Solving Number Theory: Showing Congruence Has Exactly k Distinct Solutions

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SUMMARY

The discussion focuses on solving a number theory problem related to congruences and distinct solutions. The equation presented is xk = (xp-1)m = (xp-1 - 1)(1 + xp-1 + x2(p-1) + ... + x(m-1)(p-1)). Participants note that the equation xp-1 - 1 = 0 mod p has p-1 solutions, but struggle to derive insights from the geometric series involved. The discussion emphasizes the need for clarity in manipulating these mathematical expressions.

PREREQUISITES
  • Understanding of modular arithmetic and congruences
  • Familiarity with geometric series and their properties
  • Knowledge of number theory concepts, particularly related to distinct solutions
  • Experience with algebraic manipulation of equations
NEXT STEPS
  • Study the properties of modular arithmetic in depth
  • Learn about geometric series and their applications in number theory
  • Explore techniques for solving polynomial congruences
  • Investigate the implications of distinct solutions in modular equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on number theory and algebra, will benefit from this discussion. It is also valuable for anyone looking to deepen their understanding of congruences and their solutions.

teddyayalew
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Homework Statement



http://i43.tinypic.com/fymy3l.jpg

question 22.4 (a)

Homework Equations





The Attempt at a Solution



xk=(xp-1)m = (xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1))

I know that xp-1-1 = 0 mod p has p-1 solutions but I can't make anything from the geometric sum. Can someone push me in the right direction.

 
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teddyayalew said:

Homework Statement



http://i43.tinypic.com/fymy3l.jpg

question 22.4 (a)

Homework Equations



The Attempt at a Solution



xk=(xp-1)m = (xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1))

I know that xp-1-1 = 0 mod p has p-1 solutions but I can't make anything from the geometric sum. Can someone push me in the right direction.
(xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1)) ≠ (xp-1)m

(xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1)) = (xp-1)m - 1
 

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