Difference between revisions of "2017 USAJMO Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
Let <math>x</math> be odd where <math>x>1</math>. We have <math>x^2-1=(x-1)(x+1),</math> so <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> as desired. | Let <math>x</math> be odd where <math>x>1</math>. We have <math>x^2-1=(x-1)(x+1),</math> so <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> as desired. | ||
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+ | ==Solution 3== | ||
+ | Because problems such as this usually are related to expressions along the lines of <math>x\pm1</math>, it's tempting to try these. After a few cases, we see that <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> is convenient due to the repeated occurrence of <math>4x</math> when squared and added. We rewrite the given expressions as: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.</cmath> After repeatedly factoring the initial equation,we can get: <cmath>\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).</cmath> Expanding each of the squares, we can compute each product independently then sum them: <cmath>\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},</cmath> <cmath>\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.</cmath> Now we place the values back into the expression: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.</cmath> Plugging any positive integer value for <math>x</math> into <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs <math>\left(a,b\right)</math>. <math>\square</math> | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:19, 20 October 2017
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.
Solution 3
Because problems such as this usually are related to expressions along the lines of , it's tempting to try these. After a few cases, we see that is convenient due to the repeated occurrence of when squared and added. We rewrite the given expressions as: After repeatedly factoring the initial equation,we can get: Expanding each of the squares, we can compute each product independently then sum them: Now we place the values back into the expression: Plugging any positive integer value for into yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs .
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 |