2017 USAJMO Problems/Problem 4

Problem

Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2) + 12$ is prime that properly divides the positive number $a^2 + b^2 + c^2 + abc - 2017$ ?

Let $p = (a-2)(b-2)(c-2) + 12$ and $m = a^2 + b^2 + c^2 + abc - 2017$ . We have that: $m - p = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4a - 4b - 4c - 2021 = (a+b+c-2)^2 - 45^2$ Let $a+b+c = x$ . Then $m - p$ is $(x-2)^2 - 45^2$ , which must be divisible by $p$ .

Since $m > p$ , $x > 47$ , and since $p$ divides $m - p$ , $p$ must divide either $x-47$ or $x+43$ .

It is easy to see that the minimum of $p = (a-2)(b-2)(c-2)+12$ is $x+4$ . Since $p > x+4 > x-47$ , $p$ cannot divide $x-47$ , so $p$ must divide $x+43$ . If $p \not= x+43$ , $x+43 \ge 2p$ . But $x + 43 < 2x + 8 < 2p$ , so $p = x+43$ . If $p$ is prime (p > 47 + 43 = 90), then $x$ has to be even, making one of $a,b,c$ even, making $(a-2)(b-2)(c-2) + 12$ an even number, which is a contradiction.

Thus, there are no integer triples that work.

~AopsUser101

2017 USAJMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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2017 USAJMO Problems/Problem 4

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