How to Prove a^5 ≡ a (mod 15) for Any Integer a?

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SUMMARY

The discussion focuses on proving that \( a^5 \equiv a \mod 15 \) for any integer \( a \). The key approach involves demonstrating that the expression \( a^5 - a = (a-1)a(a+1)(a^2 + 1) \) is divisible by both 3 and 5. By establishing the divisibility of this expression by these two prime factors, one can conclude that it is divisible by 15, thereby confirming the original statement.

PREREQUISITES
  • Understanding of modular arithmetic
  • Familiarity with polynomial factorization
  • Knowledge of divisibility rules for integers
  • Basic concepts of number theory
NEXT STEPS
  • Study the properties of modular arithmetic in number theory
  • Learn about polynomial factorization techniques
  • Explore the divisibility rules for 3 and 5
  • Investigate the application of the Chinese Remainder Theorem
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Students of mathematics, particularly those studying number theory, educators teaching modular arithmetic, and anyone interested in proofs involving integer properties.

sty2004
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Homework Statement


Prove that a5 \equiva (mod 15) for every integer a.


Homework Equations





The Attempt at a Solution


I do not know how to show a5-a is divisible by 15
 
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a^5 - a = (a-1)a(a+1)(a^2 + 1) can you show that this expression is divisible by 3 and by 5 individually? This would imply it's divisible by 15.
 

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