Difference between revisions of "2017 USAJMO Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
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. | 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. | ||
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Revision as of 23:50, 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.
07:34
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 |