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Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

What I Tried: We have :- $$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$ $$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$ $$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$ $$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$ $$= (250\sqrt{2009} - 1259)^5$$

However, the answer given is $32$, so there could have been more simplifications.
As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?

anantagni
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Anonymous
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3 Answers3

32

$a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$ That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe $$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$

VIVID
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Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

I see no reason for elegance. Since one of the factors in the numerator is $1$, computing $a^3$ is is straightforward.

$$a^3 = \left(\frac{1}{8}\right) \times \left[ 1 + 3\sqrt{2009} + 3(2009) + 2009\sqrt{2009} \right]$$ $$=~ \left(\frac{1}{8}\right) \times \left[6028 + 2012\sqrt{2009}\right] ~=~ \frac{1507 + 503\sqrt{2009}}{2}. $$

Therefore,

$$(a^3 - 503a - 500)$$

$$=~ \frac{1507 + 503\sqrt{2009}}{2} ~-~ \frac{503 + 503\sqrt{2009}}{2} - \frac{1000}{2} ~=~ \frac{4}{2} \implies $$

$$(a^3 - 503a - 500)^5 = 2^5 = 32.$$

user2661923
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2

Your working leads to the answer. Here is the correction -

$a = \frac{1 + \sqrt{2009}}{2}$

$a^2 - 3 = \frac{999 + \sqrt{2009}}{2}$

$a (a^2 - 3) = 752 + 250 \sqrt {2009}$

$500 (a + 1) = 250 ( \sqrt {2009} + 3)$

$a(a^2-3) - 500(a+1) = 2$

So $(a^3 - 503a - 500)^5 = 32$

Math Lover
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