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.

  • 1
    $\begingroup$ I find $a=5,b=3$ as a solution. No others below $b=250$ I just put your equation into the quadratic formula and asked when $\sqrt{1+4b+4b^3}$ was an integer. $\endgroup$ – Ross Millikan Sep 6 '19 at 21:20
  • $\begingroup$ Could you please show how it will be an integer only at b=3? $\endgroup$ – Anonymous123 Sep 6 '19 at 21:26
  • $\begingroup$ No, I can't. I just used a spreadsheet and copied down to check through $b=249$. I found $a=b=0, a=b=1$ as well, but they fail $a \lt b$. I suspect there are no more, but don't know that. That is why I made it a comment. $\endgroup$ – Ross Millikan Sep 6 '19 at 21:29
  • $\begingroup$ LMFDB entry of this elliptic curve. Only integer solutions are $(a,b)\in\{(0,0),(1,1),(5,3)\}$. $\endgroup$ – Jyrki Lahtonen Nov 27 '19 at 11:34
  • $\begingroup$ Oopsie. Forgot about the negatives of those points. That is $(-1,0)$, $(-2,1)$ and $(-6,3)$. Anyway, when restricted to positive integers and $b<a$ the only solution is $(5,3)$. $\endgroup$ – Jyrki Lahtonen Nov 27 '19 at 11:47

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 $$b^2-c^2b+c+1=0,$$ 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 in yields $$b^3+b=a^2+a=(2b-1)^2+(2b-1)=4b^2-2b,$$ which we can rearrange to get the cubic $$b^3-4b^2+3b=0\qquad\text{ and hence }\qquad b^2-4b+3=0.$$ Then either $b=1$ or $b=3$, corresponding to $a=1$ and $a=5$, respectively. Hence the only solution with $a>b$ is $(a,b)=(5,3)$.

  • 2
    $\begingroup$ Not my downvote, but it's worth mentioning that the original equation does not a priori imply that $b$ divides $a+1$, so this argument does not actually solve the problem. $\endgroup$ – Mike Bennett Sep 7 '19 at 19:59
  • $\begingroup$ The fact that $a \cdot b=c \cdot d$ with $\gcd(a,b)=\gcd(c,d)=1$ does not imply that $a$ must divide one of $c$ or $d$. Consider, for example, $6 \cdot 35 = 10 \cdot 21$. $\endgroup$ – Mike Bennett Sep 8 '19 at 15:26
  • $\begingroup$ The corresponding elliptic curve here has rank $1$ and I'm really not sure that a simple elementary argument will work. $\endgroup$ – Mike Bennett Sep 8 '19 at 15:28
  • $\begingroup$ Ah yes, I didn't notice the original assumption that $a$ and $b$ are coprime. My mistake. $\endgroup$ – Mike Bennett Sep 9 '19 at 2:22

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