arithmetic?

Tip:

Always verify results by plugging in specific examples of odd and even numbers to gain deeper insights into proofs.