Find all pairs $(p, q)$ of prime numbers satisfying \[ p^3+7q=q^9+5p^2+18p. \]
Problem
Source: Turkey JBMO TST 2016 P7
Tags: number theory, prime numbers
brdkc43
23.05.2016 02:06
There isn't any solution ! We can take $\pmod 3$, which gives us $p\equiv 2\pmod 3$ after trying $p=3k+2,k\in \mathbb{R^{+}}$ , we found $7>q^8$ which is contradiction! Am ı wrong ?
crazyfehmy
23.05.2016 02:11
$p=29, q=3$ is a solution.
mathguy623
23.05.2016 02:19
Here is some partial progress? We have $p(p^2-5p-18)=q(q^8-7).$
Suppose $p=q.$ Then $p^8=p^2-5p-11,$ which clearly does not hold for any positive integer.
So $p \neq q.$
Consider when $q=3.$ We get $p(p^2-5p-18)=19662,$ and prime factorizing the right side we get only a few possible $p.$ It's narrowed down to $29,113$ and miraculously $p=29$ works so $(p,q)=\boxed{(29,3)}$ is a solution. Now consider $p=3,$ there is clearly no satisfactory $q.$
Now, the only case left is when $p \neq q \neq 3 \neq p.$ Take mod 9 and using $q^9 \equiv q \mod 9$ by Fermat we have $$p^3 \equiv 3q+5p^2 \pmod 9,$$which is convenient because $p^3 \equiv -1,1 \pmod 9.$ Rearrange as $$3q \equiv p^2(p+4) \pmod 9,$$and note that then $p \equiv 2 \mod 3.$ But the LHS is divisible by $3$ but NOT $9,$ so $p\equiv -1,2 \mod 9.$ But since primes are $1,-1 \mod 6$ then we have $p\equiv 11,17 \pmod {18}.$
Note $q~|~p^2-5p-18.$
^This may seem kind of all over the place oops. The goal is to prove no remaining solns $p,q>3.$
rmrf
23.05.2016 04:10
this one is just an exercise in bounding...
Write as $q^9 - 6q = p^3-5p^2-18p$.
First note that $(q^3)^3 > q^9-7q > (q^3-1)^3$ holds for all primes $q$ other than $q=2,3$.
Also, the inequality $(p-1)^3>p^3-5p^2-18 >(p-2)^3$ holds for all primes $p$ other than $p=2,3,5,7,11$.
Suppose $q>3, p>11$. Then both sides of the equation are bounded by $(q^3-1)^3=(p-2)^3$ and $(q^3)^3 = (p-1)^3$.
Thus we would have $q^3=p-1$ or $p=(q+1)(q^2-q+1)$. This is clearly impossible as $p,q$ are primes so it cannot be the case.
Now we must verify by hand when $q=2,3$ or $p=2,3,5,7,11$. This can be a lot of work so here's a shortcut: For $p=2,3,5$ the LHS is negative but $q^8-7>0$. Thus either $p=7,11$ or $q=2,3$.
Clearly $p\neq q$ so $p|q^8-7$. Then $p=7\implies q=7$ a contradiction. If $p=11$ then $q^8\equiv 7$ mod $11$ but $7$ is not a quadratic residue so this fails.
Finally, for $q=3$ we get a solution $p=29$ and for $q=2$ we check that no solution exists.
hkl_student
21.07.2016 18:07
Quote: Take mod 9 and using $q^9 \equiv q \mod 9$ by Fermat I think that you made mistake while using Fermat here,because we use Fermat with primes,and I'm absolutely sure that 9 is not a prime number.
tudordarius
26.10.2016 23:04
If I'm not wrong, I think the statement is true given the fact that gcd(q,9)=1 since q is a prime other than 3
Math_CYCR
27.10.2016 03:16
For $q>3$ we get:
$(q^3+1)^3-5(q^3+1)^2-18(q^3+1)<q^9-7q< (q^3+2)^3-5(q^3+2)^2-18(q^3+2)$
For $q=2$ we get not solutions. And $q=3$ yields to $p=29$
The only solution is $(29, 3)$
bin_sherlo
19.04.2024 23:44
Answer: $(p,q)=(29,3)$ Proof: \[n=p^3-5p^2-18p=q^9-7q\]When $q=2$, we have $p^3-5p^2-18p=498\implies p|498\implies p\in\{2,3,83\}$ which gives no solution. When $q=3$, we have $p^3-5p^2-18p=19662\implies p\in\{2,3,29,113\} \implies p=29\implies \boxed{(29,3)}$ When $q\geq 5,$ we have $p\geq 31$ thus \[(p-2)^3<n<(p-1)^3\iff p^2-30p+8>0,\ 2p^2+21p-1>0\]\[(q^3-1)^3<n<(q^3)^3\iff 3q^6-3q^3-7q+1>0,\ 7q>0\]Hence $q^3=p-1$ which gives a contradiction because of $(mod \ 2)$.