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 $a^2 + b^2 + c^2 + abc - 2017$ ?

Solution 1

The answer is no. Substitute $x=a-2,y=b-2,z=c-2$ . This means that $x,y,z\geq -1$ . Then $a^2+b^2+c^2+abc-2017=(x+y+z-41)(x+y+z+49)+xyz+12.$ It is given in the problem that this is positive. Now, suppose for the sake of contradiction that $xyz+12$ is a prime. Clearly $x,y,z\neq 0$ . Then we have $\frac{(x+y+z-41)(x+y+z-49)}{xyz+12}$ is an integer greater than or equal to $1$ . This also implies that $x+y+z > 41$ . Since $xyz+12$ is prime, we must have $xyz+12\mid x+y+z-41\text{ or } xyz+12\mid x+y+z+49.$ Additionally, $x, y, z$ must be odd, so that $xyz+12$ is odd while $x+y+z-41,x+y+z+49$ are even. So, if $xyz+12\mid x+y+z-41\text{ or }xyz+12\mid x+y+z+49,$ we must have $2(xyz+12)\leq x+y+z-41\text{ or }2(xyz+12)\leq x+y+z+49.$ Now suppose WLOG that $x=-1$ and $y,z>0$ . Then we must have $yz\leq 10$ , impossible since $x+y+z>41$ . Again, suppose that $x,y=-1$ and $z>0$ . Then we must have $2(z+12)\leq z-43\text{ or }2(z+12)\leq z+47,$ and since in this case we must have $z>43$ , this is also impossible. Then the final case is when $x,y,z$ are positive odd numbers. Note that if $xyz>x+y+z$ for positive integers $x,y,z$ , then $abc>a+b+c$ for positive integers $a,b,c$ where $a>x,b>y,c>z$ . Then we only need to prove the case where $x+y+z=43$ , since $x+y+z$ is odd. Then one of $2(xyz+12)\leq 2\text{ and/or }2(xyz+12)\leq 92$ is true, implying that $xyz\leq -11$ or $xyz\leq 34$ . But if $x+y+z=43$ , then $xyz$ is minimized when $x=1,y=1,z=41$ , so that $xyz\geq 41$ . This is a contradiction, so we are done.

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

Problem

Solution 1

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