- Published on
Power Difference Divisible by x Minus y
- Authors

- Name
- Vu Hung
Problem Statement
Prove by induction that is divisible by for all positive integers , where are integers with .
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
Proof by Mathematical Induction
Base Case ():
For :
Since is divisible by , the statement holds for . \checkmark
Inductive Hypothesis:
Assume the statement is true for , where is a positive integer:
This means we can write:
for some integer .
Inductive Step:
We must prove the statement for :
Start with and manipulate to introduce :
Substitute the inductive hypothesis :
Since are all integers, is an integer.
Therefore:
This shows is divisible by . \checkmark
Conclusion:
By mathematical induction, is divisible by for all positive integers .
\hfill
\noindentNote: This problem demonstrates the crossover between Induction and Proofs topics in HSC Extension 2.
Takeaways
- Strategic Addition/Subtraction: Add and subtract to create terms involving and
- Factorization: Extract common factor after rearrangement
- Induction Crossover: Problem appears in both HSC-Induction and HSC-Proofs collections, showing technique overlap
- Alternative Formula: Result leads to
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 Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
- HSC Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/
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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 Substack. Let's tackle these challenging math problems together! You can also catch my daily math content on Instagram.
