If a³ ≡ b³ (mod n) then a ≡ b (mod n)

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Discussion Overview

The discussion centers on the mathematical statement regarding congruences: whether the equivalence \( a^3 \equiv b^3 \mod n \) implies \( a \equiv b \mod n \). Participants explore this concept through examples and counterexamples, examining both specific cases and general conditions.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that the statement is false but struggles to provide a counterexample.
  • Another participant provides a counterexample: \( 2^3 \equiv 4^3 \mod 8 \).
  • A further example is presented showing \( 1^3 \equiv 2^3 \mod 7 \), indicating that the statement can fail even with prime moduli.
  • Another participant cites \( 4^3 \equiv 10^3 \mod 13 \) as a non-trivial counterexample.
  • One participant introduces a condition under which the statement holds: if 3 is coprime to \( \phi(n) \) and both \( a \) and \( b \) are coprime to \( n \), then \( a \equiv b \mod n \) follows, providing \( n = 17 \) as an example.

Areas of Agreement / Disagreement

Participants generally agree that the statement does not hold universally, as evidenced by multiple counterexamples. However, there is a conditional agreement on specific cases where the statement may hold true.

Contextual Notes

Participants note that the validity of the statement can depend on the properties of \( n \) and the relationship between \( a \) and \( b \) with respect to \( n \). There are unresolved aspects regarding the implications of the conditions mentioned.

KOO
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Let $a, b \in Z$ and $n \in N$ . Is the following necessarily true?
If $a^3 ≡b^3$(mod n) then $a ≡ b$ (mod n)

I know it's false but I can't think of an counterexample.
 
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Re: If $a^3 ≡b^3$(mod n) then $a ≡ b$ (mod n)

KOO said:
Let $a, b \in Z$ and $n \in N$ . Is the following necessarily true?
If $a^3 ≡b^3$(mod n) then $a ≡ b$ (mod n)

I know it's false but I can't think of an counterexample.

No

$2^3 = 4^3$ mod 8
 
It can even happen with a prime modulus: $1^3 = 2^3\pmod7$.
 
Even in non-trivial cases : $4^3 = 10^3\pmod{13}$
 
Hi,
Here's an easy result in the positive direction.

If 3 is prime to $$\phi(n)$$ and both a and b are prime to n, then the conclusion does follow.

Example: n=17 or any prime p with 3 not dividing p-1
 

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