2020 CIME II Problems/Problem 10

First suppose that $c=1$ . Then $a+b+1=ab\Rightarrow (a-1)(b-1)=2$ from whence we have $(a,b,c)\in\{(2,3,1),(3,2,1)\}$ .

Now suppose that $c\geq 2$ . Since $a$ is a positive integer, from the equation we have $bc-1\leq a(bc-1)=b+c^2$ . Hence $b(c-1)\leq c^2+1$ , and since $c\geq 2$ we have $b\leq \frac{c^2+1}{c-1}=c+1+\frac{2}{c-1}\leq c+3$ . Since the original equation is symmetric in $(a,b)$ it follows that $a\leq c+3$ as well. Adding the inequalities gives $a+b\leq 2c+6$ . From the original equation we know that $c|a+b$ ; hence $a+b$ is a multiple of $c$ which is no more than $2c+6$ . It follows that $a+b\in\{c,2c,3c,4c,5c\}$ , for if $a+b\geq 6c$ we have $6c\leq 2c+6\Rightarrow c\leq3/2\Rightarrow c=1$ ; a contradiction since $c\geq 2$ .

We now check each of these 5 cases using the original equation, keeping in mind the two solutions already found.

Case I) $a+b=c\Rightarrow c+c^2=abc\Rightarrow 1+c=ab\Rightarrow1+a+b=ab\Rightarrow (a-1)(b-1)=2$ $\Rightarrow (a,b,c)\in\{(3,2,5),(2,3,5)\}$ .

Case II) $a+b=2c\Rightarrow 2+c=ab\Rightarrow4+a+b=2ab\Rightarrow(2a-1)(2b-1)=9$ $\Rightarrow(a,b,c)\in\{(5,1,3),(1,5,3),(2,2,2)\}$ .

Case III) $a+b=3c\Rightarrow (3a-1)(3b-1)=28\Rightarrow (a,b,c)\in\{(5,1,2),(1,5,2)\}$ .

Case IV) $a+b=4c\Rightarrow (4a-1)(4b-1)=65$ for which there are no solutions.

Case V) $a+b=5c\Rightarrow (5a-1)(5b-1)=126$ for which there are 2 solutions (corresponding to the factors 9 and 14) however they have $c=1$ ; already covered.

Computing $a^3+b^2+c$ for each of the 9 solutions and adding the results we have $28+29+14+18+22+32+36+128+129=436$ .

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2020 CIME II Problems/Problem 10