Divisibility by 6: Criterion and Consequence

Problem Statement

For $n \in \mathbb{Z}$, prove that:

• $n$ is divisible by 6 if and only if $n$ is divisible by 2 and 3.
• $n^3 - n$ is divisible by 6.

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

Solutions

Part (a): Biconditional Proof

Forward ($\implies$): If $6 \mid n$, then $2 \mid n$ and $3 \mid n$.

Proof:

Assume $6 \mid n$. Then $n = 6k$ for some integer $k$.

Write $n = 6k = 2(3k)$. Since $3k$ is an integer, $2 \mid n$.

Write $n = 6k = 3(2k)$. Since $2k$ is an integer, $3 \mid n$. \blacksquare

Reverse ($\impliedby$): If $2 \mid n$ and $3 \mid n$, then $6 \mid n$.

Proof:

Assume $2 \mid n$ and $3 \mid n$.

Then $n = 2a$ and $n = 3b$ for some integers $a, b$.

From $n = 2a$, $n$ is even. From $n = 3b$, we have $2a = 3b$.

Since LHS is even, RHS $3b$ must be even. Since 3 is odd, $b$ must be even.

Write $b = 2c$ for some integer $c$.

Then:

Since $c$ is an integer, $6 \mid n$. \blacksquare

Part (b): Using Part (a)

Goal: Prove $6 \mid (n^3 - n)$.

By part (a), sufficient to show $2 \mid (n^3 - n)$ and $3 \mid (n^3 - n)$.

Step 1: Factor

This is the product of three consecutive integers.

Step 2: Prove $2 \mid (n^3 - n)$

Among any three consecutive integers, at least one is even.

Therefore, $(n-1)n(n+1)$ contains an even factor, so $2 \mid (n^3 - n)$. \checkmark

Step 3: Prove $3 \mid (n^3 - n)$

Among any three consecutive integers, exactly one is divisible by 3.

Therefore, $(n-1)n(n+1)$ contains a factor divisible by 3, so $3 \mid (n^3 - n)$. \checkmark

Conclusion

Since $2 \mid (n^3 - n)$ and $3 \mid (n^3 - n)$, and $\gcd(2,3) = 1$, by part (a):

\hfill $\square$

Takeaways

• Coprime Divisibility: If $\gcd(a,b) = 1$ and both $a \mid n$ and $b \mid n$, then $ab \mid n$
• Consecutive Integer Properties: Among $k$ consecutive integers, exactly one is divisible by $k$
• Multi-part Strategy: Part (b) leverages part (a) to simplify proof (check divisibility by 2 and 3 separately)
• Factorization: $n^3 - n = (n-1)n(n+1)$ reveals structure as consecutive integer product

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Combinatorics: https://vumaths.com/booklets/hsc-combinatorics/
• HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/

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