C (gcc), 50 bytes

a,b,c;f(n){for(a=0,b=1;n>(c=a);b=c)a+=b;n=(n==c);} 

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Clojure, 61 bytes

(def s(lazy-cat[0 1](map +(rest s)s)))#((set(take(inc %)s))%) 

This actually constructs the Fibonacci sequence s, grabs enough elements from it and checks if the input is found. Returns nil for falsy and the input number for truthy.

Javascript, 89 bytes

function f(n){s=Math.sqrt;c=Math.ceil;x=5*n**2;a=s(x-4);b=s(x+4);return c(a)==a||c(b)==b} 

This uses the fact that all fibonacci numbers have the property that 5x^2+4 or 5x^2-4 must be square. It takes the square root of these numbers and checks if they equal their ceiling value.

Javascript, 69 bytes (if it doesn't need to be a function)

s=Math.sqrt;c=Math.ceil;x=5*n**2;a=s(x-4);b=s(x+4);n=c(a)==a||c(b)==b 

This one does the exact same thing, except instead of calling a function, you set n to the number to test, and n is set to true/false based on the result.

This is my first code golf entry, so let me know if there's anything to improve here. :)

QBIC, 24 bytes

≈g<:|g=p+q┘p=q┘q=g]?g=a 

Explanation

≈g<:| WHILE g (running fibonacci total) is less than input g=p+q Get the next fib by adding p (n-2, starts as 0) and q (n-1, starts as 1) ┘ (Syntactic linebreak) p=q increase n-2 ┘ (Syntactic linebreak) q=g increase n-1 ] WEND ?g=a PRINT -1 if g equals input (is a fib-number), or 0 if not. 

Swift, 72 bytes

func f(i:Int,a:Int=0,b:Int=1)->Bool{return a<i ?f(i:i,a:b,b:a+b) :i==a} 

Un-golfed:

func f(i:Int, a:Int=0, b:Int=1)->Bool{ return a<i ? f(i: i, a: b, b: a+b) : i==a } 

I am recursively calling f until a is equal to or greater then i. Then, I check to see if i and a are equal.

You can try it here

Python 3, 56 53 50 bytes

• Thanks to @Fedone for 3 bytes: as a function

def f(m): a=b=1 while a<m:b,a=a,a+b print(a==m) 

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Python 3, 59 Bytes

f=5*int(input())**2 print(not((f+4)**0.5%1and(f-4)**0.5%1)) 

[Python 3], 48 bytes

m=0;k=1;exec("k,m=m,m+k\nif k==n:print(1)\n"*n) 

If n is the input, we generate the first n Fibonacci numbers and check if our input is one of them.

MathGolf, 4 bytes

⌠rf╧ 

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Explanation

We need to increment the input twice before creating the range because we know that \$F_{n+1} \geq n,\forall n \geq 0\$.

⌠ increment twice r range(0, n) f pop a, push fibonacci(a) (maps to the range) ╧ pop a, b, a.contains(b) (check if input is in list) 

brev, 46 bytes

(c(fn(or(= x z)(and(< x z)(f y(+ x y)z))))0 1) 

AWK, 43 bytes

$0=(5*$1^2+4)^.5!~/\./||(5*$1^2-4)^.5!~/\./ 

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Prints 1 for truthy or nothing for falsey.

Slightly longer version will print 0 for falsey:

1,$0=(5*$1^2+4)^.5!~/\./||(5*$1^2-4)^.5!~/\./ 

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