Odd Squares and Divisibility by 8

Problem Statement

Prove that if $a$ is any odd integer, then $a^2 - 1$ is divisible by 8.

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

Direct Proof

Step 1: Express $a$ as an odd integer

Since $a$ is odd, we can write:

a = 2k + 1

for some integer $k$ .

Step 2: Expand $a^2 - 1$

\begin{aligned} a^2 - 1 &= (2k + 1)^2 - 1 \\ &= 4k^2 + 4k + 1 - 1 \\ &= 4k^2 + 4k \\ &= 4k(k + 1) \end{aligned}

Step 3: Analyze $k(k+1)$

Note that $k$ and $(k+1)$ are consecutive integers.

Therefore, one of them must be even, which means their product $k(k+1)$ is divisible by 2.

Write $k(k+1) = 2m$ for some integer $m$ .

Step 4: Substitute and conclude

\begin{aligned} a^2 - 1 &= 4 \cdot k(k+1) \\ &= 4 \cdot 2m \\ &= 8m \end{aligned}

Since $a^2 - 1 = 8m$ , we conclude that $a^2 - 1$ is divisible by 8.

\hfill $\square$

Takeaways

Odd Integer Form: Any odd integer can be written as $2k+1$ for some integer $k$
Consecutive Integer Property: Product of consecutive integers $k(k+1)$ is always even (one must be even)
Factor Extraction: From $4k(k+1) = 4 \cdot 2m = 8m$ , directly see divisibility by 8
Alternative View: Can also factor $a^2-1 = (a-1)(a+1)$ , both even for odd $a$ , with one divisible by 4

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 Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/
HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/

Connect with me

If you're eager for more HSC Maths insights, be sure to check out my Website - Vu's Maths Hub. For deeper dives and regular tips, join my LinkedIn. Let's tackle these challenging math problems together! You can also catch my daily math content on Instagram.