If $p$, $q$ are naturals, solve $p^3-q^5=(p+q)^2$.

Solution 1:

I came up with a solution for $\gcd(p, q)=1$. Note that $p>q$. Looking mod $p$ gives $q^2(q^3+1) \equiv 0 \pmod p$, thus $p|q^3+1$. Take modulo $q$ of both sides of the equation to get $p^2(p-1) \equiv 0 \pmod q$, hence $q|p-1$. So we have $p=qr+1$ for some positive integer $r$. It follows that$$p|q^3+1-p=q^3-qr=q(q^2-r)$$Hence there is some non-negative integer $s$ for which $q^2-r=sp=s(qr+1)$ and consequently$$q^2-rsq-(r+s)=0$$In order to get integer values for $q$ discriminant of this quadratic must be a perfect square, but notice that if $r>1$ and $s>1$ $$(rs)^2<r^2s^2+4(r+s)<(rs+2)^2$$and we must have $r^2s^2+4(r+s)=(rs+1)^2$, which leads to$$r=\frac{4s-1}{2s-4},$$Which is never an integer. Contradiction. Therefore, we must have $r =1$ or $s \le1$.

$r=1$ gives $p=q+1$ and equation becomes $(q+1)^3-q^5=(2q+1)^2$. One can simplify this and get $-q^4+q^2-q-1=0$, so $q|1$ and $q=1$, which fails to satisfy equation.

$s=0$ gives $p=q^3+1$ and equation becomes $(q^3+1)^3-q^5=(q^3+q+1)^2$. So $q$ divides the constant term of polynomial equation, that is $q|2$, and $q=1, 2$, which again fail to satisfy equation.

$s=1$ gives $p=q^2-q+1$ and equation becomes $(q^2-q+1)^3-q^5=(q^2+1)^2$. So $q$ divides the constant term of polynomial equation, that is $q|3$, and $q=1, 3$. $q=1$ fails to satisfy equation. $q=3$ satisfies the equation and gives $p=7$. Therefore $(p, q)=(7, 3)$ is the only solution when $\gcd(p, q)=1$.