1. **Problem statement:** Prove that if the sum of two integers $a + b$ is even, then at least one of the summands $a$ or $b$ is even. 2. **Recall definitions:** - An integer is **even** if it can be written as $2k$ for some integer $k$. - An integer is **odd** if it can be written as $2k + 1$ for some integer $k$. 3. **Approach:** We will prove the contrapositive: If both $a$ and $b$ are odd, then their sum $a + b$ is odd. 4. **Express $a$ and $b$ as odd integers:** $$a = 2m + 1, \quad b = 2n + 1$$ where $m,n$ are integers. 5. **Calculate the sum:** $$a + b = (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1)$$ 6. **Simplify:** $$a + b = 2(m + n + 1)$$ Since $m + n + 1$ is an integer, $a + b$ is even. 7. **Contradiction:** Our assumption that both $a$ and $b$ are odd leads to $a + b$ being even, which contradicts the original statement that if $a + b$ is even, then one summand is even. 8. **Conclusion:** Therefore, if $a + b$ is even, at least one of $a$ or $b$ must be even. This completes the proof.