I'm stuck on the last question of my homework paper and even my friend who's a maths genius can't do it:

22) Here are the first five terms of an arithmetic sequence.

7 13 19 25 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

Well this one is, at least, mildly evil....

OK, so we have a sequence so the first thing, even before we really read the question, is to find the nth term..

7,13,19,25,31

Common difference is 6 and start number 7 so nth term is 6n + 1

So the nth term is 6n + 1.

The question talks about the difference between the SQUARES of ANY tw0 terms. So we need two different terms: 6n + 1 and 6m +1

Then we need to square them both and find the difference:

Now this is obviously a multiple of 12, but we need a multiple of 24, so we need to show that inside the bracket is always even.

We can rewrite the inside of the bracket as:

3(n + m)(n - m) + (n - m) = (n - m)(3(n + m) + 1)

So n - m is a factor.

If n and m are even, n - m is even. If n and m are odd then n - m is even. If one of n is odd and one even, then n + m is odd so 3(n + m) + 1 is even.

So this is always even.

QED

Ouch - that was hard!!!!