Difference between revisions of "1959 IMO Problems/Problem 1"

Revision as of 09:59, 30 May 2012

Problem

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$ .

Solutions

First Solution

We observe that

$3(14n+3) = 2(21n+4) + 1.$

Since a multiple of $14n+3$ differs from a multiple of $21n+4$ by 1, we cannot have any postive integer greater than 1 simultaneously divide $14n+3$ and $21n+4$ . 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 $a, b$ as $(a,b)$ , we use the Euclidean algorithm as follows:

$( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1$

As in the first solution, it follows that $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Third Solution

Proof by contradiction:

Let's assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is a divisor of both the numerator and the denominator:

$14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}$

$21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}$

Subtracting the second equation from the first equation we get $1\equiv 0\pmod{p}$ which is clearly absurd.

Hence $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Fourth Solution

Let $g = {\rm gcd}(21n + 4, 14n + 3)$ . Then $g|h$ where $h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1$ . Thus, $g = h = 1$ . 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.

@@ Line 50: / Line 50: @@
 {{alternate solutions}}
+== See Also ==
 {{IMO box|year=1959|before=First question|num-a=2}}
 [[Category:Intermediate Algebra Problems]]
 [[Category:Intermediate Number Theory Problems]]

1959 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

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