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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>?
  
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==Solution==
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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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<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>.
  
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(This solution is incorrect. The correct answer is no.)
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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>.
  
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Yes. Let <math>p = (a-2)(b-2)(c-2)+12 = abc - 2(ab+ac+bc)+4(a+b+c)+4</math>. Also define <math>\alpha=a+b+c</math>.
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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.
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We want <math>p</math> to divide the positive number <math>a^2+b^2+c^2+abc-2017=(\alpha-47)(\alpha+43)+p</math>. This equality can be verified by expanding the righthand side.
 
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Because <math>a^2+b^2+c^2+abc-2017</math> will be trivially positive if <math>(\alpha-47)(\alpha+43)</math> is non-negative, we can just assume that <math>\alpha=47</math>.
 
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Analyzing the structure of <math>p</math>, we see that <math>a-2</math>,<math>b-2</math>, and <math>c-2</math> must be <math>1</math> or <math>5</math> mod <math>6</math>, or <math>p</math> will not be prime (divisibility by <math>2</math> and <math>3</math>).
 
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Thus, we can guess any <math>a</math>,<math>b</math>, and <math>c</math> which satisfies those constraints.
 
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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>.
 
  
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Thus, there are no integer triples that work.
  
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{{MAA Notice}}
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~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 14:12, 19 February 2020

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

Let $p = (a-2)(b-2)(c-2) + 12$ and $m = a^2 + b^2 + c^2 + abc - 2017$. We have that: \[m - p = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4a - 4b - 4c - 2021 = (a+b+c-2)^2 - 45^2\]Let $a+b+c = x$. Then $m - p$ is $(x-2)^2 - 45^2$, which must be divisible by $p$.

Since $m > p$, $x > 47$, and since $p$ divides $m - p$, $p$ must divide either $x-47$ or $x+43$.

It is easy to see that the minimum of $p = (a-2)(b-2)(c-2)+12$ is $x+4$. Since $p > x+4 > x-47$, $p$ cannot divide $x-47$, so $p$ must divide $x+43$. If $p \not= x+43$, $x+43 \ge 2p$. But $x + 43 < 2x + 8 < 2p$, so $p = x+43$. If $p$ is prime (p > 47 + 43 = 90), then $x$ has to be even, making one of $a,b,c$ even, making $(a-2)(b-2)(c-2) + 12$ an even number, which is a contradiction.

Thus, there are no integer triples that work.

~AopsUser101

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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