If m and n are odd positive integers, then m2 + n2 is even, but not divisible by 4. Justify.

Sure, let's break this down:

1. First, we need to understand what odd numbers are. Odd numbers are integers which are not divisible by 2. They can be written in the form 2k+1, where k is an integer.

2. So, if m and n are odd positive integers, we can write them as m = 2a+1 and n = 2b+1, where a and b are integers.

3. Now, let's square m and n. We get m^2 = (2a+1)^2 = 4a^2 + 4a + 1 and n^2 = (2b+1)^2 = 4b^2 + 4b + 1.

4. Adding m^2 and n^2, we get m^2 + n^2 = (4a^2 + 4a + 1) + (4b^2 + 4b + 1) = 4(a^2 + a + b^2 + b) + 2.

5. We can see that the first part of the sum, 4(a^2 + a + b^2 + b), is divisible by 4 because it's a multiple of 4.

6. The second part of the sum, 2, is not divisible by 4. Therefore, the whole sum, m^2 + n^2, is not divisible by 4.

7. However, since both parts of the sum are even (any number multiplied by 4 is even, and 2 is even), the whole sum, m^2 + n^2, is even.

So, we have justified that if m and n are odd positive integers, then m^2 + n^2 is even, but not divisible by 4.