Question

0x0=?

Answer 1

Find all prime numbers m,n such that n divides m^2 - 1 and m divides n^2 -4.

Answer 2

lmao

Answer 3

XD

Answer 4

well I guess it boils down to the definition of a prime\[p\mid ab \Rightarrow p\mid a \vee p\mid b \]\[n \mid (m^2 -1) \Rightarrow n\mid m-1\vee n \mid m+1 \]\[m\mid (n^2 - 4) \Rightarrow m \mid n -2 \vee m \mid n+2\]

Answer 5

If \(n\gt m \) then we can forget \(n \mid m-1\) and for the second "or" case there is only one possibility: (m, n) = (2, 3). If we plug that in into the second case, then that's not valid.

So yes, it's definitely not possible when \(n> m\).

Answer 6

If \(m>n\) then for the second-case we can forget the first "or", and the second is just not possible, unless \(m = n+2\), so we're now considering twin-primes under this case, where \(m\) is the greater of the two twin primes.

Answer 7

Hey, btw, do we say that 3 divides 0? If so, then (3, 2) is also a valid solution.

Answer 8

So anyway, we're looking at twin-primes \((m,n)\) where \(m>n\). We can forget \(n \mid m - 1\). We're left with \(n \mid m+1\) and I believe (3, 5) is the only twin-prime pair which satisfies that property.

Answer 9

yes exactly :)

Answer 10

Therefore, (2, 3) and (3, 5) appear to be the only solutions.

Answer 11

yes (:

Answer 12

Great, is this an RMO problem or what?

Answer 13

yea its RMO prblm which prof gave in class :)

Answer 14

we hav some kinda Olympiad practice tests this was one of the prblms

Answer 15

Aren't there infinitely many solutions when n=2

Answer 16

o-o seems legit