@Pi Han Goh
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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 \( c = 0 \).

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## 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.Log in to reply

What about \( n = 1 \)?

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

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

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

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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 \( c = 0 \).

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Approach 1:Simplify the problem by setting \( c = 0 \).Approach 2:Use Quadratic Diophantine Equations - Pell's Equation.Log in to reply