1. Problem: Prove algebraically that the sum of the squares of two odd integers is always even. Step 1: Let the two odd integers be $2m+1$ and $2n+1$, where $m$ and $n$ are integers. Step 2: Square each odd integer: $$ (2m+1)^2 = 4m^2 + 4m + 1 $$ $$ (2n+1)^2 = 4n^2 + 4n + 1 $$ Step 3: Sum the squares: $$ (2m+1)^2 + (2n+1)^2 = (4m^2 + 4m + 1) + (4n^2 + 4n + 1) = 4m^2 + 4m + 4n^2 + 4n + 2 $$ Step 4: Factor out 2: $$ 4m^2 + 4m + 4n^2 + 4n + 2 = 2(2m^2 + 2m + 2n^2 + 2n + 1) $$ Step 5: Since $2m^2 + 2m + 2n^2 + 2n + 1$ is an integer, the sum is $2 \times$ an integer, which means it is even. Therefore, the sum of the squares of two odd integers is always even. 2. Problem: The sum of two prime numbers is always an even number. Give a counter example to prove Maria wrong. Step 1: Understand the claim: Maria says the sum of two prime numbers is always even. Step 2: Find a counter example where the sum of two prime numbers is odd. Step 3: Consider the prime numbers 2 and 3. Step 4: Sum them: $$ 2 + 3 = 5 $$ Step 5: Since 5 is odd, this is a counter example disproving Maria's claim. Final answers: - The sum of the squares of two odd integers is always even. - The sum of two prime numbers is not always even; for example, $2 + 3 = 5$ (odd).