# Sum $\text{Square}^2$

Let $n=42$ (Input)

Then divisors are : 1, 2, 3, 6, 7, 14, 21, 42

Squaring each divisor : 1, 4, 9, 36, 49, 196, 441, 1764

Since $50\times 50=2500$ therefore we return a truthy value. If it is not a perfect square, return a falsy value.

# Examples :

42  ---> true
1   ---> true
246 ---> true
10  ---> false
16  ---> false


This is so shortest code in bytes for each language wins

Thanks to @Arnauld for pointing out the sequence : A046655

• Can the program output 0 if the result is true, and any other number if the result is false? – JosiahRyanW Sep 11 at 0:25

# R, 39 37 bytes

!sum((y=1:(x=scan()))[!x%%y]^2)^.5%%1


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Uses the classic "test if perfect square" approach, taking the non-integral part of the square root S^.5%%1 and taking the logical negation of it, as it maps zero (perfect square) to TRUE and nonzero to FALSE.

Thanks to Robert S for saving a couple of bytes!

• Could you use scan() to save a few bytes? – Robert S. Sep 10 at 18:31
• @RobertS. doh! I've been doing too much "real" R coding lately! – Giuseppe Sep 10 at 18:33

# JavaScript (ES7),  46 44  42 bytes

Saved 1 byte thanks to @Hedi

n=>!((g=d=>d&&d*d*!(n%d)+g(d-1))(n)**.5%1)


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## Commented

n =>             // n = input
!(             // we will eventually convert the result to a Boolean
(g = d =>    // g is a recursive function taking the current divisor d
d &&       //   if d is equal to 0, stop recursion
d * d      //   otherwise, compute d²
* !(n % d) //   add it to the result if d is a divisor of n
+ g(d - 1) //   add the result of a recursive call with the next divisor
)(n)         // initial call to g with d = n
** .5 % 1    // test whether the output of g is a perfect square
)              // return true if it is or false otherwise

• You can save one byte with d going from n to 0 instead of 2 to n like this: n=>!((g=d=>d?d*d*!(n%d)+g(d-1):0)(n)**.5%1) – Hedi Sep 11 at 22:17

# Python 2, 55 bytes

lambda n:sum(i*i*(n%i<1)for i in range(1,n+1))**.5%1==0


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# 05AB1E, 5 bytes

ÑnOÅ²


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### How?

ÑnOÅ²
Ñ     - divisors
n    - square
O   - sum
Å² - is square?


# Shakespeare Programming Language, 434428 415 bytes

,.Ajax,.Ford,.Puck,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy.Scene V:.Ajax:You be the sum ofyou a cat.Ford:Is the remainder of the quotient betweenyou I worse a cat?[Exit Ajax][Enter Puck]Ford:If soyou be the sum ofyou the square ofI.[Exit Puck][Enter Ajax]Ford:Be you nicer I?If solet usScene V.[Exit Ford][Enter Puck]Puck:Is the square ofthe square root ofI worse I?You zero.If notyou cat.Open heart


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-13 bytes thanks to Jo King!

Outputs 1 for true result, outputs 0 for false result.

# C (gcc), 676360 59 bytes

-1 bytes thanks to @JonathanFrech

i,s;f(n){for(s=i=0;i++<n;)s+=n%i?0:i*i;n=sqrt(s);n=n*n==s;}


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• Can ++i<=n be i++<n? – Jonathan Frech Sep 10 at 22:01
• @JonathanFrech that seems to work, thanks. – cleblanc Sep 11 at 13:36

# Brachylog, 12 8 bytes

f^₂ᵐ+~^₂


-4 bytes thanks to Fatelize cause i didn't realize brachylog has a factors functions

# explanation

f^₂ᵐ+~^₂            #   full code
f                   #       get divisors
^₂ᵐ                #           square each one
~^₂           #       is the result of squaring a number


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• f^₂ᵐ is 4 bytes shorter than ḋ{⊇×^₂}ᵘ – Fatalize Sep 12 at 7:14

# MathGolf, 5 4 bytes

─²Σ°


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## Explanation

─     Get all divisors as list (implicit input)
²    Square (implicit map)
Σ   Sum
°  Is perfect square?


Very similar to other answers, compared to 05AB1E I gain one byte for my "is perfect square" operator.

• You know, something called "MathGolf" really should have a norm operator... that would have gotten you down to 3 bytes :) – Misha Lavrov Oct 6 at 16:26
• @MishaLavrov that's not a bad idea! Right now I don't have as many vector operations as I'd like, one of these days I'll change that – maxb Oct 6 at 19:55

# Neim, 5 bytes

𝐅ᛦ𝐬q𝕚


Explanation:

𝐅      Factors
ᛦ      Squared
𝐬     Summed
𝕚   is in?
q    infinite list of square numbers


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# PowerShell, 68 56 bytes

param($n)1..$n|%{$a+=$_*$_*!($n%$_)};1..$a|?{$_*$_-eq$a}  Try it online! Seems long ... -12 bytes thanks to mazzy Does exactly what it says on the tin. Takes the range from 1 to input $n and multiplies out the square $_*$_ times whether it's a divisor or not !($n%$_). This makes divisors equal to a nonzero number and non-divisors equal to zero. We then take the sum of them with our accumulator $a. Next, we loop again from 1 up to $a and pull out those numbers where |?{...} it squared is -equal to $a. That is left on the pipeline and output is implicit. Outputs a positive integer for truthy, and nothing for falsey. • the rare case where $args[0] is shorter :) 1..$args[0]|%{$a+=$_*$_*!($n%$_)};1..$a|?{$_*$_-eq$a} – mazzy Sep 10 at 19:00
• @mazzy It's not, because you need $n inside the loop for !($n%$_). But, your rewrite of the sum saved 12 bytes, so thanks! – AdmBorkBork Sep 10 at 19:11 • what a shame. so I would like to find a case where $args[0] is shorter :) – mazzy Sep 10 at 19:18

# Japt, 119 7 bytes

-2 bytes from @Giuseppe and another -2 from @Shaggy

â x²¬v1


â x²¬v1             Full program. Implicity input U
â                   get all integer divisors of U
x²                square each element and sum
¬               square root result
v1           return true if divisible by 1


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# APL (Dyalog Unicode), 18 bytes

0=1|.5*⍨2+.*⍨∘∪⍳∨⊢


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Anonymous lambda. Returns 1 for truthy and 0 for falsy (test cases in TIO are prettified).

Shoutouts to @H.PWiz for 4 bytes!

### How:

0=1|.5*⍨2+.*⍨∘∪⍳∨⊢   ⍝ Main function, argument ⍵ → 42
∨⊢   ⍝ Greatest common divisor (∨) between ⍵ (⊢)
⍳      ⍝ and the range (⍳) [1..⍵]
∪      ⍝ Get the unique items (all the divisors of 42; 1 2 3 6 7 14 21 42)
∘        ⍝ Then
⍨         ⍝ Swap arguments of
2+.*          ⍝ dot product (.) of sum (+) and power (*) between the list and 2
⍝ (sums the result of each element in the vector squared)
⍨              ⍝ Use the result vector as base
.5*               ⍝ Take the square root
1|                  ⍝ Modulo 1
0=                    ⍝ Equals 0

• Can you do the equivalent of not rather than 0= to save a byte? – streetster Sep 11 at 6:58
• @streetster unfortunately, I cannot for 2 reasons. First, APL's not operator (~), when used monadically, only works with booleans (either 0 or 1). Since any number modulo 1 never equals 1, if I used ~ instead of 0=, I'd get a domain error on any number that's not a perfect square, since decimal values are out of ~'s domain. Furthermore, I cannot simply omit the 0=, since APL's truthy value is 1, not 0, and it wouldn't have a consistent output for falsy values. – J. Sallé Sep 11 at 12:55

# K (oK), 2625 22 bytes

Solution:

{~1!%+/x*x*~1!x%:1+!x}


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Explanation:

{~1!%+/x*x*~1!x%:1+!x} / the solution
{                    } / lambda taking x as input
!x  / range 0..x-1                        \
x%:      / x divided by and save result into x |
1!         / modulo 1                            | get divisors
~           / not                                 |
x*            / multiply by x                       /
x*              / multiply by x (aka square)          > square
+/                / sum up                              > sum up
%                  / square root                         \
1!                   / modulo 1                            | check if a square
~                     / not                                 /


Notes:

• -1 bytes taking inspiration from the PowerShell solution
• -3 bytes taking inspiration from the APL solution

# Pari/GP, 23 bytes

n->issquare(sigma(n,2))


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## Matlab, 39 37 bytes

@(v)~mod(sqrt(sum(divisors(v).^2)),1)


Unfortunately, it doesn't work on Octave (on tio) so no tio link.

Note As @LuisMendo stated, divisors() belongs to Symbolic Toolbox.

• It looks like divisors belongs to the Symbolic Toolbox. You should state that in the title. Also, you can use ~··· instead of ···==0 – Luis Mendo Sep 10 at 21:39

-14 bytes thanks to Ørjan Johansen. -11 bytes thanks to ovs.

f x=sum[i^2|i<-[1..x],xmodi<1]elemmap(^2)[1..x^2]


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Hey, it's been a while since I've... written any code, so my Haskell and golfing might a bit rusty. I forgot the troublesome Haskell numeric types. :P

• It's shorter (but slower) to avoid those conversions by searching for the square root with another list comprehension. Try it online! – Ørjan Johansen Sep 11 at 2:36
• Shorter: f x|s<-sum[i^2|i<-[1..x],mod x i<1]=round(sqrt$toEnum s)^2==s – Damien Sep 11 at 9:22 • Building up on Ørjan Johansen's suggestion, this should work for 53 bytes. – ovs Sep 11 at 13:16 # Pyt, 7 bytes ð²ƩĐř²∈  Try it online! ### Explanation  Implicit input ð Get list of divisors ² Square each element Ʃ Sum the list [n] Đ Duplicate the top of the stack ř² Push the first n square numbers ∈ Is n in the list of square numbers? Implicit output  ð²Ʃ√ĐƖ=  Try it online! ### Explanation  Implicit input ð Get list of divisors ² Square each element Ʃ Sum the list [n] √ Take the square root of n Đ Duplicate the top of the stack Ɩ Cast to an integer = Are the top two elements on the stack equal to each other? Implicit output  ð²Ʃ√1%¬  Try it online! ### Explanation  Implicit input ð Get list of divisors ² Square each element Ʃ Sum the list [n] √ Take the square root of n 1% Take the square root of n modulo 1 ¬ Negate [python typecasting ftw :)] Implicit output  # Husk, 6 bytes £İ□ṁ□Ḋ  Try it online! ### Explanation £İ□ṁ□Ḋ -- example input 12 Ḋ -- divisors: [1,2,3,4,6,12] ṁ -- map the following .. □ -- | square: [1,4,9,16,36,144] -- .. and sum: 210 £ -- is it element of (assumes sorted) İ□ -- | list of squares: [1,4,9,16..196,225,..  # Mathematica, 32 bytes IntegerQ@Sqrt[2~DivisorSigma~#]&  Pure function. Takes a number as input and returns True or False as output. Not entirely sure if there's a shorter method for checking perfect squares. # Octave / MATLAB, 43 bytes @(n)~mod(sqrt(sum(find(~mod(n,1:n)).^2)),1)  Try it online! # Perl 6, 34 bytes -1 byte thanks to nwellnhof {grep($_%%*,1..$_)>>².sum**.5%%1}  Try it online! • **.5 is one byte shorter than .sqrt. – nwellnhof Sep 11 at 14:42 ## F#, 111 bytes let d n=Seq.where(fun v->n%v=0){1..n} let u n= let m=d n|>Seq.sumBy(fun x->x*x) d m|>Seq.exists(fun x->x*x=m)  Try it online! So d gets the divisors for all numbers between 1 and n inclusive. In the main function u, the first line assigns the sum of all squared divisors to m. The second line gets the divisors for m and determines if any of them squared equals m. # Perl 5, 47 bytes $a+=$_*$_*!($n%$_)for 1..$n;$a=!($a**.5=~/\D/);  Returns 1 for true and nothing for false. ### Explanation: $a+=              for 1..$n; sum over i=1 to n$_*$_ square each component of the sum *!($n%$_) multiply by 1 if i divides n.$a=                   a equals
(\$a**.5           whether the square root of a
!       =~/\D/);   does not contain a non-digit.


# Groovy, 47 bytes

A lambda accepting a numeric argument.

n->s=(1..n).sum{i->n%i?0:i*i}
!(s%Math.sqrt(s))


Explanation

(1..n) creates an array of the values 1 to n

n%i is false (as 0 is falsy) if i divides n without remainder

n%i ? 0 : i*i is the sum of the square of the value i if it divides n without remainder, otherwise is 0

sum{ i-> n%i ? 0 : i*i } sums the previous result across all i in the array.

s%Math.sqrt(s) is false (as 0 is falsy) if the sqrt of s divides s without remainder

!(s%Math.sqrt(s)) returns from the lambda (return implicit on last statement) !false when the sqrt of s divides s without remainder

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# Java 8, 75 70 bytes

n->{int s=0,i=0;for(;++i<=n;)s+=n%i<1?i*i:0;return Math.sqrt(s)%1==0;}


-5 bytes thanks to @archangel.mjj.

Try it online.

Explanation:

n->{             // Method with integer parameter and boolean return-type
int s=0,       //  Sum-integer, starting at 0
i=0;       //  Divisor integer, starting at 0
for(;++i<=n;)  //  Loop i in the range [1, n]
s+=n%i<1?    //   If n is divisible by i:
i*i      //    Increase the sum by the square of i
:         //   Else:
0;       //    Leave the sum the same by adding 0
return Math.sqrt(s)%1==0;}
//  Return whether the sum s is a perfect square

• Hi, you can cut 5 bytes by removing the t variable (do the eval and assignment within the body of the for loop), like so: n->{int s=0,i=0;for(;++i<=n;)s+=n%i<1?i*i:0;return Math.sqrt(s)%1==0;} – archangel.mjj Sep 13 at 8:26
• @archangel.mjj Ah, of course. Not sure how I missed that. Thanks! :) – Kevin Cruijssen Sep 13 at 8:36

# Pyth, 26 bytes

Iq0%^sm^d2*M{yPQc1h1hZ1.?0


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I'm new to Pyth (and golfing in general) so this is pretty bad. Pyth doesn't have a built-in to get divisors, so I had to do that. Up until the perfect-square check, it was pretty alright, but after that it became a bit cluttered because of the if statement. Criticism is welcome.

Explanation:

               Q              #get user input
*M{yP               #get the divisors
m^d2                    #square each divisor using map
s                        #sum the list
^           c1h1          #get square root
Iq0%                Z1        #if the sqrt is a whole number
1.?0    #then print 1, else print 0

• A few savings for you - m^d2 can be ^R2, and @ functions as the root operator, so ^...c1h1 can be @...2. The return value is if the root mod 1 is non-zero - that is, should return True if the result of the mod is 0, and False otherwise - this means we can just use ! to invert the truthiness of the mod. Full program is then !%@s^R2*M{yPQ2 1 - demonstration – Sok Oct 9 at 8:26

# C# (.NET Core), 115 74 bytes

Thanks to user Misha Lavrov for teaching me how to properly do code golf! :)

a=>{int s=0;for(int i=0;i++<=a;)s+=a%i==0?i*i:0;return Math.Sqrt(s)%1==0;}


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Ungolfed:

a => {                              // reads input
int s = 0;                      // holds sum of squares
for(int i = 0; i++ <= a;)       // index from 1 to input, inclusive
{
s +=                        // add to sum
a % i == 0 ?            // check if remainder of input divided by current index is zero
i * i : 0;          // if true, add square of current index, else add zero
}
return Math.Sqrt(s) % 1 == 0;   // prints "True" if square root is whole number, "False" otherwise
}

• I think it's standard to do input and output via a pure function, something like this. So you don't have to count the length of Console.ReadLine() and such, that way. – Misha Lavrov Oct 5 at 20:26
• Good to know, thank you. I'll keep this in mind for the future! – Meerkat Oct 5 at 20:33

# Taxi, 4430 4305 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.0 is waiting at Starchild Numerology.0 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 l 1 l 1 l 2 l.Pickup a passenger going to Rob's Rest.Pickup a passenger going to Addition Alley.Go to Rob's Rest:w 1 r 2 l 1 r.Go to Cyclone:s 1 l 1 l 2 l.[B]Pickup a passenger going to Divide and Conquer.1 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 r 3 l.Pickup a passenger going to Addition Alley.Go to Addition Alley:w 1 r 3 r 1 r 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:n 2 r 2 r 1 r.Pickup a passenger going to Cyclone.Go to Sunny Skies Park:e 1 l 1 l 2 l 1 l.Go to Zoom Zoom:n 1 r.Go to Cyclone:w.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to Equal's Corner.Pickup a passenger going to Trunkers.Go to Sunny Skies Park:n 1 r.Go to Trunkers:s 1 l.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:w 1 l.Switch to plan C if no one is waiting.Pickup a passenger going to Riverview Bridge.Go to Sunny Skies Park:n.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to Cyclone.Go to Riverview Bridge:n 1 r 1 r.Go to Cyclone:w 2 l.Pickup a passenger going to Cyclone.Go to Sunny Skies Park:n 1 r.Go to Cyclone:n 1 l.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to Addition Alley.Go to Rob's Rest:s 1 r 2 l 1 l 1 r 1 r.Pickup a passenger going to Addition Alley.Go to Addition Alley:s 1 l 1 l 2 r 1 r 1 r.Pickup a passenger going to Rob's Rest.Go to Rob's Rest:n 1 l 1 l 1 l 2 r 1 r.Go to Cyclone:s 1 l 1 l 2 l.Pickup a passenger going to Magic Eight.Go to Sunny Skies Park:n 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l.Pickup a passenger going to Cyclone.Pickup a passenger going to Magic Eight.Go to Magic Eight:s 1 l 2 r.Switch to plan D if no one is waiting.Pickup a passenger going to Addition Alley.Go to Cyclone:n 1 l 2 r.Switch to plan B.[C]Go to Sunny Skies Park:n.Pickup a passenger going to Cyclone.Pickup a passenger going to Addition Alley.Go to Cyclone:n 1 l.Switch to plan B.[D]Go to Cyclone:n 1 l 2 r.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to Sunny Skies Park.Go to Sunny Skies Park:n 1 r.Go to Rob's Rest:s 2 r 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:s 1 l 1 l 2 l.Pickup a passenger going to Cyclone.Pickup a passenger going to Cyclone.Go to Zoom Zoom:n.Go to Cyclone:w.[E]Pickup a passenger going to Joyless Park.Pickup a passenger going to Divide and Conquer.Go to Joyless Park:n 2 r 2 r 2 l.Go to Cyclone:w 1 r 2 l 2 l.Pickup a passenger going to Cyclone.Pickup a passenger going to Joyless Park.Go to Joyless Park:n 2 r 2 r 2 l.Go to Cyclone:w 1 r 2 l 2 l.Pickup a passenger going to Divide and Conquer.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:n 2 r 2 r 1 r.Pickup a passenger going to Magic Eight.1 is waiting at Starchild Numerology.Go to Starchild Numerology:e 1 r 3 r 1 l 1 l 2 l.Pickup a passenger going to Magic Eight.Go to Magic Eight:w 1 r 2 r 1 r.Switch to plan F if no one is waiting.Pickup a passenger going to Riverview Bridge.Go to Joyless Park:e 2 l 4 r.Pickup a passenger going to Cyclone.Pickup a passenger going to The Underground.Go to The Underground:w 1 l.Pickup a passenger going to Cyclone.Go to Fueler Up:s.Go to Riverview Bridge:n 3 l.Go to Cyclone:w 2 l.Switch to plan E.[F]Go to Joyless Park:e 2 l 4 r.Pickup a passenger going to What's The Difference.Pickup a passenger going to Cyclone.Go to Cyclone:w 1 r 2 l 2 l.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to What's The Difference.Go to What's The Difference:n 2 l 1 r 3 l.Pickup a passenger going to Knots Landing.Go to Knots Landing:e 4 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:w 1 l.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.


Try it online!

I went to a lot of places in Townsburg that I've never visited before for this program. Because this program is HUGE (mainly due to the fact that there isn't a one-stop way to take square roots; I still don't actually do that, though), I'm going to attempt to sketch out an explanation of this program.

Go to Post Office:w 1 l 1 r 1 l.

[Take an input line from STDIN.]

Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.

[Take it to The Babelfishery, which converts the string to a double.]
[If you were to give it a double, it gives back a string.]

Pickup a passenger going to Cyclone.

[We're going to clone this number by sending it to the Cyclone.]

0 is waiting at Starchild Numerology.

[In Taxi, numeric passengers are introduced once they wait at Starchild Numerology.]
[This passenger will be our running squares total.]

0 is waiting at Starchild Numerology.

[This passenger will be our counter, going from 1 to the inputted number.]
[Yes, it's starting at 0 here, but it is incremented at the start of our loop.]

Go to Starchild Numerology:n 1 l 1 l 1 l 2 l.
Pickup a passenger going to Rob's Rest.
Pickup a passenger going to Addition Alley.
Go to Rob's Rest:w 1 r 2 l 1 r.
Go to Cyclone:s 1 l 1 l 2 l.

[Drop off the squares total at Rob's Rest. We'll pick it back up later when we need it.]
[Also, do the actual cloning of the inputted number.]

[At the start of each loop, our only passenger is the counter, going to Addition Alley.]

[B]
Pickup a passenger going to Divide and Conquer.

[Pick up one of the clones of the inputted number.]
[We will leave the other clone here for later.]

1 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 r 3 l.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:w 1 r 3 r 1 r 1 r.
Pickup a passenger going to Cyclone.

[Pick up the number 1, and add it to the counter, then pick up the result.]
[This new passenger is our new counter.]

Go to Cyclone:n 1 l 1 l.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to Divide and Conquer.

[First we pick up the other clone of the inputted number, going to Sunny Skies Park.]
[Sunny Skies Park is basically a First-In-First-Out queue, where we will stick it for later.]
[Then we pick up a clone of the counter which we just dropped off, so we can divide the input by it.]
[We will leave the other clone here for later.]

Go to Divide and Conquer:n 2 r 2 r 1 r.
Pickup a passenger going to Cyclone.

[Divide the inputted number by the counter, and prepare to clone it.]

Go to Sunny Skies Park:e 1 l 1 l 2 l 1 l.

[Drop off our clone of the inputted number at Sunny Skies Park.]

Go to Zoom Zoom:n 1 r.

[Can't forget to get gas!]
[Zoom Zoom is a close gas station, and we have so many credits that one stop here every iteration fills us up completely.]

Go to Cyclone:w.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to Equal's Corner.
Pickup a passenger going to Trunkers.

[First we pick up the other clone of the counter, going to Sunny Skies Park for later.]
[Then we pick up a clone of the division result, going to Equal's Corner.]
[Equal's Corner takes multiple numeric passengers, and returns the value of one of them if they are equal, but otherwise it returns no one.]
[Finally we pick up the other clone of the division result, going to Trunkers.]
[Dropping off a passenger at Trunkers is equivalent to the floor function. We need it here because division results are "exact".]
[(Well, as exact as double precision arithmetic can be.)]
[We will compare the result of this to the exact result of the division.]

Go to Sunny Skies Park:n 1 r.

[Drop off our clone of the counter at Sunny Skies Park.]

Go to Trunkers:s 1 l.
Pickup a passenger going to Equal's Corner.
Go to Equal's Corner:w 1 l.

[Compare the exact division result with the floored division result.]
[If they are equal, then the value of one of them will be waiting.]
[This also means that the counter is a divisor of the input, and we should add the square of the counter to our running squares total.]
[If they are not equal, no one will be waiting, the counter is not a divisor of the input, and nothing should happen.]

Switch to plan C if no one is waiting.

[This is Taxi's only conditional operator.]
[It jumps to a bracketed label (in this case, C) if there is no one waiting at this stop.]
[Of course, there is an unconditional version of this, too.]

[If we are on this path, we are adding the square of the counter to our running squares total.]

Pickup a passenger going to Riverview Bridge.

[We don't actually need the passenger that's waiting here (and leaving it there will not work at all).]
[So what do you do when you have a pesky passenger you don't want? You take them to Riverview Bridge!]
[Riverview Bridge has a lovely view, but passengers dropped off there always seem to fall in the river.]
[This means you don't collect your fare for bringing them there, but at least the pesky passenger is gone.]

Go to Sunny Skies Park:n.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to Cyclone.

[We want to pick up the counter (so we can multiply it by itself). That dang input is in the way, but we still need it.]
[So what do we do? We just take it back to where it was waiting!]
[Yup, even though one would never do this in real life, a passenger's current location is a perfectly legal destination in Taxi.]

Go to Riverview Bridge:n 1 r 1 r.

[So long, Equal's Corner result!]

Go to Cyclone:w 2 l.
Pickup a passenger going to Cyclone.
Go to Sunny Skies Park:n 1 r.

[We clone the counter so we can still use it, then we drop the inputted number back at Sunny Side Park.]
[Again, we will leave the other clone here for later.]

Go to Cyclone:n 1 l.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station:s 1 l 2 r 4 l.

[Square the counter.]

Pickup a passenger going to Addition Alley.
Go to Rob's Rest:s 1 r 2 l 1 l 1 r 1 r.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:s 1 l 1 l 2 r 1 r 1 r.
Pickup a passenger going to Rob's Rest.
Go to Rob's Rest:n 1 l 1 l 1 l 2 r 1 r.

[Pick up the running squares total from Rob's Rest, add it to the square of the counter, then bring it back.]

Go to Cyclone:s 1 l 1 l 2 l.
Pickup a passenger going to Magic Eight.

[Pick up the counter, going to Magic Eight.]
[Magic Eight takes two numeric passengers, and returns the first passenger if it is less than the second, but otherwise it returns no one.]
[We will compare the counter to the inputted number...]

Go to Sunny Skies Park:n 1 r.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Magic Eight.

[...who we take now, so we can make a clone of it and compare it to the counter.]

Go to Magic Eight:s 1 l 2 r.
Switch to plan D if no one is waiting.

[Compare.]
[If the counter is waiting here, the loop is to continue.]
[If not, switch to plan D, end the loop, and start detecting whether or not the running squares total is a square itself.]

Pickup a passenger going to Addition Alley.
Go to Cyclone:n 1 l 2 r.
Switch to plan B.

[Reset everything, and loop back to the start of plan B.]

[C]

[If we are on this path, the counter is not a divisor of the inputted number, but the loop should continue anyways.]

Go to Sunny Skies Park:n.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Addition Alley.
Go to Cyclone:n 1 l.
Switch to plan B.

[Reset everything, and loop back to the start of plan B.]

[D]

[We are finally out of the loop. Now it's time for the real fun.]

Go to Cyclone:n 1 l 2 r.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to Sunny Skies Park.
Go to Sunny Skies Park:n 1 r.

[We don't need the inputted number anymore, but we can't change its destination from the Cyclone.]
[So we go to the Cyclone, so we can take the two clones and bring them to Sunny Skies Park forever.]
[We could send them to Riverview Bridge, but that will take more bytes.]

Go to Rob's Rest:s 2 r 1 r.

[Pick up our running squares total from Rob's Rest, so it can go out and see the town.]

Pickup a passenger going to Cyclone.
Go to Cyclone:s 1 l 1 l 2 l.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Cyclone.
Go to Zoom Zoom:n.
Go to Cyclone:w.

[We need to end up with 4 clones of the total.]
[Two of them will actually be used as the counter in this loop.]

[At the start of each loop, we have no passengers.]

[E]
Pickup a passenger going to Joyless Park.
Pickup a passenger going to Divide and Conquer.
Go to Joyless Park:n 2 r 2 r 2 l.

[We will leave one clone of the total at Joyless Park (like Sunny Side Park, but on the opposite side of town).]
[The other clone will be used in a division.]

Go to Cyclone:w 1 r 2 l 2 l.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Joyless Park.
Go to Joyless Park:n 2 r 2 r 2 l.

[We will leave one clone of the counter at Joyless Park.]
[We will clone the other clone (clone-ception?).]

Go to Cyclone:w 1 r 2 l 2 l.
Pickup a passenger going to Divide and Conquer.
Pickup a passenger going to Divide and Conquer.
Go to Divide and Conquer:n 2 r 2 r 1 r.

[We will use both clones of the counter in a division.]
[Divide and Conquer can take 3 arguments, and the result here will be equivalent to total / counter^2 .]
[If the square of the counter is equal to the total, the result will be 1.]
[If the square of the counter is less than the total, the result will be greater than 1.]
[If the square of the counter is greater than the total, the result will be less than 1.]

Pickup a passenger going to Magic Eight.
1 is waiting at Starchild Numerology.
Go to Starchild Numerology:e 1 r 3 r 1 l 1 l 2 l.
Pickup a passenger going to Magic Eight.
Go to Magic Eight:w 1 r 2 r 1 r.
Switch to plan F if no one is waiting.

[We compare the result of the division against 1 to decide our next course of action.]
[If it is greater or equal, the loop should end, and the counter is equal to the floor of the square root.]
[If it is less, the loop should continue.]

Pickup a passenger going to Riverview Bridge.

[Whoops, we don't actually need this anymore.]

Go to Joyless Park:e 2 l 4 r.
Pickup a passenger going to Cyclone.

[We will clone the total later, so we can reuse it in the next loop iteration.]

Pickup a passenger going to The Underground.
Go to The Underground:w 1 l.

[Instead of taking the number 1 and subtracting it from the counter, we can do an optimization here.]
[The Underground is a destination that takes 1 passenger, and subtracts 1 from it.]
[If the result is positive, it returns the result. Otherwise, it returns no one.]
[The latter result will not happen for us, however.]

Pickup a passenger going to Cyclone.

[We will clone the counter later, so we can reuse it in the next loop iteration.]

Go to Fueler Up:s.

[Gotta remember to stop for gas!]

Go to Riverview Bridge:n 3 l.

[So long, Magic Eight result!]

Go to Cyclone:w 2 l.
Switch to plan E.

[Reset everything, and loop back to the start of plan E.]

[F]

Go to Joyless Park:e 2 l 4 r.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to Cyclone.

[Once we've figured out the square root of the total, we're still not done.]

Go to Cyclone:w 1 r 2 l 2 l.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station:s 1 l 2 r 4 l.

[We actually square the square root...]

Pickup a passenger going to What's The Difference.
Go to What's The Difference:n 2 l 1 r 3 l.

[...and subtract it from the total at What's The Difference (which subtracts dropped-off passengers).]

Pickup a passenger going to Knots Landing.
Go to Knots Landing:e 4 r 1 l.

[If the result is 0, the total is a perfect square, and a truthy value should be output.]
[If the result is nonzero, the total is not a perfect square, and a falsy value should be output.]
[Luckily, Knots Landing does our work for us.]
[Knots Landing inverts the boolean logic of numeric passengers (i.e. 0 becomes 1, nonzero becomes 0).]

Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:w 1 l.

[Convert it to a string...]

Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

[...and output it.]

[We would exit the program by going to the Taxi Garage, but that takes bytes.]
[The boss fires us because we didn't bring the taxi back to the garage, but that gets output to STDERR, so that's OK.]


# Jelly, 6 bytes

ÆD²SÆ²


Try it online! Or see the test-suite.

### How?

ÆD²SÆ² - Main Link: integer
ÆD     - divisors
²    - square
S   - sum
Æ² - is square?


# Gaia, 5 bytes

ds¦Σụ


Try it online!