Difference between revisions of "2003 IMO Problems/Problem 6"
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== Solution == | == Solution == | ||
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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 | ||
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| + | ==See Also== | ||
| + | {{IMO box|year=2003|num-b=5|after=Last Problem}} | ||
Revision as of 23:49, 18 November 2023
2003 IMO Problems/Problem 6
Problem
p is a prime number. Prove that for every p there exists a q for every positive integer n, so that 
can't be divided by q.
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 | ||
