Difference between revisions of "1959 IMO Problems/Problem 1"
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=== First Solution === | === First Solution === | ||
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− | + | For this fraction to be reducible there must be a number <math>x</math> such that <math>x*(14n+3) = (21n+4)</math>, and a <math>1/x</math> such that <math>1/x*(21n+4) = (14n+3)</math>. Since <math>x</math> can only be one number (<math>x</math> is a linear term) we only have to evaluate x for one of these equations. Using the first one, <math>x</math> would have to equal <math>3/2</math>. However, <math>3*3/2</math> results in <math>9/2</math>, and is not equal to our desired <math>4</math>. Since there is no <math>x</math> to make the numerator and denominator equal, we can conclude the fraction is irreducible | |
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=== Second Solution === | === Second Solution === |
Revision as of 23:35, 6 August 2016
Contents
[hide]Problem
Prove that the fraction is irreducible for every natural number .
Solutions
First Solution
For this fraction to be reducible there must be a number such that , and a such that . Since can only be one number ( is a linear term) we only have to evaluate x for one of these equations. Using the first one, would have to equal . However, results in , and is not equal to our desired . Since there is no to make the numerator and denominator equal, we can conclude the fraction is irreducible
Second Solution
Denoting the greatest common divisor of as , we use the Euclidean algorithm as follows:
As in the first solution, it follows that is irreducible. Q.E.D.
Third Solution
Assume that is a reducible fraction where is a divisor of both the numerator and the denominator:
Subtracting the second equation from the first equation we get which is clearly absurd.
Hence is irreducible. Q.E.D.
Fourth Solution
Let . Then where . Thus, . Note: This solution, in hindsight, is just the first solution above in a slightly different notation.
Fifth Solution
We notice that:
So it follows that and must be coprime for every natural number for the fraction to be irreducible. Now the problem simplifies to proving irreducible. We re-write this fraction as:
Since the denominator differs from a multiple of the numerator by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that is irreducible.
Q.E.D
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |