# Findind the value of $x^4 + 1/x^4$ when $x = 2+3\sqrt{3}$

This question recently came up in a competitive exam and I have been struggling to find out the easiest way to solve it.

Given that $x = 2+3\sqrt{3}$, what is the value of $x^4 + 1/x^4$.

Could someone provide me with an approach other than the direct approach of multiplying $x$ to itself $4$ times and adding it to its reciprocal.

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Hint: $\big(x+1/x\big)^2=x^2+1/x^2+2$ so $$x^2+1/x^2=(x+1/x)^2-2$$ but $$(x^2+1/x^2)^2=(x^4+1/x^4)-2$$ I think you can do the rest.

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One can both exploit the innate symmetry and preserve some of the innate structure to get a more concise result - see my answer. – Bill Dubuque Oct 13 '12 at 15:59
Helpful hint! +1 $\ddot\smile$ – amWhy Feb 28 '13 at 14:37

You the thing you need to calculate is $x^4$.

You start by isolating the square root $x = 2+3\sqrt{3} \Leftrightarrow x - 2 = 3\sqrt{3}$

Then you square both sides $(x - 2)^2 = (3\sqrt{3})^2 \Leftrightarrow x^2 - 4x + 4 = 27 \Leftrightarrow x^2 - 4x -23 = 0$

Let $P$ be $X^4$

Let $D$ be $X^2 - 4X - 23$

Now, you compute $R$, the remainder of the euclidean division of $P$ by $D$. I get $R = 248X+897$

If you name $Q$ the quotient, you have $P = D\times Q + R$. You don't need it but I get $X^2+4X+39$

But for your given $x$, $D$ evaluates to $0$ so $P$ evaluates to the same value as $R$. $( X^2+4X+39 ) \times (X^2 - 4X - 23)+ (248X+897) = X^4$

So $X^4 = 1393+744 \sqrt 3$

Then $X^4 + \frac{1}{X^4} = \frac{(X^4)^2+1}{X^4}$ and you finally get $\frac{2 \times (1800529+1036392 \sqrt 3))}{(1393+744\sqrt3)}$

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Hint $\$ Compute $\rm\: x^4 = (x^2)^2\! = \color{#C00}a+\color{#0A0}b\sqrt{d}\$ so $\rm\:\bar x^4 = a-b\sqrt{d}.\$ Let $\rm\,n = x\bar x.\:$ Therefore

$$\rm x^4 + \frac{1}{x^4}\, =\, x^4 + \frac{\bar x^4}{(x\bar x)^4}\, =\, x^4 + \frac{\bar x^4}{n^4 }\, =\, (\color{#C00}a(n^4\!+1) +\, \color{#0A0}b(n^4\!-1)\sqrt{d})/n^4$$

Here $\rm\: x^4 = (x^2)^2\! = (31+12\sqrt{3})^2\! = 1393+744\sqrt{3}\$ so $\rm\:\color{#C00}a=1393,\ \color{#0A0}b = 744,\:$ and $\rm\,n = x\bar x = -23.$

Remark $\$ As I mentioned in the closely related question, one can also employ (Newton) identities for symmetric functions (see also Babak's answer). However, in this example, that method, done rotely, will require doing more (large) integer arithmetic, and it will result in a less concise answer, since it does not preserve the $\rm\:n^4\!\pm 1\:$ structure in the result.

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