Parity of a Cubed Minus a Plus One

When working with parity problems in HSC Mathematics Extension 1, a common technique is to factor algebraic expressions to reveal the product of consecutive integers, which guarantees an even factor. In this problem, we explore two distinct proof methods for showing that $a^3 - a + 1$ is always odd, giving you practice in both algebraic manipulation and proof by cases.

Problem Statement

Prove that the expression $a^3 - a + 1$ is odd for all positive integer values of $a$.

---

Hints

Consider factoring $a^3 - a$ as a product of consecutive integers. What can you say about the parity of any product of consecutive integers?

Alternatively, try proof by cases: analyze when $a$ is even and when $a$ is odd separately.

---

Solutions

Method 1 (Factorization - Elegant):

We begin by factoring $a^3 - a$ to see its structure. Factor the expression:

$$\begin{aligned} a^3 - a + 1 &= a(a^2 - 1) + 1 \\ &= a(a-1)(a+1) + 1 \\ &= (a-1)a(a+1) + 1 \end{aligned}$$

The product $(a-1)a(a+1)$ is the product of three consecutive integers. In any set of consecutive integers, at least one must be even, so the product is even. Let $(a-1)a(a+1) = 2k$ for some integer $k$.

Therefore, $a^3 - a + 1 = 2k + 1$, which is odd by definition.

Method 2 (Cases):

Proof by cases: we consider the two possibilities for the parity of $a$.

Case 1: If $a = 2m$ (even), then

$$a^3 - a + 1 = (2m)^3 - 2m + 1 = 8m^3 - 2m + 1 = 2(4m^3 - m) + 1,$$

which is odd.

Case 2: If $a = 2m+1$ (odd), then we substitute and expand:

$$\begin{aligned} a^3 - a + 1 &= (2m+1)^3 - (2m+1) + 1 \\ &= (8m^3 + 12m^2 + 6m + 1) - 2m - 1 + 1 \\ &= 8m^3 + 12m^2 + 4m + 1 \\ &= 2(4m^3 + 6m^2 + 2m) + 1, \end{aligned}$$

which is also odd.

In both cases, the expression is odd. $\blacksquare$

---

Takeaways

• Factorising $a^3 - a$ as $(a-1)a(a+1)$ reveals the product of three consecutive integers, which is always even – a powerful shortcut for parity proofs.
• Proof by cases is a robust alternative: handling even and odd $a$ separately confirms the result without relying on factorisation.
• In general, any product of consecutive integers contains an even factor, so expressions like $n(n+1)$ or $(n-1)n(n+1)$ are always even.

---

Further Readings

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