Goldenness of an integer

Pyth - 24 23 bytes

There has to be a better way to find the divisors...

ho.a-.n3cFNfsIeTm,dcQdS

Test Suite.

Matlab, 96 81 bytes

Golfed (-15bytes), props to Luis Mendo

function w(n);a=find(~(mod(n,1:n)));[~,c]=min(abs(a./(n./a)-1.618));[a(c) n/a(c)]

Original:

function w(n)
a=find(not(mod(n,1:n)));b=abs(a./(n./a)-1.618);c=find(not(b-min(b)));[a(c) n/a(c)]

This is by far not a great solution, but my first attempt at code-golf. What fun!

Jelly, 16 15 14 bytes

Saved 1 byte thanks to @miles.

÷/ạØp
ÆDżṚ$ÇÞḢ

Try it online!

Explanation

÷/ạØp         Helper link, calculates abs(a/b - phi). Argument: [a, b]
÷/            Reduce by division to calculate a/b.
  ạØp         Calculate abs(a/b - phi).

ÆDżṚ$ÇÞḢ      Main link. Argument: n
ÆD            Get divisors of n.
  żṚ$         Pair the items of the list with those of its reverse. The reversed
              divisors of a number is the same list as the number divided by each
              of the divisors.
     ÇÞ       Sort by the output of the helper link of each pair.
       Ḣ      Get the first element [a, b] and implicitly print.