Integral solutions of $a^{2} + a = b^{3} + b$

Question

Find all pairs of coprime positive integers $(a,b)$ such that $b<a$ and $a^2+a=b^3+b$

My approach: $a(a+1)=b(b^2+1)$ so $a|(b^2+1)$ and $b|(a+1)$ Now after this I am not able to do anything.pls help.

Answer 1

As you already note, if $a$ and $b$ are coprime and $$a(a+1)=a^2+a=b^3+b=b(b^2+1),$$ then it follows that $b$ divides $a+1$. Then $a=bc-1$ for some integer $c$, where $c>1$ because $a>b$. Then $$b^3+b=a^2+a=(bc-1)^2+(bc-1)=c^2b^2-cb,$$ and since $b$ is positive we can divide both sides by $b$ and rearrange to get the quadratic $$b^2-c^2b+c+1=0,\tag{1}$$ in $b$. This shows that the integer $b$ is a root of a quadratic equation with discriminant $$\Delta=(-c^2)^2-4\cdot1\cdot(c+1)=c^4-4c-4.$$ In particular this means $c^4-4c-4$ is a perfect square. Of course $c^4$ is itself a perfect square, and the previous one is $$(c^2-1)^2=c^4-2c^2+1,$$ which shows that $-4c-4\leq-2c^2+1$, or equivalently $$2c^2-4c-5\leq0.$$ A quick check shows that this implies $c<3$, so $c=2$. Then this plugging back into $(1)$ yields $$b^2-4b+3=0,$$ and so either $b=1$ or $b=3$, corresponding to $a=1$ and $a=5$, respectively. Hence the only solution with $a>b$ is $(a,b)=(5,3)$.