Triplet Solutions for (x,y,z) in the Equation (1+1/x)(1+1/y)(1+1/z)=2

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on finding all triplets of positive integers \( (x, y, z) \) that satisfy the equation \( \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2 \). Participants share their methods and solutions, indicating that the problem has been solved through various approaches. The collaborative nature of the discussion highlights different perspectives on tackling the equation, with participants expressing gratitude for each other's contributions.

PREREQUISITES
  • Understanding of algebraic manipulation and integer solutions
  • Familiarity with the properties of positive integers
  • Knowledge of equation solving techniques
  • Basic experience with mathematical problem-solving discussions
NEXT STEPS
  • Research methods for solving polynomial equations
  • Explore integer programming techniques for optimization problems
  • Learn about combinatorial number theory
  • Investigate similar equations involving multiple variables
USEFUL FOR

Mathematicians, educators, and students interested in algebraic problem-solving, particularly those focused on integer solutions and collaborative mathematical discussions.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find all triplets of positive integers $(x, y, z)$ such that $$\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
 
Last edited:
Mathematics news on Phys.org
Re: Find all triplets of x, y and z

anemone said:
Find all triplets of positive integers $(x, y, z)$ such that $$\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.

(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them

solved as

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2
(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)
 
Re: Find all triplets of x, y and z

kaliprasad said:
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them

solved as

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2
(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Wow! I forgot to thank you explicitly for participating and also for your unique way of attacking the problem. What's worst is that I forgot completely how I approached it two months ago and I promise you to add my reply once I solved it because I could only tell at this point that I solved it differently than what you did. Sorry, kaliprasad!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K