Difference of Powers

For integers \(a, b, m, n\), prove that \[gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = {a}^{gcd(m,n)} - {b}^{gcd(m,n)}.\]

Solution

Let d=gcd(m,n)d = gcd(m,n), thus d∣md|m and d∣nd|n. We then let m=pdm=pd and n=pdn=pd.

Hence gcd(am−bm,an−bn)=gcd(apd−bpd,aqd−bqd).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({a}^{pd} - {b}^{pd},{a}^{qd} - {b}^{qd}).

Let A=adA ={a}^{d} and B=bdB = {b}^{d}, it follows that

gcd(am−bm,an−bn)=gcd(Ap−Bp,Aq−Bq).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}).

Since Ap−BpA−B=∑k=0p−1Ap−k−1Bk\frac{{A}^{p} - {B}^{p}}{A-B} = \sum _{ k=0 }^{ p-1 }{ { A }^{ p-k-1 }{ B }^{ k } } and Aq−BqA−B=∑k=0q−1Aq−k−1Bk\frac{{A}^{q} - {B}^{q}}{A-B} = \sum _{ k=0 }^{ q-1 }{ { A }^{ q-k-1 }{ B }^{ k } } we get

gcd(Ap−Bp,Aq−Bq)=A−B.gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}) = A-B.

Since A−B=ad−bdA-B = {a}^{d} - {b}^{d} and d=gcd(m,n)d =gcd(m,n), we prove that

gcd(am−bm,an−bn)=agcd(m,n)−bgcd(m,n).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = {a}^{gcd(m,n)} - {b}^{gcd(m,n)}.

Check out my other notes at Proof, Disproof, and Derivation

#NumberTheory #GreatestCommonDivisor(GCD/HCF/GCF)

Note by Steven Zheng
6 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Could you please provide a proof for the sum you have claimed to be true? Thanks

Michael Stevenson - 5 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...