1. The problem states: given any two consecutive even numbers, find the common multiples of the sum of their squares. 2. Let the first even number be $2n$ where $n$ is an integer. 3. The next consecutive even number is $2n + 2$. 4. Calculate the squares of these two numbers: $$ (2n)^2 = 4n^2 $$ $$ (2n+2)^2 = (2n+2)^2 = 4n^2 + 8n + 4 $$ 5. Now, find the sum of the squares: $$ 4n^2 + 4n^2 + 8n + 4 = 8n^2 + 8n + 4 $$ 6. Factor this expression: $$ 8n^2 + 8n + 4 = 4(2n^2 + 2n + 1) $$ 7. Since $4(2n^2 + 2n + 1)$ has a factor of 4, the sum of the squares is always divisible by 4. 8. Check divisibility by 2: since 4 includes 2, it is divisible by 2. 9. Check divisibility by 6, 8, 10, 16: - 6 = 2 * 3; factorization does not guarantee a factor of 3, so not always divisible by 6. - 8 = $2^3$; sum has factor 4 but not always factor 8 because $2n^2+2n+1$ may be odd or even. - 10 = 2 * 5; factor 5 not guaranteed. - 16 = $2^4$; factorization only guarantees 4, not 16. Therefore, the sum of the squares of any two consecutive even numbers is always divisible by 2 and 4 only. Final answer: 2 and 4.