Squares of prime numbers: why are they 24*n+1?

[lpetrich]

Squaring Primes - Numberphile - YouTube

Has a conundrum: the square of every prime number other than the first two is equal to 24*n+1 for some integer n. Can one prove it?

 

[rjh01]

Great channel, but I square 57 and add 1 and I get 3250. This is not a multiple of 24. So what have I done wrong? What is a factor of 57 (1 and 57 do not count)?
I hope that counts as your proof.

Edit. But 57 is a multiple of 3. So I was wrong. However the proof is in the video.

 

[excreationist]

I think that means that in base 6 primes from 5 onwards end in 5 or 1. In base 12 they'd end in 1, 5, 7 or 11.
In base 12 the square plus 1 ends in 1 and the second digit is even.
Another reason why I prefer base 6 or 12.

 

[lpetrich]

Here is the proof. It applies not only for prime numbers, but also for every number not divisible by 2 or 3 inclusive.

To test divisibility by 2 and 3, let us first find their least common multiple. It is 6. We can now divide the integers up into subsets using their remainders when one divides by 6:

6n -- divisible by 2 and 3
6n+1
6n+2 -- divisible by 2
6n+3 -- divisible by 3
6n+4 -- divisible by 2
6n+5

So the only ones not divisible by either 2 or 3 are 6n+1 and 6n+5. The second one is equivalent to 6n-1, after bumping n up by 1. Let us now square them.

(6n+1)^2 = 36n^2 + 12n + 1 = 12n*(3n+1) + 1
(6n-1)^2 = 36n^2 - 12n + 1 = 12n*(3n-1) + 1

So we get numbers with form 12*m+1. We are almost there. So let us consider even and odd n.

n = 2k:
12n*(3n+1) + 1 = 24k*(6k+1) + 1
12n*(3n-1) + 1 = 24k*(6k-1) + 1
n = 2k+1:
12n*(3n+1) + 1 = 24*(2k+1)*(3k+2)
12n*(3n-1) + 1 = 24*(2k+1)*(3k+1)

Thus we get the form 24*m+1