×

# Note 4

Assume that $$a|n$$ and $$b|n$$, and there exist integers $$x$$ and $$y$$ that $$ax+by=1$$.

Prove: $$ab|n$$

Note by Jason Snow
2 years, 1 month 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} $$\boxed{123}$$

Sort by:

By Bézout's identity, if there exists integer $$(x,y)$$ such that it satisfy $$ax+by = 1$$, then $$(a,b)$$ shall have the GCD as 1 (i.e. A and B is coprime)

If $$a|n$$ and $$b|n$$, then for nonzero integer $$i, j$$, it will be $$n = ai$$ and $$n = bj$$ or $$n^2 = abij$$, which means

$n^2\times\frac{1}{ab} = ij$

Which proves that $$ab|n$$, by closed-form of integer multiplication.

- 2 years, 1 month ago