Try to solve this

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

#Algebra #NumberTheory

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Math	Appears 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

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

Pi Han Goh - 5 years, 6 months ago

What about $n = 1$ ?

Calvin Lin Staff - 5 years, 6 months ago

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

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – 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).

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – 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.

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – 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 $c = 0$ .

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – HA! That's a nice interpretation. ThankYou

Pi Han Goh - 5 years, 6 months ago

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

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

Calvin Lin Staff - 5 years, 6 months ago

Math	Appears 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}`	$\boxed{123}$