Difference between revisions of "1959 IMO Problems/Problem 1"
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Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | ||
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| + | === Fourth Solution === | ||
| + | Let <math>g = {\rm gcd}(21n + 4, 14n + 3)</math>. Then <math>g|h</math> where <math>h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1</math>. Thus, <math>g = h = 1</math>. ''Note: This solution, in hindsight, is just the first solution above in a slightly different notation.'' | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 16:37, 20 May 2012
Contents
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 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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
| 1959 IMO (Problems) • Resources | ||
| Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
