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I already have two solutions for this problem, it is for high school students with an advanced level. I would like to know if there are better or more creative approaches on the problem. Here are my solutions:

  1. (1st solution): Notice that $$x^{2} = (\frac{\sqrt{111}-1}{2})^{2} = 28 - \frac{\sqrt{111}}{2}$$ $$x^{3} = x \cdot x^{2} = \left( \frac{\sqrt{111}}{2} - \frac{1}{2} \right) \left( 28 - \frac{\sqrt{111}}{2} \right) = 14 \sqrt{111} - 111/4 - 14 + \frac{\sqrt{111}}{4} = \frac{57 \sqrt{111}}{4} - \frac{167}{4} $$ $$x^{4} = (x^{2})^{2} = \frac{ 111 + 4 \cdot 784}{4} - 28 \sqrt{111} $$ $$x^{5} = x^{4} \cdot x = \left( \frac{ 111 + 4 \cdot 784}{4} - 28 \sqrt{111} \right)\left(\frac{\sqrt{111}-1}{2} \right) = \left( \frac{ 3247 }{4} - 28 \sqrt{111} \right) \left(\frac{\sqrt{111}-1}{2} \right) $$ $$ = \sqrt{111}\frac{3359}{8} - \frac{15679}{8} $$

So we have $$ 2x^{5} + 2x^{4} - 53x^{3} - 57x + 54 = $$ $$(\sqrt{111}\frac{3359}{4} - \frac{15679}{4}) + (\frac{ 111 + 4 \cdot 784}{2} - 56 \sqrt{111}) - 53 (\frac{57 \sqrt{111}}{4} - \frac{167}{4}) - 57 (\frac{\sqrt{111}-1}{2}) + 54 $$ $$ = \sqrt{111} (\frac{3359}{4} - \frac{3359}{4}) - \frac{15679}{4} + \frac{ 111 + 4 \cdot 784}{2} + \frac{53 \cdot 167}{4} + \frac{57}{2} + 54 $$ $$ = - \frac{15679}{4} + \frac{ 222 + 8 \cdot 784}{4} + \frac{53 \cdot 167}{4} + \frac{114}{4} + \frac{216}{4} $$ $$ = - \frac{15679}{4} + \frac{ 5600 + 894 }{4} + \frac{5300 + 3180 + 371 }{4} + \frac{114}{4} + \frac{216}{4} $$ $$ = - \frac{15679}{4} + \frac{ 6494 }{4} + \frac{8851}{4} + \frac{114}{4} + \frac{216}{4} $$ $$ = -4/4 = -1, $$ and the answer is $ (-1)^{2004} = 1.$ The above solution requires quite tedious calculation. Below is an alternative solution.

  1. (2nd solution): Notice that $x = \frac{\sqrt{111}-1}{2}$ is equivalent with $$ (2x + 1)^{2} = 111$$ $$ 4x^{2} + 4x + 1 = 111 $$ $$ 4x^{2} + 4x - 110 = 0$$ $$ (2x^{2} + 2x - 55) = 0 \:\: ........ \:\: (1)$$ Multiply $(1)$ with $x^{3}$ we get $$ (2x^{5} + 2x^{4} - 55x^{3}) = 0 $$ Multiply $(1)$ with $x$ we get $$ (2x^{3} + 2x^{2} - 55x) = 0 $$ Sum both of them and we get: $$ 2x^{5} + 2x^{4} - 53 x^{3} + 2x^{2} - 55x = 0 \:\: ........ \:\: (2)$$ and now we have the 1st 3 terms of the form that we want to calculate. Substract $(2)$ with $(1)$ to get: $$ 2x^{5} + 2x^{4} - 53 x^{3} - 57x + 55 = 0 $$ $$ 2x^{5} + 2x^{4} - 53 x^{3} - 57x + 54 = -1$$ So the answer is $(-1)^{2004} = 1.$
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    $\begingroup$ Your second solution is probably the best one there is. $\endgroup$ – TheSimpliFire May 18 '19 at 12:56
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    $\begingroup$ Forget the brute-force first solution... $\endgroup$ – Jean Marie May 18 '19 at 13:01
  • $\begingroup$ @JeanMarie I upvote although the 1st solution was not really that tedious. The 2nd solution does not easily come to mind, I peeked into the solution sheet. $\endgroup$ – Arief May 18 '19 at 14:25
  • $\begingroup$ No real number other than $0,1,-1$ has a neat form when raised to the power of $2004$. A ‘sneaky’ but smart student would use estimations on $x$ to guess between $0$ and $1$. $\endgroup$ – Szeto May 21 '19 at 23:32
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Need long division

Divide $$2x^5+2x^4-53x^3-57x+54$$ by $2x^2+2x-55$ to express

$$2x^5+2x^4-53x^3-57x+54=q(x)\cdot(2x^2+2x-5)+r(x)$$ where $q(x)$ is the quotient and $r(x)$ is the remainder.

$\implies2x^5+2x^4-53x^3-57x+54=r(x)$ as $2x^2+2x-5=0$

Here $r(x)=-1$

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  • $\begingroup$ you mean remainder $\endgroup$ – user10354138 May 18 '19 at 13:20
  • $\begingroup$ Correct, mathworld.wolfram.com/Residue.html $\endgroup$ – lab bhattacharjee May 18 '19 at 13:23
  • $\begingroup$ How do you know that $2x^{2} + 2x - 55$ should be the divisor? $\endgroup$ – Arief May 18 '19 at 14:15
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    $\begingroup$ @AriefAnbiya, My idea was to find $$2x^5+2x^4-53x^3-57x+54$$ as $$(2x^2+2x-55)q(x)+r(x)=r(x)$$ as $2x^2+2x-55=0$ $\endgroup$ – lab bhattacharjee May 18 '19 at 14:17
  • $\begingroup$ Well, it is shorter but it is not different than the 2nd solution.., $q(x) = x^{3} + x - 1$. $\endgroup$ – Arief May 18 '19 at 14:27
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What I did was equivalent to division, or to your second solution, but went as follows.

First I got to $2x^2+2x=55$ and multiplied this by $x^3$ so that $2x^5+2x^4=55x^3$

Now I substituted this into: $$p(x)=2x^5+2x^4-53x^3-57x+55=55x^3-53x^3-57x+54=2x^3-57x+54$$

Next I used $55x=2x^3+2x^2$ to give $$p(x)=2x^3-55x-2x+54=2x^3-2x^3-2x^2-2x+54=-2x^2-2x+54=-55+54=-1$$


I find it is sometimes handy to remember that if I am working with an $x$ such that $p(x)=q(x)$, I can substitute $p(x)$ and $q(x)$ in any expression involving $x$. Formally this works the same as polynomial division by $p(x)-q(x)$. However it is sometimes more flexible and easier to apply.

It can turn out to require more lengthy calculations than a more regular method, but it has the advantage of keeping the expressions simple at each stage. If I can't see how to make more progress, I can always resort to division, but I can start with any simpler expression I've derived.

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  • $\begingroup$ I upvote. Although still only a bit different with the 2nd solution.. $\endgroup$ – Arief May 18 '19 at 14:46

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