Find all natural solutions $(a, b)$ such that $(ab - 1) \mid (a^2 + a - 1)^2$.

Find all natural solutions $$(a, b)$$ such that $$\large (ab - 1) \mid (a^2 + a - 1)^2$$

We have that $$(ab - 1) \mid (a^2 + a - 1)^2 \implies (ab - 1) \mid [(ab)^2 - ab^2 - b^2]^2$$

$$\iff (ab - 1) \mid (ab^2 + b^2 + 1)^2 \iff (ab - 1) \mid (b^2 + b + 1)^2$$

I'm trying to prove that $$(ab - 1) \mid (a + b - 1)^2$$, yet I don't know how with the information presented.

Assuming that I know how to determine that $$(ab - 1) \mid (a + b - 1)^2$$. Let $$(a + b - 1)^2 = k(ab - 1), k \in \mathbb Z^+ \tag 1$$

where $$(a, b)$$ is the solution in which $$a + b$$ is at its minimal value.

$$\iff a^2 - [(k - 2)b + 2]a + [(b - 1)^2 + k] = 0$$

We have that the equation $$x^2 - [(k - 2)b + 2]a + [(b - 1)^2 + k] = 0$$ has two solutions $$x = a$$ and $$x = a'$$ such that $$a + a' = (k - 2)b + 2, aa' = (b - 1)^2 + k$$

It can easily be deduced that $$a' \in \mathbb Z^+ \implies (a', b)$$ is a solution to $$(1)$$

$$\implies a' + b \ge a + b \iff a' \ge a \implies \dfrac{(b - 1)^2 + k}{a} \ge a \iff (b - 1)^2 + k \ge a^2$$

It seems to me that there are infinitely many solutions, which are all consecutive elements in a sequence.

Furthermore, the assumption that $$(ab - 1) \mid (a + b - 1)^2$$ is incorrect. So I don't know what to begin from here.

• $(2,1)$, $(a,a+1)$ ($a\in {\mathbb N}$) are the obvious solutions. Commented Jan 12, 2020 at 7:16
• That is still wrong for $(a, b) = (2, 3)$ because $ab-1 = 5$ divides $(a^2+a-1)^2 = 5^2$, but it does not divide $(b^2+b+1)^2 = 13^2$. Commented Jan 12, 2020 at 10:53
• More “non-obvious” solutions: $(2, 13)$, $(74, 13)$, $(74, 433)$, $(2522, 433)$, $(2522, 14701)$. Commented Jan 12, 2020 at 11:00
• $(ab - 1) \mid (a + b - 1)^2$ does not hold either. Commented Jan 12, 2020 at 11:20
• It seems likely that your teacher miswrote the question and it should be $(ab - 1) \mid (a^2 - a + 1)^2$. In that case both conclusions $(ab - 1) \mid (b^2 - b + 1)^2$ and $(ab - 1) \mid (a + b - 1)^2$ are correct, as you can see in the AoPS link that I gave above. Commented Jan 12, 2020 at 11:35

Let $$(a,b)$$ be a solution of $$(ab - 1) \mid (a^2 \pm a - 1)^2$$ then $$(b,c)$$ is a solution of $$(bc-1) \mid (b^2 \mp b - 1)^2,$$ where $$c=\frac{(b^2\mp b-1)^2+ab-1}{b(ab-1)}.$$

Proof

Let $$N=ab-1$$. Then $$0\equiv (a^2b^2\pm ab^2-b^2)^2\equiv (b^2\mp b-1)^2$$ (mod $$N$$). Therefore $$(b^2\mp b -1)^2=NM$$ for some natural number $$M$$. Then $$M\equiv -1$$ (mod $$b$$) and so there is a natural number $$c$$ such that $$M=bc-1.$$

Application

This gives us a formula for generating solutions with every other iteration giving a solution of the original equation. Solutions $$(a,a+1)$$ just cycle round but the solution $$(2,1)$$ generates an infinite set of solutions containing those obtained by @MartinR.

In fact all solutions other than $$(a,a+1)$$ are generated from $$(2,1)$$.

A proof there are no other solutions

Let $$(a^2\pm a-1)^2=(ab-1)(ac-1).$$ If either $$b$$ or $$c$$ is less than $$a$$ then the described procedure can be used to give us a smaller solution. Otherwise we have $$b,c\ge a$$.

If $$b=a$$ then, for $$N=ab-1$$, we have $$a^2\equiv 1$$ (mod $$N$$) and then $$a\equiv 0$$ (mod $$N$$). The only possibility is $$N=1$$ and we have reached the base case.

Otherwise $$b,c\ge a+1$$ and the only possibility is $$b=c=a+1$$.

The solutions are as follows

$$\begin{matrix} (2, 13)&& (13, 74)\\ (74, 433)&& (433, 2522)\\ (2522, 14701)&& (14701, 85682)\\ (85682, 499393)&& (499393, 2910674)\\ (2910674, 16964653)&& (16964653, 98877242)\\ \end{matrix}$$

$$\cdots$$

As described in the above answer every other pair gives a solution of the original equation. The remaining pairs if reversed also give solutions but these simply form part of the other solutions. E.g. $$(2,13)$$ and $$(74,433)$$ are successive solutions. $$(74,13)$$ is also a solution which is the 'other half' of the $$(74,433)$$ one.Viz. $$(74^2+74-1)^2=(74\times 13-1)(74\times 433-1).$$

• @Piquito The error in the comment to my answer is in thinking that one can equate coefficients.
– user502266
Commented Jan 16, 2020 at 20:17
• @S. Dolan: You are right. Thank you. I delete my comment. Commented Jan 16, 2020 at 21:03