Thing practiced: math
Tools used: A Friendly Introduction to Number Theory, Python
—no one ever
AF(x)=xF1+x2F2+x3F3+…Consider the infinite polynomial series
where Fk is the kth term in the Fibonacci sequence (1, 1, 2, 3, 5, 8, ...).
For every case where x is rational and AF(x) is a positive integer, we call AF(x) a golden nugget, because these are increasingly rare – for example, the 10th golden nugget is 74,049,690.
What is the 15th golden nugget?
Solution:
Infinite series are unwieldy, so first we replace (1) with a closed-form version (derived here):
AF(x)=x1−x−x2
We need AF(x) to be a positive integer, which we’ll call n. So, rearranging (2), we get:
AF(x)=n=x1−x−x2 nx2+(n+1)x−n=0
This is a standard quadratic equation with solutions
x=−(n+1)±√(n+1)2+4n22n
As long as the square root portion of (4) is an integer, x will be rational. We can solve for this case by calling this portion m (making the discriminant m2):
m2=(n+1)2+4n2=5n2+2n+1
and doing some algebra we get:
5m2=25n2+10n+1+4=(5n+1)2+4 (5n+1)2−5m2=−4
Now, if we substitute p = 5n + 1, then (5) becomes a Pell-like equation:
p2−5m2=−4
This means that we can find solutions to (6) by finding solutions to the corresponding unit-form Pell equation x2 - 5y2 = 1 and transforming them.
To compute the solutions, we use the technique described here. We first identify the fundamental solutions to (6). Then, we transform these into solutions of the unit-form equation by the identity (pi2 - 5mi2)(xi2 - 5yi2) = -4. From these, we can generate subsequent solutions, then transform them back into solutions to (6) by the same identity. Finally, we re-substitute our original n = AF(x) back in to find the golden nuggets.
This gives us the answer: 1,120,149,658,760.