Parity of a Cubed Minus a Plus One

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):

Factor the expression:

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):

Case 1: If $a = 2m$ (even), then $a^3 - a + 1 = 8m^3 - 2m + 1 = 2(4m^3 - m) + 1$ (odd).

Case 2: If $a = 2m+1$ (odd), then expanding shows $a^3 - a + 1 = 2(\text{integer}) + 1$ (odd).

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

Takeaways

• Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
• Check edge cases and verify where each assumption is used in the argument.

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 Sequences: https://vumaths.com/booklets/hsc-sequences/
• HSC Collections: https://vumaths.com/booklets/hsc-collections/

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