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

Revision as of 14:15, 12 August 2014

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:

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.

Fifth Solution

We notice that:

$\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}$

So it follows that $7n+1$ and $14n+3$ must be coprime for every natural number $n$ for the fraction to be irreducible. Now the problem simplifies to proving $\frac{7n+1}{14n+3}$ irreducible. We re-write this fraction as:

$\frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}$

Since the denominator $2(7n+1) + 1$ differs from a multiple of the numerator $7n+1$ by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that $\frac{21n+4}{14n+3}$ 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.

@@ Line 31: / Line 31: @@
 [[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]]:
+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>
@@ Line 62: / Line 62: @@
 </center>
-Since the denominator <math> 2(7n+1) + 1 </math> differs from a multiple of the numerator <math>7n+1</math> by 1, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is irreducible.
+Since the denominator <math> 2(7n+1) + 1 </math> differs from a multiple of the numerator <math>7n+1</math> by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is irreducible.
 Q.E.D
-~ Solution by '''''z6z61im4s99'''''
+{{alternate solutions}}
-<center>
-'''SIXTH SOLUTION'''
-</center>
-we can use PRINCIPLE OF MATHEMATICAL INDUCTION TO EASILY PROVE THE ABOVE STATEMENT.
-First, we show that the statement is true for n=1. For n=1, 25/17  is irreducible because gcd(25,17) =1.
-Now, we show that '''If the statement is true for the natural number 'n', then it is also true for natural number 'n+1'.'''
-PROOF:
-Let the given statement be true for the natural number 'n', which means gcd (21n+4, 14n+3) =1
-Now we need to show that the statement is true for natural number 'n+1', which means we need to show gcd (21n+25, 14n+17) =1
- let us take a look at that,
-gcd (21n+25, 14n+17) =1
--> gcd [21n+4+(7*3), 14n+3+(7*2)] = 1, which is true since gcd (21n+25, 14n+17) =1 and adding some multiple of some number 'k=7' to both the numbers doesn't change the gcd.
-Hence, the statement is true for natural number 'n+1'.
-Hence, the statement is true for each natural number.
-For this fraction to be irreducible,
-{{alternate solutions}}
 == See Also ==
 {{IMO box|year=1959|before=First question|num-a=2}}

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