What are the possible values of m and n for which Q divides P?

Click For Summary
SUMMARY

The discussion focuses on determining the values of m and n such that the polynomial Q = x² + 5x + 7 divides the polynomial P = (X + 2)m + (X + 3)n. Through modular arithmetic, the participants derive three sets of solutions based on the modulo conditions of m and n. Specifically, they conclude that m can be expressed in terms of multiples of 3, while n is defined by specific congruences modulo 6. The final solution set is S = {{m = 3u; n = 6v + 3}, {m = 3u + 1; n = 6v + 5}, {m = 3u + 2; n = 6v + 1}}.

PREREQUISITES
  • Understanding of polynomial division and modular arithmetic
  • Familiarity with congruences and their applications in algebra
  • Basic knowledge of polynomial expressions and their properties
  • Experience with mathematical notation and problem-solving techniques
NEXT STEPS
  • Study polynomial long division techniques for deeper insights
  • Explore modular arithmetic applications in number theory
  • Learn about polynomial factorization methods
  • Investigate congruences and their role in algebraic structures
USEFUL FOR

Mathematicians, algebra students, and anyone interested in polynomial theory and modular arithmetic applications.

larry91
Messages
12
Reaction score
0
Hi everybody! I have this problem: Either P = (X+2)m+(X+3)n and Q = x2+5x+7;
Determine m, n such that Q | P;( m, n = ? (Q divide P));

May you help me please?
Thank You!
 
Last edited:
Physics news on Phys.org


We'll be happy to help you, but not unless you show some effort first...
 


Ok! I try this, but I dislike this method:
Either 'P mod Q' the rest of the P at Q rapport;
(x+2)2 mod Q = -(x+3)
(x+2)3 mod Q = (x+2)2(x+2) mod Q = (-(x+3)(x+2)) mod Q = 1
=> the debris of (x+2)m at Q are repeated from 3 to 3;
(x+3)2 mod Q = x+2
(x+3)3 mod Q = ((x+2)(x+3)) mod Q = -1
(x+3)4 mod Q = -(x+3) mod Q = -(x+3)
(x+3)5 mod Q = -(x+2)
(x+3)6 mod Q = 1
=> the debris of (x+3)n at Q are repeated from 6 to 6;

P mod Q = ((x+2)m mod Q + (x+3)n mod Q) mod Q = 0;

1) If m mod 3 = 0 => (x+2)m mod Q = 1 => (x+3)n mod Q = -1 => n mod 6 = 3 => m = 3*u; n = 6*v+3

2) If m mod 3 = 1 => (x+2)m mod Q = x+2 => (x+3)n mod Q = -(x+2) => n mod 6 = 5 => m = 3*u+1; n = 6*v+5

3) If m mod 3 = 2 => (x+2)m mod Q = -(x+3) => (x+3)n mod Q = x+3 => n mod 6 = 1 => m = 3*u+2; n = 6*v+1

=> S={{m = 3*u; n = 6*v+3}, {m = 3*u+1; n = 6*v+5}, {m = 3*u+2; n = 6*v+1}}
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
48
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K