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Prove that $$\lfloor$$$$(5 + 2\sqrt6)^{n}$$$$\rfloor$$ is Odd for $$n\in N$$.

2 years, 2 months ago

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- 2 years, 2 months ago

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

- 2 years, 1 month ago

- 2 years, 1 month ago

Comment deleted May 20, 2016

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.

- 2 years, 1 month ago

See Newton's Identities

Staff - 2 years, 1 month ago