parmenides51 wrote:

Determine positive integers $a$ and $b$ co-prime such that $a^2+b = (a-b)^3.$

Solution:
Mod $a$ analyzation: $a|b^3+b \implies a|b^2+1$ .
Mod $b$ analyzation: $b|a^2-a^3 \implies b|a-1$ .

From $a=kb+1$ we get
$kb+1|b^2+1 \implies kb+1|kb^2+k\implies kb+1|k-b$
Note that $(k-1)(b-1) \geqslant 0 \implies kb+1 \geqslant k+b > k-b$ but $kb+1|k-b \implies k-b=0 \implies k=b$ so $\boxed{a=b^2+1}-(*)$ .
$\begin{align*} &a^2+b = a^3-b^3-3ab(a-b)\\ \implies &b=a^2-a-3b(a-b)\\ \implies &b=ab^2-3b(a-b)\\ \implies &b=\frac{3a+1}{a+3}\\ \implies &a+3|8 \end{align*}$
Since $a, b \in \mathbb{Z}^+ \implies a+3 = 1,2,4,8 \implies a=1,5$ . For $a=1$ , we get $b=0$ , which is not possible. Hence $\boxed{a=5}$ , and for that we get $\boxed{b=2}$ . Hence, $(5,2)$ is the only possible solution set.