@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
·
12 months ago

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@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
·
12 months ago

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@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
·
12 months ago

## Comments

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TopNewestHint: Show that for positive even integer \(n\), \( (\sqrt 3 + \sqrt 2)^n + (\sqrt3 - \sqrt2)^n \) is always an integer. – Pi Han Goh · 12 months agoLog in to reply

– Calvin Lin Staff · 12 months ago

What about \( n = 1 \)?Log in to reply

– Pi Han Goh · 12 months ago

Ahahaha, you've found a loophole. Let me fix it.Log in to reply

Essentially what you are doing is my approach 1 but with setting \( a = 0 \) (which complicates it just slightly). – Calvin Lin Staff · 12 months ago

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Hmm... I don't see how mine is slightly harder than Approach 1, let me put my thinking cap on. – Pi Han Goh · 12 months ago

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\( ( 1 + \sqrt{2} ) ^n + ( 1 - \sqrt{2} ) ^ n \) 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 · 12 months ago

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– Pi Han Goh · 12 months ago

HA! That's a nice interpretation. ThankYouLog in to reply

Approach 1:Simplify the problem by setting \( c = 0 \).Approach 2:Use Quadratic Diophantine Equations - Pell's Equation. – Calvin Lin Staff · 12 months agoLog in to reply