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

  • Thread starter Thread starter KOO
  • Start date Start date
AI Thread Summary
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.
KOO
Messages
19
Reaction score
0
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.
 
Mathematics news on Phys.org
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top