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>. | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
==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 21:31, 9 March 2019
Problem
Are there any triples 
of positive integers such that 
is prime that properly divides the positive number 
?
Solution
Yes. Take 
then 
.
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 | ||
