2017 USAJMO Problems/Problem 4

Revision as of 18:40, 20 September 2021 by Asdf334 (talk | contribs) (Solution 1)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAJMO Problems and Solutions