Difference between revisions of "2017 USAJMO Problems/Problem 1"
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Let <math>a = 2n-1</math> and <math>b = 2n+1</math>. We see that <math>(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}</math>. Therefore, we have <math>(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}</math>, as desired. | Let <math>a = 2n-1</math> and <math>b = 2n+1</math>. We see that <math>(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}</math>. Therefore, we have <math>(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}</math>, as desired. | ||
| − | (Credits to | + | (Credits to mathmaster2012) |
==Solution 2== | ==Solution 2== | ||
Revision as of 00:23, 20 April 2017
Contents
Problem
Prove that there are infinitely many distinct pairs 
of relatively prime integers 
and 
such that 
is divisible by 
.
Solution 1
Let 
and 
. We see that 
. Therefore, we have 
, as desired.
(Credits to mathmaster2012)
Solution 2
Let 
be odd where 
. We have 
so 
This means that 
and since x is odd, 
or 
as desired.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
See also
| 2017 USAJMO (Problems • Resources) | ||
| First Problem | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
