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 19:37, 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 |