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

Home > Just Math
Primes divide expression? (Posted on 2016-07-20) Difficulty: 3 of 5
Define funcitons F, G and H:
F(x,y) = x2 - 2xy + y2 - 5x + 7y
G(x,y) = x2 - 3xy + 2y2 + x - y
H(x,y) = xy - 12x + 15y

If x and y are integers and p is an odd prime such that the values of F(x,y) and G(x,y) are both divisible by p, then prove that H(x,y) is also divisible by p.

Seventeen divides expression? is a specific instance of this puzzle where p=17.

No Solution Yet Submitted by Brian Smith    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Other similar way Comment 2 of 2 |
Broll has solved it full, anyway it's worth also the direct approach.

As broll says G=(x-y)(x-2y+1). To have G multiple of p, one of both factors should be multiple of p. 

a) (x-y) multiple of p. 
F=(x-y)^2-5x+7y. Then to have F also divisible by p for values (x-y)=0 (mod p) it should be that 7y-5x =0 (mod p). From both in bold, 2y is 0 (mod p) (=> y is 0 mod p) and x is 0 (mod p).

=>(x,y) are both 0 (mod p). H is divisible by p in this case.

b) (x-2y+1) multiple of p

Playing a little, see that: 

F=G + H/2 +y(x-2y+1)/2 

To have F also divisible by p (for values (x-2y+1) = 0 (mod p)) is necessary that also H is 0 (mod p) for that values. 
  • This is because considering the three summands: G is multiple of p and y(x-2y+1)/2 is a multiple of p or at least half a multiple of p. Only if also H is multiple of p, F will also be.  
  • Note also that F is always and integer (for x, y integers)

Conclusion: On the basis of G divisible by p, to have F divisible by p implies necessarily H divisible by p. As it is a given data that F is multiple of p for certain x,y values, the same values for which G is divisible by p, then, for that values, necessarily H should be divisible by p.

================================ 

The seventeen divides expression puzzle is not fully solved as broll's solution considers a wide set of answers. 

I can add that if p=17 and I restricted (x-2y+1) to be not only a p multiple, but just p, I mean (x-2y+1)=p*1=17 it's easy to obtain the solution x=20 y=2, not considered in the above mentioned puzzle.

F(20,2)=238=17*14
G(20,2)=306=17*18
H(20,2)=-170=-17*10

Edited on July 22, 2016, 11:17 am
  Posted by armando on 2016-07-22 04:10:53

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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information