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We take $m=12$. What follows is a standard argument: given a solution $(a,b,c)$ with $a\le b\le c$, we show that we can find a solution $(b,c,a')$ with $b\le c\le a'$. Start with $(a_0,b_0,c_0)=(1,1,1)$ (this is a solution of the equation). Let $P(n)$ be the property $a_n|b_nc_n+1, b_n|c_n+a_n, c_n|a_nb_n+1$ and let $Q(n)$ be the property $a_n\le b_n\le c_n$. Furthermore, let $R(n)$ be $(a_n,b_n)=(b_n,c_n)=(c_n,a_n)=1$. It’s clear that $P(0),Q(0),R(0)$ hold. Let’s now take $(a_1,b_1,c_1)=(b_0,c_0,\frac{b_0c_0+1}{a_0})$. $Q(1)$ holds because we already know $b_0\le c_0$ from $Q(0)$, and $b_1\le c_1$ because otherwise we would have $c_0a_0>b_0c_0+1$, which is false. $R(1)$ is equally easy to check. Let’s now prove $P(1)$: $c_1|a_1b_1+1$ because of the definition of $c_1$, $a_1|b_1c_1+1\iff a_0b_0|b_0c_0^2+c_0+a_0$, which is true because $a_0|b_0c_0+1,b_0|c_0+a_0,(a_0,b_0)=1$, and $b_1|c_1+a_1\iff c_0a_0|a_0b_0+b_0c_0+1$, which is true because $a_0|b_0c_0+1,c_0|a_0b_0+1,(c_0,a_0)=1$. We have thus shown that $P(1),Q(1),R(1)$ hold, and, in particular, we can define $(a_2,b_2,c_2)=(b_1,c_1,\frac{b_1c_1+1}{a_1})$, and so on. We use the exact same arguments to prove that $P(n),Q(n),R(n)$ are true, so we can define $(a_{n+1},b_{n+1},c_{n+1})=(b_n,c_n,\frac{b_nc_n+1}{a_n})$. On the other hand, a simple computation reveals that if $(a_n,b_n,c_n)$ is a solution, then so is $(a_{n+1},b_{n+1},c_{n+1})$, so we have found infinitely many solutions (that’s because the inequalities in $Q(n)$ become strict for $n\ge 2$, due to $R(n)$).

Consider the sequence x[0] = x[1] = 1 and, for k>1, x[k+1] = 3.x[k]-x[k-1]. For k>1, we have x[k+1]>x[k]. As x[k+1]^2 -3.x[k+1].x[k] + x[k]^2 + 1 = 0, it follows, that the given relation is verified with m = 12 for a = x[k], b = x[k+1], c = x[k] + x[k+1].

I think $ m=12$ is the only solution. Assume $ a\ge b\ge c$, then \[ a^2+a(b+c+\dfrac{1-(m-1)bc}{b+c})+bc+1=0\] We have $ a|bc+1$. Suppose $ a+b+c$ has the smallest sum, and let $ a_1$ be the other solution to this equation. We have $ aa_1=bc+1$, but $ a_1\ge a\ge b$. So either $ a^2=bc+1$ or $ a=b=c=1$. The former is impossible. Therefore the only possible $ m$ is $ 12$. We can use root flipping to generate the other solutions. $ (a,b,c)$ to $ (b,c, (bc+1)/a)$. Finally, I think root flipping basically guarantees $ (bc+1)/a$ to be an integer from the quadratic equation itself.

HEY! we need a solution with m=12 to start root flipping. In the case (1,1,1), m=9, and the root flipping gives nothing...

When $m=12$,I think we may prove: $2a_{2n}=a_{2n+1}+a_{2n-1}$ $3a_{2n+1}=a_{2n}+a_{2n+2}$

nealth wrote: I think $ m=12$ is the only solution. According to the official solutions here, "using a little more theory on quadratic forms, it can be shown that if the equation is soluble for a given value of $m$ then there are infinitely many solutions for that value of $m$." They also construct solutions for infinitely many $m$, but don't provide any details on how to actually use quadratic forms for fixed $m$ when there exists a solution. In particular, does anyone see how to construct infinitely many solutions for $m = 15$ ($(a,b,c) = (1,4,5)$ works)? The root flipping argument also generates $(1,4,1)$, $(21,4,5)$, and $(21,4,17)$, but nothing else from there...

Note that we can expand to get $(a+b+c)(ab+bc+ca+1)=mabc$, so rearranging gives a quadratic in $a$: $(b+c)a^2+((b+c)^2+(1-m)bc+1)a+(b+c)(bc+1)=0$. Using a Vieta Jumping argument, if $(a,b,c)$ is a solution then so is $\left(\dfrac{bc+1}{a},b,c\right)$. So we take $a=b=c=1$, so then $m=12$. We define the sequence $(a_n)_{n\ge 1}$ by $a_1=a_2=a_3=1$ and $a_n=\dfrac{a_{n-1}a_{n-2}+1}{a_{n-3}}$ for all $n\ge 4$. It is easy to show that $a_n>a_{n-1}$ for $n\ge 4$, so if we can show the sequence only contains integers, then note that $(a_n,a_{n+1},a_{n+2})$ gives us a different solution to the equation for $m=12$ for all positive integers $n$. We will induct to show that $a_n=4a_{n-2}-a_{n-4}$ for all $n\ge 5$, from which the result will follow. Note that $a_4=2$ and $a_5=3$, so the base case holds. Now, assume $a_k=4a_{k-2}-a_{k-4}$ for some $k\ge 5$, and we will show that $a_{k+1}=4a_{k-1}-a_{k-3}$: \begin{align*} a_k&=4a_{k-2}-a_{k-4}\\ \implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-a_{k-1}a_{k-4}\\ \implies a_{k-1}a_k&=4a_{k-1}a_{k-2}-(a_{k-2}a_{k-3}+1)\\ \implies a_{k-1}a_k+1&=a_{k-2}(4a_{k-1}-a_{k-3})\\ \implies a_{k+1}a_{k-2}&=a_{k-2}(4a_{k-1}-a_{k-3})\\ \implies a_{k+1}&=4a_{k-1}-a_{k-3}, \end{align*} as desired.

My solution is quite different from others; therefore I wanna share it with Aops users. Granted, we take $m=12$ and choose $a=b+c$. After some algebra, it remains to show that there are infinitely many positive integers $b,c$ such that $b^2$ $+$ $c^2$ $+$ $1$ $=$ $3bc$. As this problem is very famous and quite easy to deal with it, we can even find that the solution set of aforementioned equation is increasing, giving us solution set of $a,b,c$ infinite.

Systematicworker wrote: My solution is quite different from others; therefore I wanna share it with Aops users. Granted, we take $m=12$ and choose $a=b+c$. After some algebra, it remains to show that there are infinitely many positive integers $b,c$ such that $b^2$ $+$ $c^2$ $+$ $1$ $=$ $3bc$. As this problem is very famous and quite easy to deal with it, we can even find that the solution set of aforementioned equation is increasing, giving us solution set of $a,b,c$ infinite. How famous? Never heard of the equation $(b-c)^2 = bc-1$ before.

In fact, for the equation to have integer solutions, $5c^2-4$ would have to be a square, which occurs infinitely often by Pell's equations and the initial solution $(1,1)$.

benstein wrote: In fact, for the equation to have integer solutions, $5c^2-4$ would have to be a square, which occurs infinitely often by Pell's equations and the initial solution $(1,1)$. Can you explain more about why it is connected with Pell’s function?

By Pell theory, any equation of the form $px^2-y^2=q$ has infinitely many solutions iff it has $1$. In this case, the equation is $5c^2-y^2=4$, which has the initial solution $(c,y)=(1,1)$.

The answer is yes with $m=18$. Lemma. The equation $3c^2-11k^2=4$ has infinitely many solutions in positive integer. Proof. Let $c=4u+22v$ and $y=2u+12v$ the equation become $$u^2-33v^2=1$$Notice that $(u,v)=(23,4)$ is a solution. Therefore by Pell's equation we obtain infinitely many solution of it. Hence there are infinitely many solutions of $3c^2-11k^2=4$. Now letting $m=18$. We show that a solution exists for infinitely many values of $c$. After clearing denominator, the equation transform into $$18abc=(a+b+c)(ab+bc+ca+1)$$Subsitiute $x=ab$ and $y=a+b$. Notice that for any pair of integer $x,y$, there exists positive integers satisfying $x=ab$ and $y=a+b$ if and only if $y^2-4x=k^2$ for some $k\in\mathbb N$. Now treating $c$ as a constant we have \begin{align*} 18xc&=(y+c)(x+cy+1)\\ x(17c-y)&=(y+c)(cy+1) \end{align*}We now make the crucial step: there exists a solution which satisfies $c=2y$ for infinitely many $y$. Indeed, substituting $c=2y$ we have \begin{align*} 33xy&=3y(2y^2+1)\\ x&=\frac{2y^2+1}{11} \end{align*}Hence \begin{align*} k^2&=y^2-4x\\ &=\frac{3c^2-4}{11}\\ 3c^2-11k^2&=4 \end{align*}From the lemma there exists infinitely many solutions, each of which induces a pair of $(a,b,c)$. This completes the proof.

benstein wrote: By Pell theory, any equation of the form $px^2-y^2=q$ has infinitely many solutions iff it has $1$. In this case, the equation is $5c^2-y^2=4$, which has the initial solution $(c,y)=(1,1)$. I see. Thank for your explaining.

Good problem. It takes some time to get an equation where you can use vieta jumping. The answer is yes, and the example is $m=12$. First the equation is the same as: $$ \frac{(a+b+c)(ab+bc+ca+1)}{abc}=m$$So $c \mid (a+b)(ab+1)$ after seeing that the degree of $c$ is $1$ (This was my motivation). We can guess that $c=a+b$, so if $m=12, c=a+b$ we have: $$\frac{2(ab +(a+b)^2+1}{ab}=12 \implies \frac{a^2+b^2+1}{ab}=3$$Now we need to prove that exists infinites pairs such that: $a^2 + b^2+1-3ab=0$, the trival solution is $(1,1)$ and if $(a,b)$ is a solution with $a\leq b$ implies that $(b,3b-a)$ is also a solution but with largest maximum number so there are infinites pairs. $\blacksquare$

We claim that $m=12$ works. Let $c=a+b$, then we will find infinitely many solutions to: $$\frac1a+\frac1b+\frac1{a+b}+\frac1{ab(a+b)}=\frac6{a+b}$$$$\Leftrightarrow a^2-3ab+b^2+1=0.$$We start with the initial solution $(a,b)=(1,1)$. Using the method of Vieta jumping, suppose a solution $(a,b)$ exists with $a+b$ maximal. WLOG $a\le b$, then $(3b-a,b)$ is another solution but $3b-a+b\ge a+b$, a contradiction. So infinitely many solutions must exist.

Consider the following sequences: \[a_0=1,a_1=1,a_2=1,a_3=2,a_n=\frac{a_{n-1}a_{n-2}+1}{a_{n-3}}\]\[b_0=1,b_1=1,b_2=1,b_3=2,b_n=4b_{n-2}-b_{n-4}\]Suppose $a_i=b_i$ for all $i\le n$ then \begin{align*} a_n &= \frac{a_{n-1}a_{n-2}+1}{a_{n-3}} \\ &= \frac{(4a_{n-3}-a_{n-5})a_{n-2}+1}{a_{n-3}} \\ &= \frac{4a_{n-2}a_{n-3}-a_{n-5}a_{n-2}+1}{a_{n-3}} \\ &= \frac{4a_{n-2}a_{n-3}-a_{n-5}\frac{a_{n-3}a_{n-4}+1}{a_{n-5}}+1}{a_{n-3}} \\ &= \frac{4a_{n-2}a_{n-3}-a_{n-3}a_{n-4}}{a_{n-3}} \\ &= 4a_{n-2}-a_{n-4} \end{align*}as desired. Since $a_{i},a_{i+1},a_{i+2}$ being a solution implies $a_{i+1},a_{i+2},a_{i+3}$ being a solution, we are done.

Motivation. Observe the degree of each variable after multiplying both sides by $abc(a+b+c)$. Once you know that the degree is $2$, or equivalently we have a quadratic equation wrt each variable individually, Vieta Jumping idea hits instantly. So, if we have at least one triple $(a, b, c)$, we can construct the "larger" one. For one solution, just take $(a,b,c)=(1,1,1)$ for which $m=12$. Solution. $m=12$ works. Proof. WLOG assume that $\min\{a, b, c\} = a$. Rewrite the equation as $(b+c)a^2 + (b^2 - 9bc + c^2 - 1)a + (b+c)(bc+1) = 0$. Hence, $a$ is a solution to the quadratic equation $(b+c)x^2 + (b^2 - 9bc + c^2 - 1)x + (b+c)(bc+1) = 0$. Let $a'$ be the second root. by Vieta, $a'a=bc+1 \implies a' = \frac{bc+1}{a} \ge \frac{a^2 + 1}{a} > a$. So we can construct the triple $(a', b, c)$ from $(a, b, c)$, having larger minimum, so we can construct an infinite number of new triples and we are done.

Very nice solution!

We'll prove that for $m = 12$, there are infinitely many triples $(a, b, c)$ such that $a = b + c$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc} = \frac{12}{a + b + c}$. Note that the given equality is equivalent to $a^2 + a(b + c - \frac{11bc - 1}{b + c}) + (bc + 1) = 0$. Substituting $a = b + c$, we get $b^2 - b \cdot 3c + (c^2 + 1) = 0$. Thus we only need to prove that the above equation has infinitely many solutions. Note that $b = c = 1$ is a solution of above equation. WLOG, assume $b \le c$. Then root of the quadratic equation $x^2 - x \cdot 3c + (c^2 + 1) = 0$ is $b$ and let $b_1$ be the other root. Then $b + b_1 = 3c$ and $bb_1 = (c^2 + 1)$, hence $b_1$ is integer and $b_1 = \frac{c^2 + 1}{b} > b$. Hence swapping $(b, c) \to (b_1, c)$, we see that there are infinitely many solutions. $\blacksquare$

Yeah, consider $m = 12$, and: $a = r (2 - \sqrt{3})^n + (1-r) (2+\sqrt{3})^n$, where $r = \frac{\sqrt{3} + 1}{2\sqrt{3}} = \frac{3 + \sqrt{3}}{6}$. $b = \frac{(2-\sqrt{3})^n +(2 + \sqrt{3})^n}{2} $. $c = r(2 - \sqrt{3})^{n+1} + (1-r) (2 + \sqrt{3})^{n+1}$ for any positive integer $n$