MHB If a³ ≡ b³ (mod n) then a ≡ b (mod n)

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The statement "If a³ ≡ b³ (mod n) then a ≡ b (mod n)" is false, as demonstrated by counterexamples such as 2 and 4 modulo 8, and 1 and 2 modulo 7. Additionally, it is noted that for certain conditions, specifically when 3 is coprime to φ(n) and both a and b are coprime to n, the conclusion holds true. Examples include n=17 or any prime p where 3 does not divide p-1.

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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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