We have had two good solutions to this problem. Both noticed that if you take any prime number greater than 3, square it and subtract 1 you always get a multiple of 24.

Biren Patel (Age:13) from The Heathland School went on to write: I tried this with as many prime numbers as I could and it always worked. I then used algebra to try and prove it.

P ²-1 is the same as (p+1)(p-1). As 24 was a large number to handle, I wrote it as the product of its prime factors ie. 3 x 2 x 2 x 2 = 3 x 2 ³.

Now we have a statement stating:
(p+1)(p-1) is divisible by 3 x 2 ³.

We have to prove that 3 will go into either p+1 or p-1 as this would then make (p+1)(p-1) a multiple of 3 which we want. Any prime number p greater than 3 when divided by 3 can give two remainders: 1 or 2.

If p left a remainder of 1 when divided by 3, then p-1 would leave a remainder of 0, which would mean that it was a multiple of 3. So therefore (p+1)(p-1) would be divisible by 3 as p-1 is divisible by 3.

If p left a remainder of 2 when divided by 3 then p+1 would leave a remainder of 0, which would mean that it was a multiple of 3, excellent! So therefore (p+1)(p-1) would be divisible by 3 as p+1 is divisible by 3.

Now I have to prove that 2 ³ will go into (p+1)(p-1). If a multiple of n was multiplied by a multiple of x then the answer would be a multiple of nx. (eg. Multiple of 4 x multiple of 7 is a multiple of 28 ie. 8 x 14 = 112 which is divisible by 28). Since p is odd, p+1 is a multiple of 2 and p-1 is the previous multiple of 2, so (p+1)(p-1) is a multiple of 8, since either p+1 or p-1 is a multiple of 4 and the other is a multiple of 2 (if all the even numbers were written out: 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32... the first is a multiple of 2, the next is a multiple of 4, the next is a multiple of 2, the next is a multiple of 4, etc.)

And so since (p+1)(p-1) is both a multiple of 8 and 3 it is also a multiple of 24.

On a similar vein, Ling Xiang Ning, Allan (Age: 13) from Raffles Institution in Singapore wrote:

I notice that the numbers are always divisible by 24.
To prove that this will go on infinitely, I show that the numbers are always divisible by 8 and 3, which means that they will be divisible by 24.

As the starting number is a prime greater than 3, it must be 2(modulo 3) or 1(modulo 3). The square of the prime number is always 4(modulo 3) or 1(modulo 3). Therefore, the square of the prime number - 1 is divisible by 3. The prime number at first is either 1(modulo 8), 3(modulo 8), 5(modulo 8), or 7(modulo 8), as it is odd. The square of a 1(modulo 8) is still 1(modulo 8), the square of a 3(modulo 8) is 9(modulo 8), or 1(modulo 8), the square of a 5(modulo 8) is 25(modulo 8), or 1(modulo 8), and the square of a 7(modulo 8) is 49(modulo 8), or 1(modulo 8). Therefore, the square of the prime number - 1 is divisible by 8.

Therefore, the final numbers would be divisible by 24 to infinity.

Thankyou both for such good solutions.