Number of values between input and next highest square

Question

Given a positive square number as input. Output the number of values between the input and next highest square.

Example

Input: 1

Output: 2

Reason: The numbers 2 and 3 are between 1 and 4, the next highest square

Input: 4

Output: 4

Reason: The numbers 5, 6, 7, 8 are between 4 and 9

Answer 1

S.I.L.O.S, 26 bytes

readIO
i=2*i^.5
printInt i

Try it online!

Straightforward port of Martin Ender's Mathematica answer.

Answer 2

C#, 24 bytes

n=>System.Math.Sqrt(n)*2

Answer 3

Perl 6, 12 bytes

*.sqrt.Int*2

Try it

---

If it had to work with positive numbers that aren't squares:

{+($_^..^(.sqrt.Int+1)²)}

Try it

Answer 4

CJam, 11 10 bytes

ri_mQ)_*-~

Explanation:

ri e# Read integer: | 16
_  e# Duplicate:    | 16 16
mQ e# Square root:  | 16 4
)  e# Increment:    | 16 5
_* e# Square:       | 16 25
-  e# Subtract:     | -9
~  e# Bitwise NOT:  | 8

Alternative Solution, 6 bytes

rimQ2*

Explanation:

ri e# Read integer: | 16
mQ e# Square root:  | 4
2* e# Times 2:      | 8

Answer 5

Pyth, 4 bytes

y@Q2

Explanation:

     # Implicit copy input to Q
y    # Multiply by 2
 @Q2 # Square-root of Q
     # Implicit print

Answer 6

Charcoal, 7 bytes

ＩײＸＮ·⁵

Try it online!

Explanation

Ｉ         Cast (int -> str)
  ײ       Multiplied by 2
    ＸＮ·⁵ N to the power of .5 (Square root of N)

Answer 7

C#, 42 bytes

int b=(int)Math.Sqrt(n);b=(b+1)*(b+1)-n-1;

Try it online!

Answer 8

Dyalog APL, 6 bytes

2×*∘.5

I can't believe this hasn't been posted yet. A variant with the same byte count:

+⍨*∘.5

Both of them work by doubling the square root.

Answer 9

NewStack, 3 bytes

2√ᵢ

The breakdown:

 √ᵢ    Append the square root of the input.
2      Multiply it by two.

Why this works:

To find how many numbers N are in between the square number x and the next square number m, we first need to find a formula for m.

Since x is a square number, √x will be an integer. And √x + 1 will equal √m.

Therefore m = (√x + 1)^2 = x + 2√x + 1

The amount of numbers in between x and m will just equal their difference minus 1.

N = m - x - 1 = (x + 2√x + 1) - x - 1

So we've concluded

N = 2√x

Answer 10

Plain English 129 98 bytes

To put a n number's d into a r number:
Put the square root of the n in the r.
Add the r to the r.

Ungolfed:

To put a number's delta into a result number:
Put the square root of the number in the result.
Add the result to the result.

The Plain English IDE is available at github.com/Folds/english. The IDE runs on Windows. It compiles to 32-bit x86 code.

Answer 11

J, 4 bytes

+:%:

    e.g. +:%: 9  ====>  6

Answer 12

Excel, 9 bytes

=2*A1^0.5

---

To handle all cases (not only squares): 21 bytes

=INT(A1^0.5+1)^2-A1-1

Answer 13

Cubix, 22 18 bytes

+@UOI0),c?;^..u;;/

Try it online!

Watch it online!

A port of Martin Ender's solution. Since Cubix only supports integer arithmetic, computing the square root is literally just trying every number until we hit the square root, so this approach only works since the input is a perfect square.

Expands to the following cube:

    + @
    U O
I 0 ) , c ? ; ^
. . u ; ; / . .
    . .
    . .

I       # read in input [n^2]
0       # push 0 [n^2, 0]
Main loop:
)       # increment [n^2, 1]
,       # divide, rounding to zero [n^2, 1, n^2]
c       # bitwise xor -- zero if equal, positive if not
?       # turn right if positive, go straight if zero [n^2,1,n^2,___]
 nonzero branch:
  /      # reflect IP direction
  ;;     # pop top two off stack [n^2, 1]
  u      # right-hand U-turn
  (re-enter main loop)
 zero branch: stack is [n^2, n, n, 0]
  ;      # pop top of stack [n^2, n, n]
  ^      # go north
  +      # add top two elements [n^2, n, n, 2n]
  U      # left-hand u-turn
  O      # output as int
  @      # terminate program