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2008 JBMO Problems/Problem 3

Revision as of 23:22, 2 January 2019 by KRIS17 (talk | contribs) (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...")

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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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