Odd Squares and Divisibility by 8

#hsc-maths#proofs#medium

This problem explores a classic number theory result: the square of any odd integer, when reduced by 1, is always a multiple of 8. For HSC Mathematics Extension 1 students, it’s a great exercise in algebraic manipulation and logical reasoning, reinforcing how to represent odd numbers and use properties of consecutive integers. By working through this proof, you’ll gain confidence in constructing direct proofs and spotting factor tricks that simplify divisibility arguments.

Problem Statement

Prove that if aa is any odd integer, then a21a^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 aa as an odd integer

To start the proof, we express the odd integer aa using its definition. Since aa is odd, we can write:

a=2k+1a = 2k + 1

for some integer kk.

Step 2: Expand a21a^2 - 1

Now we substitute this expression into a21a^2 - 1 and simplify.

a21=(2k+1)21=4k2+4k+11=4k2+4k=4k(k+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)k(k+1)

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

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

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

Step 4: Substitute and conclude

Finally, we substitute this factorization back into our expression for a21a^2-1.

a21=4k(k+1)=42m=8m\begin{aligned} a^2 - 1 &= 4 \cdot k(k+1) \\ &= 4 \cdot 2m \\ &= 8m \end{aligned}

Since a21=8ma^2 - 1 = 8m, we conclude that a21a^2 - 1 is divisible by 8.


Takeaways

  • Odd Integer Form: Any odd integer can be written as 2k+12k+1 for some integer kk
  • Consecutive Integer Property: Product of consecutive integers k(k+1)k(k+1) is always even (one must be even)
  • Factor Extraction: From 4k(k+1)=42m=8m4k(k+1) = 4 \cdot 2m = 8m, directly see divisibility by 8
  • Alternative View: Can also factor a21=(a1)(a+1)a^2-1 = (a-1)(a+1), both even for odd aa, 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, HSC Last Resorts, HSC Mechanics

Written by Vu Hung Nguyen

Mathematics Educator · LinkedIn · About