Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2008 JBMO Problems/Problem 3

Revision as of 16:40, 3 March 2024 by Dskull16 (talk | contribs) (→‎Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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: $(7, 3, 2), (3, 2, 7), (5, 3, 5)$


$Kris17$

Solution 2 (similar to Solution 1, credit to dskull16)

The equation can be rearranged into this form

$\frac{p(r+1)-4q}{q(r+1)} = 1$ $\implies p(r+1)-4q = q(r+1)$ $\implies (p-q)(r+1) = 4q$

By the nature of primes this means that either q divides (p-q) or (r+1) and if q divides (p-q) then that means that q divides p which is a contradiction since any two primes have a GCD of 1 meaning that q divides $(r+1)$. We can easily deal with the case where r is 2 but this gives a contradiction since we get that $p=6$ where it needs to be prime.

$(r+1) = kq, k \in \mathbb{N}$ $\implies (p-q)(kq) = 4q$ $\implies (p-q)k = 4$

In the case k is 4, we get that p and q have a difference of 1 meaning that $p=3, q=2$ and therefore $r=7$ which clearly works.

In the case that k is 2, we get that p and q have a difference of 2 meaning that $p=5, q=2$ and therefore $r=3$ which also works.

In the case that k is 1, we get that p and q have a difference of 4 meaning that $p=7, q=3$ and therefore $r=5$ which works too and is our final solution.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_JBMO_Problems/Problem_3&oldid=216433"