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

Revision as of 21:44, 25 February 2018

Problem

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

Solutions

First Solution

For this fraction to be reducible there must be a number $x$ such that $x \cdot (14n+3) = (21n+4)$ , and a $1/x$ such that $1/x \cdot (21n+4) = (14n+3)$ . Since $x$ can only be one number ( $x$ is a linear term) we only have to evaluate $x$ for one of these equations. Using the first one, $x$ would have to equal $3/2$ . However, $3*3/2$ results in $9/2$ , and is not equal to our desired $4$ . Since there is no $x$ to make the numerator and denominator equal, we can conclude the fraction is irreducible.

EDIT: It appears to me that this solution is incorrect because it assumes that if $ax + b = cx + d$ , then $a = c$ and $b = d$ -- the problem says irreducible for ALL $n$ . If you agree, please remove this solution. 12/17/2017

EDIT 2: Disregard the first edit, because when comparing two polynomials (linear functions in this case), the coefficients for each term must be equal to eachother. For example, if $x$ can be any number, then in $ax^2+4=6x^2+b$ , we must have $(a,b)=(6,4)$ . Thus, if $ax+b=cx+d$ for all values $x$ , then $a$ must equal $c$ and $b$ must equal $d$ .

Clarification for edit 2: In this problem, we have $14nx+3x=21n+4$ , with an unknown $n$ , so we must have $14x=21$ and $3x=4$ , which cannot be true for any $1$ value of $x$ .

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

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.

Alternate Solution by pokemonduel:

Using Euclidean's Algorithm, you get $21n+4-(14n+3)$ and grouping like terms, you get $7n+1$ . Then, you use Euclidean's Algorithm again, and you get $14n+3-2(7n+1)$ or $1$ . Therefore, since the GCD of $21n+4$ and $14n+3$ is $1$ , the fraction $\frac{21n+4}{14n+3}$ is irreducible.

@@ Line 65: / Line 65: @@
 {{alternate solutions}}
+Alternate Solution by pokemonduel:
+Using Euclidean's Algorithm, you get <math>21n+4-(14n+3)</math> and grouping like terms, you get <math>7n+1</math>. Then, you use Euclidean's Algorithm again, and you get <math>14n+3-2(7n+1)</math> or <math>1</math>. Therefore, since the GCD of <math>21n+4</math> and <math>14n+3</math> is <math>1</math>, the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible.
 == 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

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