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

Find all pairs of coprime positive integers $$(a,b)$$ such that $$b and $$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.

• 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. – Ross Millikan Sep 6 '19 at 21:20
• Could you please show how it will be an integer only at b=3? – Anonymous123 Sep 6 '19 at 21:26
• 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. – Ross Millikan Sep 6 '19 at 21:29
• LMFDB entry of this elliptic curve. Only integer solutions are $(a,b)\in\{(0,0),(1,1),(5,3)\}$. – Jyrki Lahtonen Nov 27 '19 at 11:34
• 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 $(5,3)$. – 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 $$(a,b)=(5,3)$$.
• 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. – Mike Bennett Sep 7 '19 at 19:59
• 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, $6 \cdot 35 = 10 \cdot 21$. – Mike Bennett Sep 8 '19 at 15:26
• The corresponding elliptic curve here has rank $1$ and I'm really not sure that a simple elementary argument will work. – Mike Bennett Sep 8 '19 at 15:28
• Ah yes, I didn't notice the original assumption that $a$ and $b$ are coprime. My mistake. – Mike Bennett Sep 9 '19 at 2:22