Integer solutions to $ \frac{1}{a} + \frac{1}{b}=\frac{1}{2018} $

Asked Jun 30 '22 at 14:38

Active Jun 30 '22 at 17:10

Viewed 72 times

We need to find integer solutions to

$ \frac{1}{a} + \frac{1}{b}=\frac{1}{2018} $

and this is equivalent to the problem $ \frac{a+b}{ab}=\frac{n}{2018n} $

so i must solve the sistem $ a+b=n $ and $ ab=2018n$ for some integers 'n'

so i must find only integers value roots of the polynomial

$ a(n-a)=2018n $

and i can not go further

edited Jun 30 '22 at 17:10

asked Jun 30 '22 at 14:38

Jose Garcia

1

Well, $2018 = 2\times 1009$ and $1009$ is prime... so if you know that $a(n-a)=2018$ what can you say about $a$? – JMoravitz Jun 30 '22 at 14:40
Your equation should be $a(n-a)=2018,n$. – wasn't me Jun 30 '22 at 14:44
Using both of the above comments, both $(2)$ and $(1009)$ must be present in the collective prime factorizations of $(a)$ and $(n-a)$. – user2661923 Jun 30 '22 at 15:07
+1 : to your question. I know that this is self-indulgence, but it is actually a pleasure to read a posting where the original poster shows work. – user2661923 Jun 30 '22 at 15:09
Please use more descriptive titles. – Shaun Jun 30 '22 at 15:11
2

https://artofproblemsolving.com/community/c4h1779351p11715323. Does this help? You can use Approach$0$ to search your queries before posting to avoid duplicates. – insipidintegrator Jun 30 '22 at 15:17
$a=b=4036$ will work – John Douma Jun 30 '22 at 15:18
A general method for equations like this is to complete the product (rectangle) - see the linked dupe. – Bill Dubuque Jun 30 '22 at 15:33

0 Answers0