It's obvious that $q$ is even, so $q=2$, then $p=3$.

Murad.Aghazade wrote: $\boxed{\text{N4}}$ Find all primes $p,q$ such that $2p^3-q^2=2(p+q)^2$. Checking odd and even parity, we've $q^2$ is even $\implies q=2$ Thus $2p^3-4=2(p+2)^2$ $\implies p^3=p^2+4p+6$ Hence $p|6$ so $p$ is $2$ or $3$, of which only $p=3$ works Hence only solution $(p,q)=(3,2)$

Abhinandan18 wrote: Murad.Aghazade wrote: $\boxed{\text{N4}}$ Find all primes $p,q$ such that $2p^3-q^2=2(p+q)^2$. Checking odd and even parity, we've $q^2$ is even $\implies q=2$ Thus $2p^3-4=2(p+2)^2$ $\implies p^3=p^2+4p+6$ Hence $p|6$ so $p$ is $2$ or $3$, none of which is a solution. Hence no solution $(p,q)$ $p=3,q=2$ works.

Murad.Aghazade wrote: $\boxed{\text{N4}}$ Find all primes $p,q$ such that $2p^3-q^2=2(p+q)^2$. $q = 2$ and by replacing in the equation we get: $(p-3)((p+1)^2+1)=0$ which means $p=3$.

l1090107005 wrote: $p=3,q=2$ works. Yeah, made a minor error, thanks for pointing out

My solution it is easy to see that $q=2$. Our problem is equialent to $p^{2}(p-1)=4p+6$ Suppose $p>3$ $\boxed{case1}$ $p\equiv 1 mod 3$ $4p+6\equiv 10\equiv 0 mod 3$ which is contradiction. $\boxed{case2}$ $p\equiv 2 mod 3$ ---->> $1\equiv 2 mod 3$ which is contradiction. So $(p,q)=(3,2)$

My solutions. $\text{Clearly from the given equation we have q must be even.As q is prime,we have q=2}$ $\text{Putting the value of q in the equation we get}$ $2p^3-2^2=2(p+2)^2 \implies p^3-p^2-4p-6=0$ $\implies (p-3)(p^2+2p+2)=0$ $\boxed{Case1:}$ $p=3$ $\text{then the solution is}$ $\boxed{(3,2)}$ $\boxed{Case2:}$ $p^2+2p+2=0$ $\text{This case no solutions}$

As a quick alternative, note that $-q^2 \equiv 2q^2 \mod p \Rightarrow 3q^2 \equiv 0 \mod p,$ and since $p,q$ are primes, it follows $p=3, q=2.$

Wasn't that a really easy problem or am I wrong???

tenplusten wrote: $\boxed{\text{N4}}$ Find all primes $p,q$ such that $2p^3-q^2=2(p+q)^2$. This question must be given in HSM and not here.

For parity reasons, we must have $2\mid q$, hence $q=2$. The new equation is $p^3-p^2-4p-6=0$. By RRT, we get the solution $(p,q)=\boxed{(3,2)}$, otherwise, we would need $p^2+2p+2=0$, but since $p^2+2p+2=(p+1)^2+1$, we have no more solutions.