Twin Primes Problem

Prove that when you multiply a pair of twin primes you get a number that has remainder 8 after division by 9. With one exception.

I'm already getting some good solutions. It's interesting to see people do it in slightly different ways.

I'll kill some space here just in case people can see the beginning of the description if it gets reposted.

OK, I think the most succinct answers go along these lines:

Twin primes must be of the form 3n-1 and 3n+1. Multiplying these two primes gives us 9n^2 - 1 = 9(n^2 - 1) + 8. So we have a remainder of 8 after division by 9. The exception is 3 and 5.

I've summarised answers there, but Alienturnedhuman was the first commenter who said something like that.

Some people used the same argument using 6n-1 and 6n+1. It is true that all primes larger than 3 are of this form. That works as well, but it's not quite as neat as above.

Some people used modular arithmetic. I can't assume all viewers know modular arithmetic, but one argument is:
All primes are 1, 2, 4, 5, 7, 8 mod 9 The possible pairs are
2x4 = 8 mod 9
5x7 = 35 = 8 mod 9
8x1 = 8 mod 9.

That's actually how I did it when given the problem. It's not the neatest solution.

Видео Twin Primes Problem канала singingbanana