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

Revision as of 10:12, 6 April 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 $a^2+b^2+c^2+abc-2017=2311$ .

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 3: / Line 3: @@
 ==Solution==
-Yes. Take <math>a=b=c=1</math> then <math>11 \mid -2013 = 1^2 + 1^2 + 1^2 + 1\times1\times1 - 2017</math>.
-==Solution 2==
 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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Difference between revisions of "2017 USAJMO Problems/Problem 4"

Revision as of 10:12, 6 April 2019

Problem

Solution

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