If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.

Question

If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.

I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?

Answer 1

Here is a general statement:

If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m \equiv b^m \bmod p$ iff $a \equiv b\bmod p$.

This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v \in \mathbb N$.

In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10\cdot (-2) + 3\cdot(7)=1$. Write this as $1 + 10 \cdot 2 = 3\cdot 7 $. Then $$ a \equiv a \cdot a^{10 \cdot 2} = a^{1 + 10 \cdot 2} = a^{3\cdot7} \equiv b^{3\cdot7} = b^{1 + 10 \cdot 2} = b \cdot b^{10 \cdot 2} \equiv b \bmod 11 $$

Answer 2

If $11\mid ab$ we are done.

Say $11\not{\mid} ab$. Then by Fermat we have $$a^{10}\equiv b^{10}\equiv 1\pmod {11}$$

Since $$a^{9}\equiv b^{9}\pmod {11}$$ we have $$a^{10}\equiv ba^{9}\pmod {11}$$

so $$ 11\mid a^9(a-b)\implies 11\mid a-b$$

and we are done.