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

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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>?
  
Let <math>p = (a-2)(b-2)(c-2) + 12</math> and <math>m = a^2 + b^2 + c^2 + abc - 2017</math>. We have that:
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==Solution 1==
<cmath>m - p = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4a - 4b - 4c - 2021 = (a+b+c-2)^2 - 45^2</cmath>Let <math>a+b+c = x</math>. Then <math>m - p</math> is <math>(x-2)^2 - 45^2</math>, which must be divisible by <math>p</math>.
 
  
Since <math>m > p</math>, <math>x > 47</math>, and since <math>p</math> divides <math>m - p</math>, <math>p</math> must divide either <math>x-47</math> or <math>x+43</math>.
 
  
It is easy to see that the minimum of <math>p = (a-2)(b-2)(c-2)+12</math> is <math>x+4</math>. Since <math>p > x+4 > x-47</math>, <math>p</math> cannot divide <math>x-47</math>, so <math>p</math> must divide <math>x+43</math>. If <math>p \not= x+43</math>, <math>x+43 \ge 2p</math>. But <math>x + 43 < 2x + 8 <  2p</math>, so <math>p = x+43</math>. If <math>p</math> is prime (p > 47 + 43 = 90), then <math>x</math> has to be even, making one of <math>a,b,c</math> even, making <math>(a-2)(b-2)(c-2) + 12</math> an even number, which is a contradiction.
 
 
Thus, there are no integer triples that work.
 
 
~AopsUser101
 
  
 
==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 08:21, 18 January 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

See also

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