Difference between revisions of "2003 IMO Problems/Problem 6"

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Which means there exists q which is a prime factor of n that doesn't satisfy <math>q\equiv{1}\pmod{p^2}</math>.
 
Which means there exists q which is a prime factor of n that doesn't satisfy <math>q\equiv{1}\pmod{p^2}</math>.
 
\\unfinished
 
\\unfinished
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== Solution 2 ==
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For <math>p</math> prime and <math>gcd(n, p) = 1</math>, <math>n^{p}\equiv{n}\pmod{p}</math>.
  
 
==See Also==
 
==See Also==
 
{{IMO box|year=2003|num-b=5|after=Last Problem}}
 
{{IMO box|year=2003|num-b=5|after=Last Problem}}

Revision as of 21:11, 29 March 2025

2003 IMO Problems/Problem 6

Problem

Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.

Solution

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Let N be $1 + p + p^2 + ... + p^{p-1}$ which equals $\frac{p^p-1}{p-1}$ $N\equiv{p+1}\pmod{p^2}$ Which means there exists q which is a prime factor of n that doesn't satisfy $q\equiv{1}\pmod{p^2}$. \\unfinished

Solution 2

For $p$ prime and $gcd(n, p) = 1$, $n^{p}\equiv{n}\pmod{p}$.

See Also

2003 IMO (Problems) • Resources
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Problem
All IMO Problems and Solutions