×

# Try to solve this

Prove that always exist integer number a,b,c such as $0< \left | a+b\sqrt{2} +c\sqrt{3}\right |< \frac{1}{1000}$

Note by Ms Ht
2 years 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:

Hint: Show that for positive even integer $$n$$, $$(\sqrt 3 + \sqrt 2)^n + (\sqrt3 - \sqrt2)^n$$ is always an integer.

- 2 years ago

What about $$n = 1$$?

Staff - 2 years ago

Ahahaha, you've found a loophole. Let me fix it.

- 2 years ago

Nope, not fixed yet. It's not true for odd integers.

Essentially what you are doing is my approach 1 but with setting $$a = 0$$ (which complicates it just slightly).

Staff - 2 years ago

Awhhh!! I should have known! Fixed again~

Hmm... I don't see how mine is slightly harder than Approach 1, let me put my thinking cap on.

- 2 years ago

The slightly harder part being why we have to use $$2n$$ instead.

$$( 1 + \sqrt{2} ) ^n + ( 1 - \sqrt{2} ) ^ n$$ is an integer for all $$n$$, because odd powers of 1 is still an integer. Applying your approach results in $$c = 0$$.

Staff - 2 years ago

HA! That's a nice interpretation. ThankYou

- 2 years ago

Approach 1: Simplify the problem by setting $$c = 0$$.

Approach 2: Use Quadratic Diophantine Equations - Pell's Equation.

Staff - 2 years ago