Proof verification: If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$

Question

Would someone be willing to verify the following proof?

Theorem: Suppose $a, b \in \mathbb{Z}; n \in \mathbb{N}$. If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$.

Proof:

$a \equiv b (mod$ $n) \rightarrow xn = a - b; x \in \mathbb{Z}$

$\rightarrow a = xn + b$

Then, $a^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) + b^3$.

This gives us $a^3 - b^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) = n(x^3n^2 + 3bx^2n + 3b^2x)$

$n | n(x^3n^2 + 3bx^2n + 3b^2x) \rightarrow n | (a^3 - b^3)$

Therefore $a^3 \equiv b^3 (mod$ $n)$.

Answer 1

Your proof is correct.

You could have stopped at $a^3=b^3+kn$, it doesn't matter whether $k$ has a complicated expression, if it's an integer, then $a^3\equiv b^3\pmod n$.

However, notice that you can factorize $xn$ actually, and that $xn=a-b$.

Thus $a^3-b^3\equiv(a-b)(a^2+ab+b^2)\equiv 0\pmod n$ since $(a-b)\equiv 0\pmod n$.

You are not forced to get back to definition everytime, try to use modular calculus directly as well, it is powerful.

Answer 2

The posted proof is correct. It might be even easier, however, to first prove the more general:

$$ a \equiv b \pmod{n} \quad\text{and}\quad a' \equiv b' \pmod{n} \quad \implies \quad a \cdot a' \equiv b \cdot b' \pmod{n}\quad $$

The proof goes the same: $a\cdot a' = (xn+b)\cdot(x'n+b')=b\cdot b'+n\cdot (x\cdot b'+ x' \cdot b+n\cdot x\cdot x')\,$.

Then, using the proposition with $a'=a, b'=b$ gives $a \equiv b \implies a^2 \equiv b^2 \implies a^3 \equiv b^3 \pmod{n}$.