Question

Prove the following statements using either direct or contrapositive proof. Let \(a, b \in \mathbb{Z}\) and \(n \in \mathbb{N}\). If \(a \equiv b(\bmod n)\), then \(a^{3} \equiv b^{3}(\bmod n)\).

Short answer

We proved that if \(a \equiv b (\bmod n)\), then \(a^{3} \equiv b^{3} (\bmod n)\) holds true by showing that \(a^{3}\) could be expressed as an addition of \(b^{3}\) and terms which are all equivalent to 0 modulo \(n\).

Step-by-step solution

Apply the definition of modular equivalence

The given condition is \(a \equiv b (\bmod n)\). By definition of modular equivalence, we know that for some integer \(k\), we have \(a = b + nk\).

Begin the direct proof

Take the expression for \(a\) and cube it, giving: \(a^{3} = (b+nk)^{3}\). This simplifies to: \(a^{3} = b^{3} + 3b^{2}nk + 3bn^{2}k^{2} + n^{3}k^{3}\).

Show that \(a^{3} \equiv b^{3} (\bmod n)\)

Observe that each term except \(b^{3}\) in \(a^{3} = b^{3} + 3b^{2}nk + 3bn^{2}k^{2} + n^{3}k^{3}\) has \(n\) as a factor, meaning each of those terms is equivalent to 0 modulo \(n\). Therefore, we can say: \(a^{3} \equiv b^{3} (\bmod n)\), as required.