Team Selection Test for EGMO 2023

We are to find all prime numbers $p,q$ such that $$p(p^4+p^2+10q)=q(q^2+3).$$

First, note that both sides are positive for positive primes $p,q$.

Let us analyze the equation: $$p(p^4+p^2+10q)=q(q^2+3).$$

Expand the left side: $$p^5+p^3+10pq=q^3+3q.$$

Bring all terms to one side: $$p^5+p^3+10pq-q^3-3q=0.$$

Group terms: $$p^5+p^3+10pq-q^3-3q=0.$$

Let us try small values for $p$.

Try $p=2$: $$2(2^4+2^2+10q)=q(q^2+3)$$ $$2(16+4+10q)=q(q^2+3)$$ $$2(20+10q)=q(q^2+3)$$ $$40+20q=q^3+3q$$ $$40+20q-q^3-3q=0$$ $$40+17q-q^3=0$$ $$q^3-17q-40=0$$

Try small prime values for $q$: - $q=2$: $8-34-40=-66$ - $q=3$: $27-51-40=-64$ - $q=5$: $125-85-40=0$

So $q=5$ works with $p=2$.

Now try $p=3$: $$3(3^4+3^2+10q)=q(q^2+3)$$ $$3(81+9+10q)=q(q^2+3)$$ $$3(90+10q)=q(q^2+3)$$ $$270+30q=q^3+3q$$ $$270+30q-q^3-3q=0$$ $$270+27q-q^3=0$$ $$q^3-27q-270=0$$

Try small prime values for $q$: - $q=2$: $8-54-270=-316$ - $q=3$: $27-81-270=-324$ - $q=5$: $125-135-270=-280$ - $q=7$: $343-189-270=-116$ - $q=11$: $1331-297-270=764$ - $q=13$: $2197-351-270=1576$

No solution for $p=3$ and small $q$.

Try $q=2$: $$p(p^4+p^2+20)=2(4+3)=14$$ Try $p=2$: $2(16+4+20)=2(40)=80\ne 14$ Try $p=3$: $3(81+9+20)=3(110)=330\ne 14$

Try $q=3$: $$p(p^4+p^2+30)=3(9+3)=36$$ Try $p=2$: $2(16+4+30)=2(50)=100\ne 36$ Try $p=3$: $3(81+9+30)=3(120)=360\ne 36$

Try $q=5$: $$p(p^4+p^2+50)=5(25+3)=140$$ Try $p=2$: $2(16+4+50)=2(70)=140$ So $p=2$, $q=5$ is a solution (already found).

Try $q=7$: $$p(p^4+p^2+70)=7(49+3)=364$$ Try $p=2$: $2(16+4+70)=2(90)=180\ne 364$ Try $p=3$: $3(81+9+70)=3(160)=480\ne 364$

Try $q=11$: $$p(p^4+p^2+110)=11(121+3)=1364$$ Try $p=2$: $2(16+4+110)=2(130)=260\ne 1364$ Try $p=3$: $3(81+9+110)=3(200)=600\ne 1364$

Now, consider the degree of the equation. For large $p$, the left side grows much faster than the right side, so only small values are possible.

Now, check for $p=q$: $$p(p^4+p^2+10p)=p(p^2+3)$$ $$p^5+p^3+10p^2=p^3+3p$$ $$p^5+10p^2=3p$$ $$p^5+10p^2-3p=0$$ $$p(p^4+10p-3)=0$$ So $p=0$ or $p^4+10p-3=0$, which has no positive integer solution for $p$.

Now, try to check modulo $3$ for possible contradictions for $p>3$: If $p>3$, $p$ is odd and $p\equiv 1$ or $2(\mathrm{mod}3)$.

Compute $p(p^4+p^2+10q)(\mathrm{mod}3)$: - $p^4\equiv 1(\mathrm{mod}3)$ if $p\not\equiv 0(\mathrm{mod}3)$ - $p^2\equiv 1(\mathrm{mod}3)$ - $10q\equiv q(\mathrm{mod}3)$ So $p^4+p^2+10q\equiv 1+1+q=q+2(\mathrm{mod}3)$ So $p(p^4+p^2+10q)\equiv p(q+2)(\mathrm{mod}3)$

The right side: $q(q^2+3)\equiv q(q^2)\equiv q^3(\mathrm{mod}3)$ But for $q\not\equiv 0(\mathrm{mod}3)$, $q^3\equiv q(\mathrm{mod}3)$ So $q(q^2+3)\equiv q(\mathrm{mod}3)$

So $p(q+2)\equiv q(\mathrm{mod}3)$ If $p\equiv 1(\mathrm{mod}3)$: $q+2\equiv q(\mathrm{mod}3)⟹2\equiv 0(\mathrm{mod}3)$, contradiction. If $p\equiv 2(\mathrm{mod}3)$: $2(q+2)\equiv q(\mathrm{mod}3)⟹2q+4\equiv q(\mathrm{mod}3)⟹q+1\equiv 0(\mathrm{mod}3)⟹q\equiv 2(\mathrm{mod}3)$ But $q$ is a prime $>3$, so $q\equiv 1$ or $2(\mathrm{mod}3)$. Try $q=2$: But $q=2$ already checked.

Thus, for $p>3$, there is a contradiction modulo $3$.

Therefore, the only possible values are $p=2$ or $p=3$.

For $p=2$, $q=5$ is a solution. For $p=3$, as above, $q^3-27q-270=0$ has no integer solution for $q$.

Therefore, the only solution is $(p,q)=(2,5)$.

Answer: The only pair of prime numbers $p,q$ satisfying the equation is $(p,q)=(2,5)$.