There is typo in the question. Original question was: Find all pairs of $p,q$ prime numbers that satisfy the equation $p(p^4+p^2+10q)=q(q^2+3)$ We proceed to the solution. Notice $p\neq q$ otherwise $LHS>RHS$ While $p,q>3$ $p\nmid q\Rightarrow p|q^2+3\Rightarrow p\equiv 1$ (mod $3$) $(q+2) \equiv RHS \equiv LHS \equiv q $ (mod $3$), contradiction. So checking $p$ or $q \leq3$, only solution is $(p,q) = (2,5)$

please do not ruin this masterpiece by typo...

Corrected. Thanks.

AlperenINAN wrote: Find all pairs of $p,q$ prime numbers that satisfy the equation $p(p^4+p^2+10q)=q(q^2+3)$ This one is atleast only modular 6 bash

p<q is obvious 1.p=2 q=5 2.p=3 no solution 3.p=6k+1 by mod 3 no solution 4.p=6k-1 then q=2 mod 3 and 0=2 mod 3 conclusion

Let's examine the equation according to $mod$ $5$. $RHS \equiv LHS (mod 5) $ Let's look at the values of the $p(p^4+p^2+10q)$ expression. $p(p^4+p^2+10q) \equiv p(p^4+p^2) (mod5)$ If $p\equiv 0$, $p(p^4+p^2)\equiv 0 (mod5)$ If $p\equiv 1$, $p(p^4+p^2)\equiv 2 (mod5)$ If $p\equiv 2$, $p(p^4+p^2)\equiv 0 (mod5)$ If $p\equiv 3$, $p(p^4+p^2)\equiv 0 (mod5)$ If $p\equiv 4$, $p(p^4+p^2)\equiv 3 (mod5)$ Now, Let's look at the values of the $q(q^2+3)$ expression. If $q\equiv 0$, $q(q^2+3)\equiv 0 (mod5)$ If $q\equiv 1$, $q(q^2+3)\equiv 4 (mod5)$ If $q\equiv 2$, $q(q^2+3)\equiv 4 (mod5)$ If $q\equiv 3$, $q(q^2+3)\equiv 1 (mod5)$ If $q\equiv 4$, $q(q^2+3)\equiv 1 (mod5)$ $RHS \equiv LHS (mod 5) $ $q$ that satisfies this equivalence can only be 0 in $mod 5$. $q\equiv 0$ $(mod5)$ and q is a prime number. $\Rightarrow$ $q=5$. Let's write 5 instead of q in the main equation. $p(p^4+p^2+50)=5(5^2+3)=140$ $p(p^4+p^2+50)=140$ Only $q=2$ satisfies this equation. Then the only solution pair is $(p,q)=(2,5)$.

looking to mod 5 gives us the only answer $(p, q) = (2, 5)$