Prove that the sum of the squares of three consecutive odd numbers plus one is divisible by 12 but not by 24.

Question

I tried to find a similar problem to help me with this question, but I couldn’t find anything. The only relevant thing I may know is that this may be a quadratic word problem; any ideas how to solve this?

Answer 1

Hint:

$(2k-1)^2+(2k+1)^2+(2k+3)^2+1$

$=4k^2-4k+1+4k^2+4k+1+4k^2+12k+9$

$=12k^2+12k+12=12(k(k+1)+1)$

and $k(k+1)$ is even.

Answer 2

Let the numbers be $x-2$ , $x$ and $x+2$. Now ,

$$\begin{align}S &= (x-2)^2 + x^2 + (x+2)^2+1\\ &= x^2 +4 -4x+x^2+x^2+4+4x +1\\ & = 3x^2 + 9 \end{align}$$

Since $x$ is odd , we have $x=2k+1$

$$S = 3(2k+1)^2 + 0 \implies \color{#d05}{S = 12k^2 + 12k+12}$$

Clearly $S $ is divisible by $12$ .

Since $(k^2+k)$ is always divisible by $2$ , we take $(k^2+k) = 2\lambda$.So ,

$$S = 12(2\lambda) + 12 \implies \color{#2bf}{S = 24\lambda + 12}$$

So $S $ always leaves $12$ as the remainder when divided by $24.$