Difference between revisions of "2003 IMO Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | p | + | Let <math>p</math> be a prime number. Prove that there exists a prime number <math>q</math> such that for every integer <math>n</math>, the number <math>n^p-p</math> is not divisible by <math>q</math>. |
== Solution == | == Solution == | ||
| + | {{solution}} | ||
| + | |||
Let N be <math>1 + p + p^2 + ... + p^{p-1}</math> which equals <math>\frac{p^p-1}{p-1}</math> | Let N be <math>1 + p + p^2 + ... + p^{p-1}</math> which equals <math>\frac{p^p-1}{p-1}</math> | ||
<math>N\equiv{p+1}\pmod{p^2}</math> | <math>N\equiv{p+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>. | 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 | ||
| + | |||
| + | ==See Also== | ||
| + | {{IMO box|year=2003|num-b=5|after=Last Problem}} | ||
Latest revision as of 08:39, 5 July 2024
2003 IMO Problems/Problem 6
Problem
Let 
be a prime number. Prove that there exists a prime number 
such that for every integer 
, the number 
is not divisible by .
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let N be 
which equals 
Which means there exists q which is a prime factor of n that doesn't satisfy 
.
\\unfinished
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 | ||
