Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 USAJMO Problems/Problem 4"

(→‎Solution)
(→‎Solution 1)
(9 intermediate revisions by 7 users not shown)
Line 2: Line 2:
 
Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the positive number <math>a^2 + b^2 + c^2 + abc - 2017</math>?
 
Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the positive number <math>a^2 + b^2 + c^2 + abc - 2017</math>?
  
−
==Solution==
+
==Solution 1==
−
 
+
The answer is no. Substitute <math>x=a-2,y=b-2,z=c-2</math>. This means that <math>x,y,z\geq -1</math>. Then <cmath>a^2+b^2+c^2+abc-2017=(x+y+z-41)(x+y+z+49)+xyz+12.</cmath> It is given in the problem that this is positive. Now, suppose for the sake of contradiction that <math>xyz+12</math> is a prime. Clearly <math>x,y,z\neq 0</math>. Then we have <cmath>\frac{(x+y+z-41)(x+y+z-49)}{xyz+12}</cmath> is an integer greater than or equal to <math>1</math>. This also implies that <math>x+y+z > 41</math>. Since <math>xyz+12</math> is prime, we must have <cmath>xyz+12\mid x+y+z-41\text{ or } xyz+12\mid x+y+z+49.</cmath> Additionally, <math>x, y, z</math> must be odd, so that <math>xyz+12</math> is odd while <math>x+y+z-41,x+y+z+49</math> are even. So, if <cmath>xyz+12\mid x+y+z-41\text{ or }xyz+12\mid x+y+z+49,</cmath> we must have <cmath>2(xyz+12)\leq x+y+z-41\text{ or }2(xyz+12)\leq x+y+z+49.</cmath> Now suppose WLOG that <math>x=-1</math> and <math>y,z>0</math>. Then we must have <math>yz\leq 10</math>, impossible since <math>x+y+z>41</math>. Again, suppose that <math>x,y=-1</math> and <math>z>0</math>. Then we must have <cmath>2(z+12)\leq z-43\text{ or }2(z+12)\leq z+47,</cmath> and since in this case we must have <math>z>43</math>, this is also impossible.
−
Yes. Take <math>a=b=c=1</math> then <math>11 \mid -2013 = 1^2 + 1^2 + 1^2 + 1\times1\times1 - 2017</math>.
+
Then the final case is when <math>x,y,z</math> are positive odd numbers. Note that if <math>xyz>x+y+z</math> for positive integers <math>x,y,z</math>, then <math>abc>a+b+c</math> for positive integers <math>a,b,c</math> where <math>a>x,b>y,c>z</math>. Then we only need to prove the case where <math>x+y+z=43</math>, since <math>x+y+z</math> is odd. Then one of <cmath>2(xyz+12)\leq 2\text{ and/or }2(xyz+12)\leq 92</cmath> is true, implying that <math>xyz\leq -11</math> or <math>xyz\leq 34</math>. But if <math>x+y+z=43</math>, then <math>xyz</math> is minimized when <math>x=1,y=1,z=41</math>, so that <math>xyz\geq 41</math>. This is a contradiction, so we are done.
−
{{MAA Notice}}
 
  
 
==See also==
 
==See also==
 
{{USAJMO newbox|year=2017|num-b=3|num-a=5}}
 
{{USAJMO newbox|year=2017|num-b=3|num-a=5}}

Revision as of 19:40, 20 September 2021

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 (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_4&oldid=162525"