Difference between revisions of "2017 USAJMO Problems/Problem 1"
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| − | Let <math>x</math> be any odd number above 1 | + | Let <math>x</math> be any odd number above 1. We have <math>x^2-1=(x-1)(x+1).</math> Since <math>x-1</math> is even, <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+x+2^x \equiv 0 \pmod{2x+2}.</math> This means for any odd x, the ordered triple <math>(x,x+2)</math> satisfies the condition. Since there are infinitely many values of <math>x</math> possible, there are infinitely many ordered triples, as desired. |
Revision as of 18:37, 19 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 laegolas)
Solution 2
Let 
be any odd number above 1. We have 
Since 
is even, 
This means that 
and since x is odd, 
or 
This means for any odd x, the ordered triple 
satisfies the condition. Since there are infinitely many values of possible, there are infinitely many ordered triples, 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 | ||
