Why is the difference between two odd squares multiples of $3$ is divisible by $72$?

Question

Why is the difference between two odd squares multiples of $3$ is divisible by $72$?

Here is my solution and I am not sure what should be next.

Note that any odd number is in the form of $2m+1$

Here we take $m$, $n$ so that $m>n$

$(2m+1)^2−(2n+1)^2$

$=4(m^2−n^2+m−n)$

$=4(m-n)(m+n+1)$

Since $m$ and $n$ are odd, the difference will be even. Hence, $m-n$ is divisible by $2$, which makes the difference between two odd squares divisible by $8$. Now, since the given difference between two odd squares must be multiples of $3$, $(m+n+1)$ must be divisible by $3$. This makes the given difference between two odd squares divisible by $24$.

From here, I am not sure how to show that it is divisible by $72$. Please feel free to share your ideas on how to solve this problem. Thank you in advance.

Answer 1

Hint:

A square multiple of $3$ is necessarily a multiple of $9$ by the uniqueness of the decomposition into prime factors. Hence you only have to prove the difference of odd squares is divisible by $8$.

It's simpler to do it with congruences. Note that an odd integer is congruent to $\pm 1$ or $\pm 3\bmod 8$. What can you deduce for its square?

Answer 2

I think you're asking why two quantities of the form $(3(2a+1))^2$ differ by a multiple of $72$. This is equivalent to$$8|(2a+1)^2-(2b+1)^2=4(a-b)(a+b+1).$$Compare the parities of $a-b,\,a+b+1$.

Answer 3

Just use the form $6n+3$ the way you used the form $2n+1$.

$$(6n+3)^2 - (6k+3)^2 = 36(n^2 - k^2 +n-k) = 36(n-k)(n+k+1).$$

If $n$ and $k$ have the same parity, then $n-k$ is even. If they have opposite parity, then $n+k+1$ is even.