All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
The Quartic Condition (Posted on 2024-09-01) Difficulty: 3 of 5
Let a, b, c, d be distinct integers such that (x-a)(x-b)(x-c)(x-d) - 4 = 0 has an integer root r.

Show that 4r=a+b+c+d.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts a first step | Comment 1 of 2
The expansion of the function starts with the two terms:

x^4 - (a+b+c+d)x^3 + ....

So the sum of the four roots is a+b+c+d.

The mean of the four roots is (a+b+c+d)/4.

Roots tend to occur in pairs of the form mean_of_roots ± something.

I am guessing that for this function the "something" will be complicated enough such that for the root to be an integer, the something would have to be zero making r = (a+b+c+d)/4.

  Posted by Larry on 2024-09-01 13:04:15
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information