Difference between revisions of "2017 USAJMO Problems/Problem 4"
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Thus, we can guess any <math>a</math>,<math>b</math>, and <math>c</math> which satisfies those constraints. | 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>. | 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 21:30, 16 April 2019
Problem
Are there any triples of positive integers such that
is prime that properly divides the positive number
?
Solution
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
.
This solution is wrong. No actually exist.
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 |