Primes Dividing Each Other

[@pedrojose392]

Good evening! pq because if p=q then p|1, contradiction as p is prime. By symmetry WLOG p>q If q is even then q=2 ==> p | 9 and so p=3, checking q^2=4 | 28= p^3+1 and for now we have two solutions (2,3) and (3,2) If both are odd p^2| (q+1) (q^2-q+1) As p does not divide q+1 because p > q+1 then p^2 | q^2-q +1 but p>q ==>p^2>q^2 > q^2-q+1 as q >1 so we dont have solutions for both odd. Then (2,3) and (3,2) are the only solutions.

[@venkateshsaki5980]

Very nicely explained

[@particleonazock2246]

Thanks for the great underrated vid

[@mcwulf25]

Good problem. I got the answer but couldn't prove it was the only answer.

[@prithujsarkar2010]

Was this a reupload?

[@timetraveller2818]

yea it was

[@letsthinkcritically]

Yes, as mentioned in the remark.

[@rupeshkapoor5380]

I think all that Euclidean algorithm business was unnecessary. For p to divide q+1 where p > q, p must be equal to q+1. The only two consecutive primes are 2 and 3.

[@lreactor]

This looks like overkill... p=q=2 doesn't work, which means p (the larger one, which can't be 2) is odd. But then q^3+1 is odd, so q is even. So q=2. From the LHS eq, p^2 | 9; so p=3 is the only solution. That works for the RHS equation as well, so we're done.