Question

Find all triples (p, q, r) of primes such that p^3 + q^3 + r^3-3pqr divides 2016.

Answer 1

I also knows there is exactly 36 factors of 2016.

Answer 2

Hint :
\[p^3 + q^3 + r^3-3pqr = (p+q+r)(p^2+q^2+r^2-pq-qr-rp)\]

Answer 3

I have tried bruteforcing, but I do not think that it is effective.

Answer 4

yeah looks tricky
that hint seems useless..

Answer 5

\[p^3 + q^3 + r^3-3pqr = pqr(p^2+q^2+r^2-3)\]
So p,q, and r must divide 2016 for starters, that might eliminate most possibilities

Answer 6

whoops

Answer 7

ok, pqr divides it regardless, by factorization is of course hilariously wrong

Answer 8

no wait I'm going insane this is all nonsense hahaha

Answer 9

It couldn't be 1, since p,q,r must be prime.

Answer 10

\(p+q+r\) must be a divisor of \(3^2*7\)
\[p+q+r \in \{1, 3,7, 9, 21, 63\}\]

Answer 11

right, eliminate 1

Answer 12

and It can't be 3 either

Answer 13

Nice, all other combinations must be possible.. .
recalling goldbach conjecture

Answer 14

Also 2 is a factor of 2016, so shouldn't p+q+r be a divisor of 3^2*7*2^5, making p+q+r∈{,7,9,21,63} still incomplete?

Answer 15

in above case, we're assuming \(p+q+r\) is odd

Answer 16

that is true if none of \(p, q, r\) is \(2\)

Answer 17

what we're doing is more or less bruteforce only
lets think of some other ways maybe..

Answer 18

This looks like a Diophantine problem.

Answer 19

A Diophantine is used if the answer is an integer, but in this case we need a prime number.

Answer 20

A prime is an integer. :P

Answer 21

But not all integers are prime

Answer 22

The converse of my statement, which is true.

Answer 23

The primes can't all be the same, because then the expression is 0.

What if two of the primes are the same? Then with \(p=r\) we have:
\[(2p+q)(p-q)^2\]

Since \(2016 = 2^53^27\) this restricts us to \(p-q\) or \(q-p\) being equal to 2 or 3 only. In other words, p and q are either twin primes OR 2 and 5.

Of course, this is only for the cases where we have repeated primes.

Answer 24

This could work if we could prove that there is no case in which all different primes are p,q, and r.

Answer 25

Yeah totally, that's what I'm sorta drifting towards. I just figured I'd try to prove some degenerate cases to get a feel for what kinds of stuff is going on.

On an unrelated note, if we force the inequality \(p<q<r\) can that get us anything

Answer 26

Are you attempting to use p<q<r to prove that at least two of the letters have to be same?

Answer 27

No, but that would be convenient. I was attempting to do it because it looks like the other cases where you have repeated values are pretty much handled, so we can focus solely on when the primes are all distinct

Answer 28

Im pretty sure, the primes can't be distinct.

Answer 29

Well in that case if you are sure enough we can take this approach forward when you have one prime that occurs twice. Cause all twin primes greater than 3,5 are of the form of the littler one \(6n-1\) and the larger one \(6n+1\) so that's pretty restrictive I guess. Might be pretty straightforward to prove that this has no solutions or something idk

Answer 30

nvm

Answer 31

2,3,7 disproves that

Answer 32

The good part is the fact that I could easily brute force for primes that are all distinct.

Answer 33

Yeah if you really wanted to solve it and not look for tricks, we could just write a program and be done with it haha. Then we could look for patterns from within the answer to retroactively discover solutions too lol.

Answer 34

True.

Answer 35

But because this is a math olympiad question, I know there is a trick.

Answer 36

Hmm. Since 2016 has 8 prime factors, that means when we factor that thing the total number of factors they could have gets split between the two. For instance if we can show that
\(p^2+q^2+r^2-pq-pr-qr \ne 1\) then that limits the other thing, \(p+q+r\) to having at most 7 prime factors in it.

Not terribly useful probably though haha

Answer 37

I'm pretty sure 2,3,5,7 are the only numbers that can become p,q, or r, because once one of the variables become 11, the number becomes too big.

Answer 38

Since neither one of us has any idea on this, could we instead work on Let a and b be real numbers such that
2^(a+6)-2^(a-6) / 2^(b+3)+ 2^(b-3) = 2016.
Find a and b.

Answer 39

\[\dfrac{2^{a+6}-2^{a-6}}{2^{b+3}+2^{b-3}}=2016\\~\\~\\\dfrac{2^a(2^6+2^{-6})}{2^b(2^3+2^{-3})} = 2016\\~\\~\\
2^{a-b} = \dfrac{2016(2^3+2^{-3})}{2^6+2^{-6}}
\]

Answer 40

next you may take log both sides to isolate \(a-b\)