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

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For example, <math>a = 21</math>,<math>b=13</math>, and <math>c=13</math> works. <math>p=2311</math> is prime, and it divides the positive number <math>a^2+b^2+c^2+abc-2017=2311</math>.
 
For example, <math>a = 21</math>,<math>b=13</math>, and <math>c=13</math> works. <math>p=2311</math> is prime, and it divides the positive number <math>a^2+b^2+c^2+abc-2017=2311</math>.
  
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This solution is wrong.  No <math>p</math> actually exist.
 
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:54, 17 September 2019

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

Yes. Let $p = (a-2)(b-2)(c-2)+12 = abc - 2(ab+ac+bc)+4(a+b+c)+4$. Also define $\alpha=a+b+c$. We want $p$ to divide the positive number $a^2+b^2+c^2+abc-2017=(\alpha-47)(\alpha+43)+p$. This equality can be verified by expanding the righthand side. Because $a^2+b^2+c^2+abc-2017$ will be trivially positive if $(\alpha-47)(\alpha+43)$ is non-negative, we can just assume that $\alpha=47$. Analyzing the structure of $p$, we see that $a-2$,$b-2$, and $c-2$ must be $1$ or $5$ mod $6$, or $p$ will not be prime (divisibility by $2$ and $3$). Thus, we can guess any $a$,$b$, and $c$ which satisfies those constraints. For example, $a = 21$,$b=13$, and $c=13$ works. $p=2311$ is prime, and it divides the positive number $a^2+b^2+c^2+abc-2017=2311$.


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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
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