Mathematica, 33 bytes

AtomQ@*InverseFunction[Fibonacci] 

JS (ES6), 57 bytes

n=>(y=y=>((5*(n**2)+y)**0.5),~~y(4)==y(4)|~~y(-4)==y(-4)) 

Uses carusocomputing's method. Alot golfier than my other answer.

Ungolfed

n=>{ y=y=>((5*(n**2)+y)**0.5);//carusocomputing's method in a function return ~~y(4) === y(4) || ~~y(-4) === y(-4);//~~x === Math.floor(x) } 

CJam, 37 bytes

ri1T{\1$+_3$-g"1T 0_ 1"S/=~}g]W= 

CJam has no Fibonnaci built-in. On the bright side, this does use g twice, and I think this is the first time I've ever used it!

k, 20 bytes

{*x=*|(*x>)(|+\)\1 1} 

Generates fibonacci numbers until it overshoots. Then it checks the last one it generated for equality. 1 is truthy, 0 is falsey.

Try it online.

Mathematica, 30 bytes

Or@@EvenQ[2Sqrt[5#^2+{4,-4}]]& 

Java 8, 94 bytes

x->{int i=0;for(;c(i++)<=x;);return c(i-2)==x;}int c(int n){return n<1?0:n<2?1:c(n-1)+c(n-2);} 

Explanation:

Try it here. (NOTE: It's a bit slow for very large test-cases.)

x->{ // Method (1) with integer parameter and boolean return-type int i=0; // Index for(;c(i++)<=x;); // Loop as long as the Fibonacci number is smaller or equal to the input return c(i-2)==x; // And then return if the input equals the previous Fibonacci number } // End of method (1) // Method to get `n`th Fibonacci number int c(int n){ // Method (2) with integer parameter and integer return-type return n<1? // If `n`==0: 0 // Return 0 :n<2? // Else if `n`==1 1 // Return 1 : // Else: c(n-1)+c(n-2); // Return recursive calls with `n-1` and `n-2` } // End of method (2) 

05AB1E, 7 bytes

ÅF夹_~ 

Try it online!

><>, 40 83 bytes

Added 43 bytes so that it takes the correct input

i:0(?vc4*- v&a~< +>l2(?v$&:a*&* v ~&< >10r:&1)?v1n; =?v&:&)?v>:{+::&:& >1n;n0< 

A less golfy version would be:

// Read input i:0(?vc4*- >~a&v >l2(?v$&:a*&*+ >&~04. // Determine if in Fibonacci 10r:&1)?v1n; >:{+::&:&=?v&:&)?v >1n;n0< 

AWK, 56 63 61 bytes

{for(n[++j]++;n[j]<$1;n[++j]=n[j]+n[j-1]){}$0=$0?n[j]==$1:1}1 

Try it online!

Brute force is fun. :) If you want it to work for arbitrarily large numbers, add a -M argument, but that is outside the scope of the problem.

7 bytes added to account for 0 as input, but shaved a couple off using the ternary operator.

Actually, 2 bytes

fu 

Try it online!

Pushes either a positive number for truthy or 0 for falsy.

Cubix, 22 24 bytes

0 is truthy, nothing is falsey

@0O1I!^/@.W<rq\?-p+;;u 

Try it online!

 @ 0 O 1 I ! ^ / @ . W < r q \ ? - p + ; ; u . . 

Watch it run

I may be able to get a couple more out of this ... found them with a change to the initial redirect into the loop

• I get the integer to check
• ! check for 0 input
  • ^O@ if zero, output and halt
• /01 initialise the stack for doing the sequence
• W<W change lane onto the redirect back to self, then change lane into looping section
• +p-? bring the check value to the top, subtract and check
  • /@ On a positive result reflect and halt
  • \^O@ On a zero result reflect, output and halt
  • u;\qr; Remove the check, move check value to bottom, rotate the sum, remove the low value. Continue into loop.

Java, 40 bytes

r->Math.abs((r*Math.sqrt(5)-~r)%2*r-r)<2 

This is a straight Java port of @xnor's answer.

D, 57 bytes

A nice, clean, no-nonsense solution:

int f(int n,int x=0,int y=1){return y<n?f(n,y,x+y):y==n;} 

This one is 58 bytes but doesn't use recursion, and so might be more practical for larger inputs:

alias f=(n){int x,y=1;for(;y<n;y+=x,x=y-x){}return y==n;}; 

And here's one where the function declaration itself is only 54 bytes, though it depends on the mach library.

import mach.range : r=recur, l=last; import mach.math.vector : v=vector; const z=v(0,1); // The 54-byte function alias f=n=>z.r!(a=>v(a.y,a.x+a.y),a=>a.y>n).l(z).y==n; // Exploded for readability alias f=n=>( vector(0,1) // Seed the sequence .recur!(v=>vector(v.y,v.x+v.y),v=>v.y>n) // Compute Fib numbers until N .last(vector(0,1)).y == n // If the last number was N, return true // Value in parens "last(...)" is a fallback for n==0 and empty seq. ); 

><>, 33+3 = 36 bytes

3 bytes added for the -v flag

10:{:}=?!v1n; )?v:@+10.\:{:} n0/; 

Try it online!

Or 54 bytes without using the -v flag

 0ic4*-:0(?v$a*+10. :{:}=?!v1n;\10 v:@+d1.\:{:})? \0n; 

Try it online!

Japt, 8 7 bytes

ÆMgXÃøU 

Test it

Explanation

Implicit input of integer U.

Æ Ã 

Generate an array of integers from 0 to U-1 and pass each through a function where X is the current element.

MgX 

Get the Xth Fibonacci number.

øU 

Check if the array contains (ø) the original input U. Implicitly return the boolean result.

C, 36 bytes

f(x,a,b){return x>b?f(x,b,a+b):x==b} 

It puts some warnings, and requires at least 32-bit integers. Newer C standards probably won't even compile it. It should be called as f(142857,0,1).

Bonus: it can calculate Fibonacci-ness with different initial values, too.

Ruby, 64 41 40 bytes

->n,a=b=1{a,b=b,a+b;a<n ?redo:a>n ?p: 1} 

Try it online!

cQuents, 8 bytes

=0,1?Z+Y 

Try it online!

Explanation

=0,1 Set sequence start to 0,1 ? Mode: Query (assumes increasing sequences) Z+Y Each item is the previous two summed 

Brachylog, 16 14 bytes

1;0⟨t≡+⟩ⁱhℕ↙.! 

Try it online!

Takes input through the output variable and outputs through success or failure, and in the case of success the input variable is unified with 1.

1;0 Starting with [1,0], ⁱ iterating ⟨ ≡ ⟩ replacing ⟨t ⟩ the first element of the pair with the last element ⟨ +⟩ and the last element with the sum of the pair h until the first element ℕ↙ is greater than or equal to . the output variable, ! and stopping then, h the first element of the pair is equal to the output variable. 

ℕ↙.! is necessary for it to terminate on false test cases.

k4, 30 26 bytes

-4 thanks to ngn!

{x in(x>*|:){x,+/-2#x}/!2} 

the above is a simple while iterator. (cond){func}/arg.

 {x,+/-2#x} / x join sum over last 2 elements of x (i.e. append next Fib) (x>*|:) /!2 / while outer func input is greater than last element (x>*|:) of inner func output, pass inner func output to inner func x in / check if x is in array. returns boolean 

CSASM v2.1.2.3, 259 bytes

func main: push 0 pop $1 push 1 pop $2 in "" conv i32 pop $a push $a push 1 comp.lte push $f.o brfalse c .lbl b push 1 print ret .lbl a push 0 print ret .lbl c clf.o push $1 dup push $2 add pop $1 pop $2 push $1 push $a comp.gt push $f.o brtrue a push $1 push $a comp push $f.o brtrue b br c ret end 

Commented and ungolfed:

func main: ; Seed the sequence ($1 = new value, $2 = old value) push 0 pop $1 push 1 pop $2 ; Get the input, convert it to an integer and store it in the accumulator in "" conv i32 pop $a ; If $a <= 1, print truthy (1) push $a push 1 comp.lte push $f.o brfalse loop .lbl isFib ; Print a truthy value push 1 print ret .lbl notFib ; Print a falsy value push 0 print ret ; Keep generating new Fibonacci numbers until $1 is >= $a .lbl loop ; Clear the Comparison flag clf.o ; Get the next Fibonacci pair: ; $2 = $1, $1 = $1 + $2 push $1 dup push $2 add pop $1 pop $2 ; If $1 > $a, the input wasn't a Fibonacci number push $1 push $a comp.gt push $f.o brtrue notFib ; If $1 == $a, the input was a Fibonacci number push $1 push $a comp push $f.o brtrue isFib ; Still need to generate more numbers br loop ret end 

dc, 34 bytes

d0r^+0 1[d3R+d4Rd_5R>s]dssx0r4R-^p 

Try it online!

Prints 1 if the number at the top of the stack is a fibonacci number, 0 otherwise.

Bash, 87 bytes

read z;f=0;g=1;h=1 for((;;)){ h=$(($f+$g)) if [ $z == $f ];then echo 1;fi f=$g;g=$h } 

Attention endless loop, no break - stop the code after some seconds manually

Try it online!

Prolog (SWI), 36 bytes

\N:-1+N+0. Y+N+X:-X<N,X+Y+N+Y;X=:=N. 

Try it online!

Thunno 2, 5 bytes

ĖÆF$Ƈ 

Attempt This Online!

Explanation

ĖÆF$Ƈ # Implicit input Ė # Push [0..input] ÆF # For each n, get the nth Fibonacci number # (starting at 0->0, 1->1, 2->1, etc.) $Ƈ # Is the input in this list? # Implicit output 

C (gcc), 76, 52, 71 bytes,

Thanks GPT4 with a bit of prompt engineering

Way off, just plain wrong, good for the three test inputs but...

GTP4 after some serious tutoring finally 58 bytes

f(n,a,b){for(a=b=1;a<n;b+=a,a=b-a);return n==a||n==b||!n;} 

F(n){return n<2?n:F(n-1)+F(n-2);}i;f(n){for(i=0;F(i)<n;i++);int x=F(i)==n;} 

f(n,i,j){for(i=0,j=1;j<n;i=j,j+=i);return j==n||!n;} 

Try it online!

Desmos, 58 41 bytes

-17 bytes thanks to Aiden Chow!

f(n)=0^{mod((5nn+[4,0^n4-4])^{.5},1).min} 

Try it on Desmos!

Checks if either \$5n^2+4\$ or \$5n^2-4\$ is a perfect square.

Swift, 66 bytes

func f(n:Int){var a=0,b=1,c=0;while n>a{c=a;a=b;b=c+b};print(n<a)} 

Try it out! - NOTE: Prints False as truthy and True for falsy.

JS (ES6), 78 bytes

n=>{y=n?0:1;f=x=>x<3?1:f(x-1)+f(x-2);for(x=0;x<n+2;x++)f(x)==n?y=1:0;return y} 

Ungolfed:

var f = n => { var y = n ? 0 : 1; f=x=>x<3?1:f(x-1)+f(x-2);//from this: https://codegolf.stackexchange.com/a/25142/70700 for (var x = 0; x < n + 2; x++){ if(f(x) == n){ y = 1; }else{ y = 0; } } return y; }; 

Groovy, 44 43 37 bytes

{n->[-4,4].any{!((n*n*5+it)**0.5%1)}} 

If (5*(n**2)±4)**0.5 is ever an integer, the number is a fibbonacci number.