Difference between revisions of "2017 USAJMO Problems/Problem 4"
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Yes. Take <math>a=b=c=1</math> then <math>11 \mid -2013 = 1^2 + 1^2 + 1^2 + 1\times1\times1 - 2017</math>. | 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>. | ||
+ | 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. | ||
+ | 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>. | ||
+ | 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>). | ||
+ | Thus, we can guess any <math>a</math>,<math>b</math>, and <math>c</math> which satisfies those constraints. | ||
+ | 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>. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:02, 19 March 2019
Contents
[hide]Problem
Are there any triples of positive integers such that
is prime that properly divides the positive number
?
Solution
Yes. Take then
.
Solution 2
Yes. Let . Also define
.
We want
to divide the positive number
. This equality can be verified by expanding the righthand side.
Because
will be trivially positive if
is non-negative, we can just assume that
.
Analyzing the structure of
, we see that
,
, and
must be
or
mod
, or
will not be prime (divisibility by
and
).
Thus, we can guess any
,
, and
which satisfies those constraints.
For example,
,
, and
works.
is prime, and it divides the positive number
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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 |