Find all primes $(p, q)$ such that $p^3 - q^5 = (p+q)^2$ After rearranging I got that,
$p^2(p-1)=q(q^4+q+2p)$ and clearly $p>q$
So, $q|p-1$
Thus, $p-1=kq$ where $k$ is any integer
Now putting it in the original equation,I get that
$$k^3q^2+2k^2q+k-q^4-2kq-q-2=0$$
But now I don't know how to proceed from here...please give me some hints or correct me if I am wrong
I am so sorry for my terrible mistake....it would be $q^5$ instead of $q^2$ .....really sorry(although i got some really good answers)
 A: $p^3 - q^2 = p^2+2pq + q^2$ so $p^3-p^2=2q^2+2pq$ so $p^2(p-1)=2q(q+p)$
So $p|2q(q+p)$ but $p$ is prime so either $p|2$ and $p=2$ or $p|q$ and $p = q$ or $p|q+p$ and so $p|q$ and $p = q$.
So either $p = 2$ or $p = q$.
If $p = 2$ we have $4 = 2q(q+2)$ so $2=q(q+2)$ but that equation has no integer solutions.
So $p = q$ and we have
$p^2(p-1) = 2p(p+p) = 4p^2$  so $p-1 = 4$ and $p = q = 5$ and $5^3 - 5^2 = (5+5)^2$ is the only solution.
Of course, everyone else made one calculation error or another so I probably did too.
A: $$p^3 - q^2 = (p+q)^2 \Rightarrow p^3-p^2-2pq = 2q^2 \Rightarrow p |2q^2$$
Since $p$ is prime we get that either $p=2$ or $p|q$.
If $p=2$ the equation becomes
$$
8-q^2=(2+q)^2 \Rightarrow q^2+(2+q)^2=8  
$$
which gives $2q^2 < 8$ and hence $q<2$ not possible.
If $p|q$ then since $q$ is prime we get $p=q$ and hence
$$
p^3=p^2+4p^2 \Rightarrow p=5 
$$
Thus $p=q=5$.
A: We have,
$$\begin{align}&(p+q)^2-p^3+q^2=0 \\&2q^2+2pq+(p^2-p^3)=0\end{align}$$
$$\begin{align}\Delta&=p^2-2(p^2-p^3)\\
&=p^2(2p-1)\end{align}$$
$$\begin{align}&2p-1=(2n-1)^2\\
&\implies p=2n^2-2n+1\end{align}$$
$$\begin{align}q&=\frac{-p+\sqrt {\Delta}}{2}\\
&=(2 n^2 - 2 n + 1) (n - 1)\end{align}$$
Thus we deduce $n=2$. This means,
$$q=5,~ p=5$$
