## Consider the infinite polynomial series $A_G(x) = x G_1 + x^2 G_2 + x^3 G_3 + \cdots$, where $G_k$ is the $k$th term of the second order recurrence relation $G_k = G_{k-1} + G_{k-2}$, $G_1 = 1$ and $G_2 = 4$; that is, $1, 4, 5, 9, 14, 23, \dots$.

For this problem we shall be concerned with values of $x$ for which $A_G(x)$ is a positive integer.

The corresponding values of $x$ for the first five natural numbers are shown below.

$x$$A_G(x)$

$\frac{\sqrt{5}-1}{4}$$1$

$\tfrac{2}{5}$$2$

$\frac{\sqrt{22}-2}{6}$$3$

$\frac{\sqrt{137}-5}{14}$$4$

$\tfrac{1}{2}$$5$

We shall call $A_G(x)$ a golden nugget if $x$ is rational, because they become increasingly rarer; for example, the $20$th golden nugget is $211345365$.

Find the sum of the first thirty golden nuggets.

### Let’s consider that infinite polynomial,

$A_G(x) = xG_1 + x^2G_2 + x^3G_3 + \cdots$.

We also know that $G_k = G_{k-1} + G_{k-2}$, where $G_1 = 1$ and $G_2 = 4$.

Let’s manipulate $A(x)$ with that relation.

$xA_G(x) = x^2G_1 + x^3G_2 + x^4G_3 + \cdots = 4x^2 + x^3 + 4x^4 + 5x^5 + \cdots$,

$x^2A_G(x) = x^3G_1 + x^4G_2 + x^5G_3 + \cdots = x^3 + 4x^4 + 5x^5 + 9x^6 + \cdots$.

Subtract both equations (and use $G_1 + G_2 = 5$):

$xA_G(x) – x^2A_G(x) = 4x^2 + x^3 + \cdots – x^3 – 4x^4 – \cdots = 4x^2 – x^4 + \cdots = x^2(4 – x^2 + \cdots)= x^2(4-A_G(x^2))$,

$xA_G(x)(1 – x) = 4x^2 – A_G(x^2)x^4$,

$A_G(x) = \frac{4x^2}{1 + x^3}$.

Golden nuggets are when this $A_G(x)$, where $x$ is rational, results in a natural number. Hence, we need to find all rational roots ($x$) such that $A_G(x)$ is a natural number.

Rewrite $A_G(x)$ as,

$N = A_G(x) = \frac{4x^2}{1 + x^3}$,

$x^3 + 1 = \frac{4x^2}{N}$,

$x^3 – 4Nx^2 + 1 = 0$.

The discriminant for this cubic equation must be a perfect square for there to be rational roots. Given the roots are $x$, $4N\pm \sqrt{D}$, we need $D$ to be a perfect square.

So now we focus on:

$D = (4N)^2 – 4(1)$,

$D = 16N^2 – 4$.

To be a perfect square, that is to fit the formula $(aN+b)^2$,

$16N^2 – 4 = 16N^2 – 16bN + b^2$,

$bN = b^2 + 1$,

$b(N-b) = 1$.

Notice that the last equation and both $N$and $b$ are positive integers, so $N$ and $b$ are pairs of consecutive numbers.

Hence, by selecting values $b = 1, 2, 3, \cdots, 29, 30$($30$ pairs), we have correspondingly $N = 2, 3, \cdots, 30, 31$ as $N$ are the golden nuggets,

The sum of these golden nuggets is $(2+3+\cdots+30+31)=\frac{(2+31)*30}{2}=495$.

So the sum of the first thirty golden nuggets is $495$.

##### More Answers:

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