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

Difference between revisions of "2013 JBMO Problems/Problem 1"

(Created page with "==Problem== Find all ordered pairs <math>(a,b)</math> of positive integers for which the numbers <math>\dfrac{a^3b-1}{a+1}</math> and <math>\dfrac{b^3a+1}{b-1}</math> are bot...")
 
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 12: Line 12:
 
<math>\dfrac{a^3b+a}{a+1}</math> = <math>\dfrac{a(a^2b+1)}{a+1}</math> is a positive integer
 
<math>\dfrac{a^3b+a}{a+1}</math> = <math>\dfrac{a(a^2b+1)}{a+1}</math> is a positive integer
  
−
<math> => (a+1) | (a^2b+1)</math>
+
<math> \implies (a+1) \mid (a^2b+1)</math>
−
<math> => (a+1) | (((a+1) - 1)^2b+1)</math>
+
<math> \implies (a+1) \mid (((a+1) - 1)^2b+1)</math>
−
<math> => (a+1) | (b+1)</math>
+
<math> \implies (a+1) \mid (b+1)</math>
  
 
Similarly,  
 
Similarly,  
Line 22: Line 22:
 
<math>\dfrac{b^3a+b}{b-1}</math> = <math>\dfrac{b(b^2a+1)}{b-1}</math> is a positive integer
 
<math>\dfrac{b^3a+b}{b-1}</math> = <math>\dfrac{b(b^2a+1)}{b-1}</math> is a positive integer
  
−
<math> => (b-1) | (b^2a+1)</math>
+
<math> \implies (b-1) | (b^2a+1)</math>
−
<math> => (b-1) | (((b-1) + 1)^2a+1)</math>
+
<math> \implies (b-1) | (((b-1) + 1)^2a+1)</math>
−
<math> => (b-1) | (a+1)</math>
+
<math> \implies (b-1) | (a+1)</math>
  
 
Combining above <math>2</math> results we get:
 
Combining above <math>2</math> results we get:
Line 30: Line 30:
 
<math>(b-1) | (b+1)</math>
 
<math>(b-1) | (b+1)</math>
  
−
<math> => b=2,3 </math>
+
<math>\implies b=2,3 </math>
  
 
<math>Case 1: b=2</math>
 
<math>Case 1: b=2</math>
−
<math> => a+1|3 => a=2 </math> which is a valid solution.
+
<math> \implies a+1|3 \implies a=2 </math> which is a valid solution.
  
 
<math>Case 2: b=3</math>
 
<math>Case 2: b=3</math>
−
<math> => a+1|4 => a=1,3 </math> which are valid solutions.
+
<math> \implies a+1|4 \implies a=1,3 </math> which are valid solutions.
  
 
Thus, all solutions are: <math>(2,2), (1,3), (3,3)</math>
 
Thus, all solutions are: <math>(2,2), (1,3), (3,3)</math>
Line 42: Line 42:
  
 
<math>Kris17</math>
 
<math>Kris17</math>
 +
 +
== Solution 1.5 (credit to dskull16) ==
 +
 +
To get the two results:
 +
<math> (a+1) \mid (b+1)</math>
 +
<math> (b-1) \mid (a+1)</math>
 +
 +
We can also add zero to the numerator as follows:
 +
 +
<math>\dfrac{a^3b-1}{a+1} = \dfrac{a^3b + b - (b+1)}{a+1}</math>
 +
<math>\implies (a+1) \mid (b+1)</math>
 +
 +
since <math>a^3 + b = b(a+1)(a^2-a+1)</math>
 +
 +
 +
<math>\dfrac{b^3a+1}{b-1} = \dfrac{b^3a - a + (a+1)}{b-1}</math>
 +
<math>\implies (b-1) \mid (a+1)</math>
 +
 +
since <math>b^3a - a = a(b-1)(b^2 + b + 1)</math>
 +
 +
Then proceed as above.

Latest revision as of 17:09, 10 March 2024

Problem

Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers


Solution

Adding $1$ to both the given numbers we get:

$\dfrac{a^3b-1}{a+1} + 1$ is also a positive integer so we have:

$\dfrac{a^3b+a}{a+1}$ = $\dfrac{a(a^2b+1)}{a+1}$ is a positive integer

$\implies (a+1) \mid (a^2b+1)$ $\implies (a+1) \mid (((a+1) - 1)^2b+1)$ $\implies (a+1) \mid (b+1)$

Similarly,

$\dfrac{b^3a+1}{b-1} + 1$ is also a positive integer so we have:

$\dfrac{b^3a+b}{b-1}$ = $\dfrac{b(b^2a+1)}{b-1}$ is a positive integer

$\implies (b-1) | (b^2a+1)$ $\implies (b-1) | (((b-1) + 1)^2a+1)$ $\implies (b-1) | (a+1)$

Combining above $2$ results we get:

$(b-1) | (b+1)$

$\implies b=2,3$

$Case 1: b=2$ $\implies a+1|3 \implies a=2$ which is a valid solution.

$Case 2: b=3$ $\implies a+1|4 \implies a=1,3$ which are valid solutions.

Thus, all solutions are: $(2,2), (1,3), (3,3)$


$Kris17$

Solution 1.5 (credit to dskull16)

To get the two results: $(a+1) \mid (b+1)$ $(b-1) \mid (a+1)$

We can also add zero to the numerator as follows:

$\dfrac{a^3b-1}{a+1} = \dfrac{a^3b + b - (b+1)}{a+1}$ $\implies (a+1) \mid (b+1)$

since $a^3 + b = b(a+1)(a^2-a+1)$


$\dfrac{b^3a+1}{b-1} = \dfrac{b^3a - a + (a+1)}{b-1}$ $\implies (b-1) \mid (a+1)$

since $b^3a - a = a(b-1)(b^2 + b + 1)$

Then proceed as above.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_JBMO_Problems/Problem_1&oldid=216775"