University of Cambridge Starter Mathematics Interview Question 

It’s so cool how much you can analyse with deceivingly little information. Great vid

In addition: To prove that 24 is the best you can do, look at the smallest two primes > 6, which is 7 and 11. p^2-1 is 48 and 120. These numbers have 24 as their largest common factor. Therefore there is no larger number dividing all p^2 -1 for prime p > 6

For the consecutive even numbers bit, you can also just say that (p-1) = 2q and (p+1) = 2q+2 so (p-1)(p+1) = 2q(2q+2) = 4q(q+1) and q(q+1) are consecutive numbers so always multiply to make an even number (as if q is odd then (q+1) is even, and odd x even=even, and vice versa) so (p-1)(p+1) = 4 x "some even number" so (p-1)(p+1) is divisible by 8. Similar to the mod argument

This is the best and easiest way of showing divisibility by all these numbers, good job!

Nice video, I didn’t see that trick with 3 coming but I would have got the rest

We can also show this by a simpler proof by using the fact that all primes can be written as 1(mod(6)) or -1(mod(6)) Since p isn't 2 or 3, we can write it as 6k+1 or 6k-1 , by doing p²-1 we get 36k²+12k or 36k²-12k , here by some factorization we can further get: 12k(3k+1) -> 12 is a factor and one of k or 3k+1 is even or 12k(3k-1) -> 12 is a factor and one of k or 3k-1 is even Hence we conclude that for any prime p (≠2, 3) p²-1 is divisible by 24

I love how this is so true, but to complex to remember

All prime numbers greater than 3 are either 6n + 1 or 6n - 1 e.g. 11 = 2 * 6 - 1 and 31 = 5 * 6 + 1. So (p + 1) (p - 1) is either (6n) (6n - 2) or (6n + 2) (6n) which can be simplified to 12n (3n ± 1).

Every perfect square has remainders 0 and 1 when divided by 3. Since p is not divisible by 3, then p^2 == 1 mod 3. So p^2-1 == 0 mod 3.

My solution: p is of the form 6n+1 or 6n+5 where n>0. Then p^2-1 is either 12*n(3n+1) or 12*(n+1)(3n+2). The first term is divisible by 24 and all its divisors, since n or 3n+1 is even. The second term has these In general, there need not be more divisors as the examples p=7 and p=13 show. The second term also is divisble by 24 and all its divisors for the same reason. In general there need not be more divisors as the examples p=11 and p=17 show. Thus p^2-1 is guaranteed to be divible by 24 and all its divisors that's the best we can do

it is quite obvious it must be divisable by 2, 3, 4 and 8 - this follows directy from (a+b)^2 = a^2 + 2ab + b^2; also there can be no more divisors as the first three results devided by 24 give prime numbers

I've seen this done on Numberphile with Matt Parker and they resolve that they also have the same answer.

if you watch the powers of 3 and powers of 2 in the factors of numbers, they follow a pattern: 2, 4, 6, 4, 2, 12, 2, 4, 6, 4, 2, 12...; notice since all primes are 6k+/-1 that p+1 and p-1 will be one of those multiples of 6 and then one of the adjacent evens. the product of those is always at least a multiple of 24, so p^2 = 24m + 1 for all p > 3

A better proof for divisiblility of 3 would be taking the prime as 6k +1 and 6k-1 in both cases you would get 3 as a factor

Why p>6 and not p>4? Does the proof fail for p=5 in any way?

Hi, I am from India 🇮🇳. Can you explain how 2,4 and 8 has come as divisors of p^2-1. What is mod 4 meaning in the solution, please make a detailed video on the concepts used to solve the problem. Love and regards.

These twelve integers (and their negatives) are always factors of p^2-1, (but for p

i found (i’m only listing the prime powers): 2^3, 3. i did it in one min so this is just for future me to see if i got em all

P = 1 it is ovious

Why can't p be 5? It would be (5-1)(5+1) being equal 24.

A desperately pedantic point, but you missed 6 and 12 as solutions. A don may not have faulted you because you did the thinking on 2, 3, and 4, and applied the necessary logic to get to 24.

p=7

Certainly 3, lol...because p-1, p and p+1 are 3 consecutive integers...

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