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#Python 3, 96 bytes

Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

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The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

#Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 
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#Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answerthis SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

#Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

#Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 
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#Python 3, 96 bytes

Pretty simple solution. Makes use of this SO answer.

lambda n:min([((i,n//i),abs(1.618-i/(n//i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0] 

Try it online

The same solution in Python 2 is one byte longer.

lambda n:min([((i,n/i),abs(1.618-1.*i/(n/i)))for i in range(1,n+1)if n%i<1],key=lambda x:x[1])[0]