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...") |
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So, <math>q = 3k</math> | So, <math>q = 3k</math> | ||
<math>=> k = 1 => q = 3, p = 5</math> and <math>r = 5</math> | <math>=> k = 1 => q = 3, p = 5</math> and <math>r = 5</math> | ||
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Thus we have the following solutions: <math>(7, 3, 2), (3, 2, 7), (5, 3, 5)</math> | Thus we have the following solutions: <math>(7, 3, 2), (3, 2, 7), (5, 3, 5)</math> | ||
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<math>Kris17</math> | <math>Kris17</math> | ||
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+ | == Solution 2 (similar to Solution 1, credit to dskull16) == | ||
+ | |||
+ | The equation can be rearranged into this form | ||
+ | |||
+ | <math>\frac{p(r+1)-4q}{q(r+1)} = 1</math> | ||
+ | <math>\implies p(r+1)-4q = q(r+1)</math> | ||
+ | <math>\implies (p-q)(r+1) = 4q</math> | ||
+ | |||
+ | 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 <math>(r+1)</math>. We can easily deal with the case where r is 2 but this gives a contradiction since we get that <math>p=6</math> where it needs to be prime. | ||
+ | |||
+ | <math>(r+1) = kq, k \in \mathbb{N}</math> | ||
+ | <math>\implies (p-q)(kq) = 4q</math> | ||
+ | <math>\implies (p-q)k = 4</math> | ||
+ | |||
+ | In the case k is 4, we get that p and q have a difference of 1 meaning that <math>p=3, q=2</math> and therefore <math>r=7</math> which clearly works. | ||
+ | |||
+ | In the case that k is 2, we get that p and q have a difference of 2 meaning that <math>p=5, q=2</math> and therefore <math>r=3</math> which also works. | ||
+ | |||
+ | In the case that k is 1, we get that p and q have a difference of 4 meaning that <math>p=7, q=3</math> and therefore <math>r=5</math> which works too and is our final solution. |
Latest revision as of 17:40, 3 March 2024
Problem
Find all prime numbers , such that
Solution
The given equation can be rearranged into the below form:
then we have
and and
then we have
and and
note that if , then which is a contradiction.
and
then we have
and and We have that exactly one of is a multiple of .
cannot be a multiple of since . Since is prime, then we have is a prime.
contradiction.
Also, cannot be a multiple of since, contradiction.
So,
and
Thus we have the following solutions:
Solution 2 (similar to Solution 1, credit to dskull16)
The equation can be rearranged into this form
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 . We can easily deal with the case where r is 2 but this gives a contradiction since we get that where it needs to be prime.
In the case k is 4, we get that p and q have a difference of 1 meaning that and therefore which clearly works.
In the case that k is 2, we get that p and q have a difference of 2 meaning that and therefore which also works.
In the case that k is 1, we get that p and q have a difference of 4 meaning that and therefore which works too and is our final solution.