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

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

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 $p = 2$: $2(16 + 4 + 70) = 2(90) = 180 \neq 364$
Try $p = 3$: $3(81 + 9 + 70) = 3(160) = 480 \neq 364$

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

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 $2 \pmod{3}$.

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

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

So $p(q + 2) \equiv q \pmod{3}$
If $p \equiv 1 \pmod{3}$: $q + 2 \equiv q \pmod{3} \implies 2 \equiv 0 \pmod{3}$, contradiction.
If $p \equiv 2 \pmod{3}$: $2(q + 2) \equiv q \pmod{3} \implies 2q + 4 \equiv q \pmod{3} \implies q + 1 \equiv 0 \pmod{3} \implies q \equiv 2 \pmod{3}$
But $q$ is a prime $> 3$, so $q \equiv 1$ or $2 \pmod{3}$.
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 $(p, q) = (2, 5)$.