Jump to content
BrainDen.com - Brain Teasers
  • 0
Sign in to follow this  
bonanova

Integer equation?

Question

Here's an equation that's easy to analyze with a little insight.
It's not original with me, I ran across it yesterday. (I'll provide attribution after it's solved.)

x3 - y3 = 217

Does this equation have integer solutions (pairs of {x, y} integers)?
If not, prove there are none; if so, find as many as you can.

Share this post


Link to post
Share on other sites

2 answers to this question

  • 0
Spoiler

x3 - y3 factors into (x - y)(x2 + xy + y2)

Which means if there are any integer solutions, x - y would be an integer and be a factor of 217. There are only 4 factors of 217: 1, 7, 31, 217. We also know that x3 > y3, which means x > y. With that information, we can take a "brute force" method to find all possibilities, knowing that x - y must equal either 1, 7, 31, or 217. It would be something like this:

Assume x - y = 1

  1. Solve for x: x = y + 1
  2. Substitute: 217 = (y + 1 - y)((y + 1)2 + (y + 1)y + y2) = 3y2 + 3y + 1
  3. Simplify: 216 = 3y2 + 3y...72 = y(y + 1)
  4. Solve for y: y=8 or y=-9
  5. Substitute and solve for x to find 2 solutions:
    x=9, y=8
    x=-8, y=-9

Now just repeat for x - y = 7, 31, 217 to find all solutions (note: after going through this, I realized you could probably logically drop 31 and 217 from the list of possible values from the start)...but regardless, the remaining integer solutions are:
x=1, y=-6
x=6, y=-1

 

Share this post


Link to post
Share on other sites
  • 0

I agree

Spoiler

 

x³ - y³ = 217

x = (217 + y³)   [or y = (x³ - 217)]

 

 

Only 4 pairs possible …

(x,y) = (6,-1), (1,-6), (-8,-9), (9,8)

 

 

Edited by bonanova
Spoiler

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this  

  • Recently Browsing   0 members

    No registered users viewing this page.

×