Difference between revisions of "2017 USAJMO Problems/Problem 1"
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where we use our lemma and the Euler totient theorem: <math>a^{\phi{n}} \equiv 1 \pmod{n}</math> when <math>a</math> and <math>n</math> are relatively prime. | where we use our lemma and the Euler totient theorem: <math>a^{\phi{n}} \equiv 1 \pmod{n}</math> when <math>a</math> and <math>n</math> are relatively prime. | ||
| + | ==Solution 2== | ||
| + | Let <math>x</math> be any odd number above 1 | ||
{{MAA Notice}} | {{MAA Notice}} | ||
==See also== | ==See also== | ||
{{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}} | {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}} | ||
Revision as of 18:27, 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 
and 
are relatively prime (they are consecutive positive odd integers).
Lemma: 
.
Since every number has a unique modular inverse, the lemma is equivalent to proving that 
. Expanding, we have the result.
Substituting for and , we have
![\[(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},\]](http://latex.artofproblemsolving.com/5/5/7/5572c842a8d19ea1e35b514541c85bea6a906030.png)
where we use our lemma and the Euler totient theorem: 
when and 
are relatively prime.
Solution 2
Let 
be any odd number above 1
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 | ||
