1. **Problem statement:** Prove that if the sum of two integers $a$ and $b$ 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. **Assume the sum is even:** $$a + b = 2m$$ for some integer $m$. 4. **Proof by contrapositive:** We prove the contrapositive statement: If both $a$ and $b$ are odd, then their sum $a + b$ is odd. 5. **Express $a$ and $b$ as odd integers:** $$a = 2k + 1, \quad b = 2l + 1$$ where $k$ and $l$ are integers. 6. **Calculate the sum:** $$a + b = (2k + 1) + (2l + 1) = 2k + 2l + 2 = 2(k + l + 1)$$ 7. **Simplify the sum:** $$a + b = 2(k + l + 1)$$ Since $k + l + 1$ is an integer, the sum is even. 8. **Contradiction:** Our contrapositive shows that if both summands are odd, the sum is even, which contradicts the original claim. 9. **Re-examine step 6:** Actually, the sum of two odd numbers is even, so the original statement "If the sum is even, then one summand is even" is false. 10. **Conclusion:** The statement is false because the sum of two odd integers is also even, but neither summand is even. **Final answer:** The statement is false; the sum of two odd integers is even, so it is not necessary that one summand is even if the sum is even.