Difference between revisions of "1959 IMO Problems/Problem 1"
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As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | ||
+ | === Third Solution === | ||
+ | |||
+ | [[Proof by contradiction]]: | ||
+ | |||
+ | Let's assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is a [[divisor]] of both the [[numerator]] and the [[denominator]]: | ||
+ | |||
+ | <center> | ||
+ | <math>14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}</math> | ||
+ | |||
+ | <math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math> | ||
+ | </center> | ||
+ | |||
+ | Subtracting the second [[equation]] from the first [[equation]] we get <math>1\equiv 0\pmod{p}</math> which is cleary absurd. | ||
+ | |||
+ | Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | ||
+ | <!--Solution by tonypr--> | ||
{{IMO box|year=1959|before=First question|num-a=2}} | {{IMO box|year=1959|before=First question|num-a=2}} |
Revision as of 21:44, 17 May 2009
Problem
Prove that the fraction is irreducible for every natural number .
Solutions
First Solution
We observe that
Since a multiple of differs from a multiple of by 1, we cannot have any postive integer greater than 1 simultaneously divide and . Hence the greatest common divisor of the fraction's numerator and denominator is 1, so the fraction is irreducible. Q.E.D.
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
Let's 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 cleary absurd.
Hence is irreducible. Q.E.D.
1959 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |