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 even with prime moduli, such as 13, the equivalence can fail. However, if 3 is coprime to φ(n) and both a and b are coprime to n, then the conclusion holds true, particularly in cases involving prime numbers like 17. This discussion highlights the nuances of modular arithmetic and the conditions under which the equivalence may or may not be valid.
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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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