Odd Squares and Divisibility by 8
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 is any odd integer, then 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 as an odd integer
To start the proof, we express the odd integer using its definition. Since is odd, we can write:
for some integer .
Step 2: Expand
Now we substitute this expression into and simplify.
Step 3: Analyze
Note that and are consecutive integers.
Therefore, one of them must be even, which means their product is divisible by 2.
Write for some integer .
Step 4: Substitute and conclude
Finally, we substitute this factorization back into our expression for .
Since , we conclude that is divisible by 8.
Takeaways
- Odd Integer Form: Any odd integer can be written as for some integer
- Consecutive Integer Property: Product of consecutive integers is always even (one must be even)
- Factor Extraction: From , directly see divisibility by 8
- Alternative View: Can also factor , both even for odd , 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