×

# Prove that $$x^{n}$$ + $$y^{n}$$ is divisible by x+y if n is odd

Can anybody prove it.Please.

Note by Ayush Choubey
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

Hi,@Ayush Choubey If n is odd, then $x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-....+y^{n-1})$ which can be proved easily by expanding or using induction.

- 2 years, 6 months ago

Log in to reply

Let's denote $$p(x) = x^{n}+y^{n}$$. Due to remainder theorem (little Bezout's theorem) remainder will be $$p(-y)= (-y) ^{n} + y^{n}=0$$, because n is odd. Therefore the remainder is 0, so p(x) is divisible by x+y.

- 3 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...