Kevin's integer solutions

Without loss of generality, assume $a \geq b \geq c$ .

Rearrange the equation as follows -

$\qquad(ab-1)(c-1) = (a+b)(c+1)$

This can immediately tell us that $c \neq 1$ , otherwise $(a+b)(2) = 0$ , impossible for positive $a,b$ .

The similarity of $(c-1)$ and $(c+1)$ suggest we should combine them somehow for information on $a$ and $b$ . If we divide by $(c-1)$ we get:

$ab-1 = (a+b)\frac{1+c}{c-1} = (a+b)(1 + \frac{2}{c-1})$

However, $1+\frac{2}{c-1}$ is always between $1$ and $3$ for $c \geq 2$ . This gives us a way to constrain $a,b$ . There will be some casework afterwards, which is reduced incredibly by checking the cases $c = 2,3$ now.

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For $c=2$ ,

$\qquad\qquad\qquad\;\; ab-1 = 3(a+b)\\ \qquad ab-3(a+b)-1 = 0\\ \qquad\quad\:\: (a-3)(b-3) = 10$

From this, $(13,4,2)$ and $(8,5,2)$ are the only solutions with $c=2$ .

For $c=3$ ,

$\qquad\qquad\quad\,\, 2(ab-1) = 4(a+b)\\ \qquad ab-2(a+b)-1 = 0\\ \qquad\quad\:\: (a-2)(b-2) = 5$

So $(7,3,3)$ is the only solution with $c=3$ .

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Now return to

$\qquad ab-1 = (a+b)(1+\frac{2}{c-1}) < (a+b)(1+1),$ as $\frac{2}{c-1} < 1$

$\qquad\qquad\qquad\; ab-1 < 2(a+b)\\ \qquad ab - 2a - 2b - 1 < 0\\ \qquad\quad\: (a-2)(b-2) < 5$

This is satisfied, along with $a \geq b \geq c \geq 4$ , only for $a=4,b=4$ . Putting these values into the original gives no solution for $c$ . (n.b. this is the only case to check; without having eliminated $c=2,3$ first, there were $16$ to check.)

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The only unordered solutions are therefore $(13,4,2),(8,5,2)$ and $(7,3,3)$ . There are $6$ permutations of the first two and $3$ of the last, so $6+6+3=15$ overall solutions.

If $a = 1$ , then the given equation becomes $1 + b + c + b + bc + c = bc + 1.$ But it is clear that the left-hand side is greater than the right-hand side, so there are no solutions when $a = 1$ . Similarly, there are no solutions when $b = 1$ and $c = 1$ .

Hence, $a$ , $b$ , and $c$ must all be at least 2. Let $x = a - 1$ , $y = b - 1$ , and $z = c - 1$ , so $x$ , $y$ , and $z$ are positive integers. Also, $a = x + 1$ , $b = y + 1$ , and $c = z + 1$ . Substituting into the given equation and simplifying, we get $2x + 2y + 2z + 4 = xyz. \quad (*)$ This equation is symmetric in $x$ , $y$ , and $z$ , so without loss of generality, assume that $x \le y \le z$ . We can divide both sides by $xyz$ to get $\frac{2}{xy} + \frac{2}{xz} + \frac{2}{yz} + \frac{4}{xyz} = 1.$

If $x \ge 3$ , then $y \ge 3$ and $z \ge 3$ , and $\frac{2}{xy} + \frac{2}{xz} + \frac{2}{yz} + \frac{4}{xyz} \le \frac{2}{3^2} + \frac{2}{3^2} + \frac{2}{3^2} + \frac{4}{3^3} = \frac{22}{27} < 1,$ so $x = 1$ or $x = 2$ .

If $x = 1$ , then equation $(*)$ becomes $2 + 2y + 2z + 4 = yz,$ which we can write as $(y - 2)(z - 2) = 10$ . This leads to the values $y = 3$ and $z = 12$ , and $y = 4$ and $z = 7$ .

If $x = 2$ , then equation $(*)$ becomes $4 + 2y + 2z + 4 = 2yz,$ which we can write as $(y - 1)(z - 1) = 5$ . This leads to the values $y = 2$ and $z = 6$ .

Substituting back, we find that all the solutions in $(a,b,c)$ are $(2,4,13)$ , $(2,5,8)$ , $(3,3,7)$ , and their permutations. This gives us a total of $6 + 6 + 3 = 15$ solutions.

From AM-GM, we have

$a+b+c+ab+bc+ac=abc+1\le6\sqrt[6]{a^3b^3c^3}=6\sqrt{abc}$

Letting $u=abc$ , we can bound the product $abc$ by finding the positive values that satisfy

$u+1\le6\sqrt{u}$

It turns out that $abc\le34$ . Next, assume WLOG that $c\le b\le a$ . Hence, $c\le\lfloor\sqrt{34}\rfloor=3$ . The rest is casework.

Case $c=1$ : From the original equation, $a=-b$ , a contradiction. No solutions.

Case $c=2$ : From the original equation, $a=3+\frac{10}{b-3}$ . This yields the solutions $(a,b,c)=\boxed{(13,4,2),(8,5,2)}$

Case $c=3$ : From the original equation, we have $a=2+\frac{5}{b-2}$ . This yields the solution $(a,b,c)=\boxed{(7,3,3)}$ .

Counting permutations of these three solutions, we get that there are $3!+3!+\frac{3!}{2!}=\boxed{15}$ solutions.

WLOG,we can assume that, $a \ge b\ge c$ . At first ,we will prove that c<4. if $c \ge 4$ then, $a+b+b+ab+b^2+ab \ge L.H.S$ .So, $a+2b+2ab+b^2 \ge abc+1 \ge 4ab+1$ .Simplifying, $a+2b+b^2 \ge 2ab+1$ .But since, $2ab+1 =ab+ab+1 \ge b^2 +ab+1$ .So,showing $ab+1 > a+2b$ we will get our contradiction. But we can write this as, (b-1)(a-2)>1 since $a \ge b \ge c \ge 4$ this is true. For rest of the cases,some factoring and case work gives the solutions, (2,4,13),(2,5,8),(3,3,7) and their permutations which are 6+6+3=15 in total.

Suppose that a >= b >= c >= 1; We have the equation equivalents to (c + 1) (a + b) = (c-1) ( ab -1 ). Because a >=b >= c, so we have: (c - 2)^2 <= 3, so 1<= c <= 3. c = 1, no root c = 2, a = 3 + 10 /(b - 3), so 2 roots: (10, 4, 2), (8, 5, 2) c = 3, we have 1 root (3,3,7) So, we have totally 15 roots