@Aditya Chauhan
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Take \( K = (5 + 2 \sqrt{6})^n \), and \( I = ( 5 - 2 \sqrt{6})^n \). Now, \( I < 1 \). (Easy to prove.)

Now look at \( K + I \). All the odd terms which are irrational cancel out, leaving all the integral terms. Also, each integral term comes twice, therefore, it is even. So \( K + I = 2m \implies K = 2m - I > 2m - 1. \).

Therefore \( \lfloor K \rfloor = 2m - 1 \), which is odd.

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TopNewest@Calvin Lin @Sandeep Bhardwaj @Satyajit Mohanty

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Hint: Consider the roots of the equation \(x^2-10x+1 = 0 \). Show that the sum of powers its roots are always even.

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@Aditya Chauhan By Newton's Identities!

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Comment deleted May 20, 2016

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Now look at \( K + I \). All the odd terms which are irrational cancel out, leaving all the integral terms. Also, each integral term comes twice, therefore, it is even. So \( K + I = 2m \implies K = 2m - I > 2m - 1. \).

Therefore \( \lfloor K \rfloor = 2m - 1 \), which is odd.

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Newton's Identities

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