Non-Homogenous Diophantine Equation

Rearrange this expression to get

$b+c = a(bc-a)$

Notice that the RHS should dominate as the variables get larger, especially as $b$ and $c$ get larger. Also note that if $a \geq bc$ , then there are no solutions because the $RHS$ is not positive but the LHS is.

Consider the graph of the parabola $f(x)=x(bc-x)$ . It is an upside-down parabola that crosses the $x$ -axis at $x=0$ and $x=bc$ . Thus, if $f$ is restricted to the interval $1 \leq x \leq bc-1$ , the common minimum value being $f(1)=bc-1$ . Since $1\leq a \leq bc-1$ , we have $b+c=f(a)\geq f(1)$ . Thus, we must have

$bc-1 \leq b+c.$

This certainly is a candidate for bounding! WLOG $c\geq b$ . If $b \geq 3$ , then $bc \geq c+c+c > b+c+1$ , which is a contradiction. Therefore, we must have either $b=1$ or $b=2$ .

Case 1: $b=1$

The original equation can be rearranged to

$c = \dfrac {a^2+1}{a-1} = a+1+\dfrac {2}{a-1}$

Since $\frac {2}{a-1}$ must be an integer, this leads to $a=2$ or $a=3$ . The solutions corresponding to this case are then $(a, b, c) = (2, 1, 5)$ and $(3, 1, 5)$ .

Case 2: $b=2$

The original equation can be rearranged to

$c = \dfrac {a^2+2}{2a-1}$

Here we will try to use a polynomial modulus to help simplify it.

$\begin{aligned} a^2 + 2 &\equiv 0 \pmod{2a-1}\\ 4a^2+8 &\equiv 0 \pmod{2a-1}\\ (2a)^2+8 &\equiv 0 \pmod{2a-1}\\ 1^2 + 8 &\equiv 0 \pmod{2a-1} \end{aligned}$

Thus, $2a+1 \mid 9$ . Checking the six factors of 9, including the negative ones, gives rise to six solutions, namely, $a=-4, -1, 0, 1, 2, 5$ . However, we wish $a$ to be positive, so $a=1, 2$ or $5$ . The solutions corresponding to this case are $(a, b, c) = (1, 2, 3), (2, 2, 2)$ and $(5, 2, 3)$ .

Thus, there are $5$ solutions if $b \leq c$ . We get another 5 if $c \leq b$ , but $(2, 2 ,2)$ are common. Therefore, there are 9 solutions altogether.

---

Disclaimer: This problem and solution came from Problem Solving Tactics.

$a^2+b+c=a b c \\ \implies a^2-a.(b c)+(b+c)=0$ .

Let, $a$ has two solutions in positive integers $m,n$ , so,

$(a-m)(a-n)=a^2-a.(b c)+(b+c)$ . Comparing coefficients we get

$m+n=b c \,\, \text{and} \,\, m n=b+c$ .This ordered pairs has a finite number of solutions.

Firstly ,put $b=c$ we get a unique solution $b=c=2$ .So, a ordered pair is $(2,2,2)$ . Next if $b \neq c$ then

there will be two solutions $(b,c)=(2,3);(1,5)$

If one of $m,n$ is negative then $(a-m)(a+n)=a^2-a.(b c)+(b+c)$ now

$m-n=b c$ and $b+c=-m n$ .Here $m$ is a positive solution and $n$ is a negative solution.As $bc >0$ then $m-n >0 \implies m>n$ .Again $b+c>0 \implies -mn >0 \implies mn<0 \implies n<0$ .First we assume that n is a negative solution and last we get $n<0$ so $n$ becomes positive.So both should be positive.

Lastly if both $m,n$ shouldn't be negative because solutions should be in positive integers.

So, the solutions are $(2,2,2);(1,2,3),(1,3,2);(2,1,5);(2,5,1);(3,1,5);(3,5,1);(5,2,3);(5,2,3)$

This is essentially the same as problem 4 of IMO 1994, because $c=\frac{a^2+b}{ab-1}$ Now we have $ab-1 | a^2+b$ or $ab-1 | a(a^2+b)-(ab-1)=a^3+1$ . Thus $\frac{a^3+1}{ab-1}$ must be an integer. @Wen Z

And I like this problem. Working with $(n, m)$ instead of $(a, b)$ .

First note that if $mn-1|n^3+1$ then $mn-1|m^3n^3-1+n^3+1=n^3(m^3+1)$ It follows that $mn-1|m^3+1$ because $\gcd(mn-1, n^3)=1$ , so if $(m,n)$ is an answer $(n,m)$ will be too. If $n=m$ we get $\frac{n^3+1}{mn-1}=\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ So $n-1|1$ and $n=2$ . Therefore $(2,2)$ is an answer and from now on we can suppose $m>n$ . Let $n=1$ , it gives $m=2,3$ , we get four answers from here: $(1,2),(2,1),(1,3),(3,1)$ Note that if $\frac{n^3+1}{mn-1}=r$ then $r\stackrel{n}{\equiv} -1$ , and $r$ have the form of $kn-1$ for some positive integer $k$ . Since $m>n$ $kn-1=\frac{n^3+1}{mn-1}<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1} \\ kn-1<n+\frac{1}{n-1} \\ k<1+\frac{1}{n}+\frac{1}{n(n-1)} \le 2 \\ k<2$ Thus $k=1$ and we have $(n-1)(mn-1)=n^3+1$ , hence $n-1|n^3+1$ or $n-1|n^3+1-(n^3-1)=2 \\ n-1|2$ As a result $n=2$ or $n=3$ which gives us $2m-1=9$ or $2(3m-1)=28$ and both of them leads to $m=5$ , so we get four other answers from here: $(2,5),(5,2),(3,5),(5,3)$ .