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Difference between revisions of "2008 JBMO Problems/Problem 3"

(Created page with "==Problem== Find all prime numbers <math> p,q,r</math>, such that <math> \frac{p}{q}-\frac{4}{r+1}=1</math> == Solution == The given equation can be rearranged into the be...")
(No difference)

Revision as of 00:22, 3 January 2019

Problem

Find all prime numbers $p,q,r$, such that $\frac{p}{q}-\frac{4}{r+1}=1$


Solution

The given equation can be rearranged into the below form:

$4q = (p-q)(r+1)$


$Case 1: 4|(p-q)$

then we have

$q = ((p-q)/4)(r+1)$ $=> (p-q)/4 = 1$ and $q = r + 1$ $=> r = 2, q = 3$ and $p = 7$


$Case 2: 4|(r+1)$

then we have

$q = (p-q)((r+1)/4)$ $=> (p-q) = 1$ and $q = (r + 1)/4$ $=> p = q + 1 => q = 2, p = 3$ and $r = 7$

note that if $(r+1)/4 = 1$, then $q = (p-q) => p = 2q$ which is a contradiction.


$Case 3: 2|(p-q)$ and $2|(r+1)$

then we have

$q = ((p-q)/2)((r+1)/2)$ $=> (p-q)/2 = 1$ and $q = (r+1)/2$ $=> p = q + 2$ and $r = 2q - 1$ We have that exactly one of $q, q + 1, q + 2$ is a multiple of $3$.


$q + 1$ cannot be a multiple of $3$ since $q + 1 = 3k => q = 3k - 1$. Since $r = 2q - 1$ is prime, then we have $2(3k - 1) - 1 = 3(2k-1)$ is a prime. $=> k = 1 => q = 2 => p = 4$ contradiction.


Also, $q + 2$ cannot be a multiple of $3$ since, $q + 2 = 3k => p = 3k => k = 1 => q = 1$ contradiction.


So, $q = 3k$ $=> k = 1 => q = 3, p = 5$ and $r = 5$


Thus we have the following solutions: $(7, 3, 2), (3, 2, 7), (5, 3, 5)$


$Kris17$

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