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parmenides51 wrote: Determine positive integers $a$ and $b$ co-prime such that $a^2+b = (a-b)^3$ . $\rm a-b\mid (a-b)^3=a^2+b\Leftrightarrow a-b\mid (a^2-b)-(a^2-ab)=ab+b=b(a+1)\overset {(a-b,b)=1}{\Leftrightarrow} a-b\mid a+1\,\,(1)\Leftrightarrow $ $\rm \Leftrightarrow a-b\mid (a+1)-(a-b)=b-1\Leftrightarrow a-b\mid b+1\,\,(2)$. In addition $\rm a^2+b\equiv a^3-3a^2b+3b^2a-b^3 \pmod b \Leftrightarrow a^2\equiv a^3 \pmod b\overset{(a,b)=1}{\Leftrightarrow} b\mid a-1 $. If $\rm a=1$ then $\rm 2\leq b+1=(1-b)^3\leq 0$ ,contradiction. Let $\rm a=bc+1,c \in \mathbb{N^*}$. $\rm (2)\Leftrightarrow b(c-1)+1\mid b+1\Rightarrow b+1\geq b(c-1)+1\Leftrightarrow c\leq 2$ If $\rm c=1$ then $ \rm b+1=1^3$,contradiction. If $\rm c=2$ then $\rm 4b^2+4b+1+b=b^3+3b^2+3b+1\Leftrightarrow...\overset {b>0}{\Leftrightarrow}b=2,a=5$ So the only solution is $\rm (a,b)=(5,2)$.

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.