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

Revision as of 17:35, 23 December 2013

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.

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, then 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

~ Solution by z6z61im4s99

SIXTH SOLUTION

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 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

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