Simplify a square root

Question

Given a positive integer n, simplify the square root √n into the form a√b by extracting all square factors. The outputted a,b should be positive integers with n = a^2 * b with b as small as possible.

You may output a and b in either order in any reasonable format. You may not omit outputs of 1 as implicit.

The outputs for n=1..36 as (a,b):

1 (1, 1)
2 (1, 2)
3 (1, 3)
4 (2, 1)
5 (1, 5)
6 (1, 6)
7 (1, 7)
8 (2, 2)
9 (3, 1)
10 (1, 10)
11 (1, 11)
12 (2, 3)
13 (1, 13)
14 (1, 14)
15 (1, 15)
16 (4, 1)
17 (1, 17)
18 (3, 2)
19 (1, 19)
20 (2, 5)
21 (1, 21)
22 (1, 22)
23 (1, 23)
24 (2, 6)
25 (5, 1)
26 (1, 26)
27 (3, 3)
28 (2, 7)
29 (1, 29)
30 (1, 30)
31 (1, 31)
32 (4, 2)
33 (1, 33)
34 (1, 34)
35 (1, 35)
36 (6, 1)

These are OEIS A000188 and A007913.

Related: A more complex version.

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Answer 1

Jelly, 9 bytes

ÆE;0d2ZÆẸ

Try it online! or verify all test cases.

How it works

ÆE;0d2ZÆẸ  Main link. Argument: n

ÆE         Exponents; generate the exponents of n's prime factorization.
  ;0       Append 0 since 1ÆE returns [].
    d2     Divmod by 2.
      Z    Zip/transpose to group quotients and remainders.
       ÆẸ  Unexponent; turn the exponents of prime factorizations into integers.

Answer 2

Jelly, 6 bytes

²×¥þœi

Try It Online!

TFW you outgolf Dennis by 33% :P

Outputs [b, a].

This does not use any prime factorization built-ins. Instead, it uses an array-oriented built-in œi (multi-dimensional index of), which is interestingly not found in any of APL/J/K/BQN.

How it works

²×¥þœi    Monadic link. Input: the positive integer n.
²×¥þ      Compute the table of (x^2 * y) for both x and y in 1..n
          For n = 3, this returns:
          × | 1  4  9
          --+--------
          1 | 1  4  9
          2 | 2  8 18
          3 | 3 12 27
    œi    Find the multi-dimensional index of n in the matrix above
          [3, 1]

Since it finds the first occurrence in the matrix, b (first index) is guaranteed to be minimized, and n always appears at least once in the matrix at the coordinates [n, 1].

Answer 3

Python 2, 43 bytes

k=n=input()
while n%k**2:k-=1
print k,n/k/k

Test it on Ideone.

Answer 4

PARI/GP, 12 bytes

n->core(n,1)

core returns the squarefree part of n by default, but setting the second argument flag to 1 makes it return both parts. Output order is (b, a), e.g. (n->core(n,1))(12) -> [3, 2].

Answer 5

MATL, 12 bytes

t:U\~f0)GyU/

Try it online!

Explanation

t     % Take input n implicitly. Duplicate
:U    % Push [1 4 9 ... n^2]
\~    % True for entries that divide the input
f0)   % Get (1-based) index of the last (i.e. largest) dividing number
G     % Push input again
y     % Duplicate index of largest dividing number
U     % Square to recover largest dividing number
/     % Divide input by that. Implicitly display stack

Answer 6

Julia, 32 bytes

\(n,k=n)=n%k^2>0?n\~-k:[k,n/k^2]

Try it online!

Answer 7

Mathematica 34 bytes

#/.{a_ b_^_:>{a, b},_[b_,_]:>{1,b}}&

This says to replace all the input (#) according to the following rules: (1) a number, a, times the square root of b should be replaced by {a, b} and a function b to the power of whatever should be replaced by {1,b} . Note that the function assumes that the input will be of the form, Sqrt[n]. It will not work with other sorts of input.

This unnamed function is unusually cryptic for Mathematica. It can be clarified somewhat by showing its full form, followed by replacements of the original shorter forms.

Function[
   ReplaceAll[
      Slot[1],
      List[
         RuleDelayed[Times[Pattern[a,Blank[]],Power[Pattern[b,Blank[]],Blank[]]],List[a,b]],
         RuleDelayed[Blank[][Pattern[b,Blank[]],Blank[]],List[1,b]]]]]

which is the same as

   ReplaceAll[
      #,
      List[
         RuleDelayed[Times[Pattern[a,Blank[]],Power[Pattern[b,Blank[]],Blank[]]],List[a,b]],
         RuleDelayed[Blank[][Pattern[b,Blank[]],Blank[]],List[1,b]]]]&

and

ReplaceAll[#, 
  List[RuleDelayed[
    Times[Pattern[a, Blank[]], 
     Power[Pattern[b, Blank[]], Blank[]]], {a, b}], 
   RuleDelayed[Blank[][Pattern[b, Blank[]], Blank[]], {1, b}]]] &

and

ReplaceAll[#, 
  List[RuleDelayed[Times[a_, Power[b_, _]], {a, b}], 
   RuleDelayed[Blank[][b_, _], {1, b}]]] &

and

ReplaceAll[#, {RuleDelayed[a_*b^_, {a, b}], RuleDelayed[_[b_, _], {1, b}]}]&

and

ReplaceAll[#, {a_*b^_ :> {a, b}, _[b_, _] :> {1, b}}] &

Answer 8

05AB1E, 14 bytes

Lv¹ynÖi¹yn/y‚ï

Explained

Lv              # for each x in range(1,N) inclusive
  ¹ynÖi         # if N % x^2 == 0
       ¹yn/y‚ï  # create [N/x^2,x] pairs, N=12 -> [12,1] [3,2]
                # implicitly output last found pair

Try it online

Answer 9

Python, 74 Bytes

def e(k):a=filter(lambda x:k/x**2*x*x==k,range(k,0,-1))[0];return a,k/a**2

Straightforward enough.

Answer 10

Pyth, 15 bytes

/Q^
ef!%Q^T2SQ2

Test suite.

Answer 11

Matlab, 51 bytes

x=input('');y=1:x;z=y(~rem(x,y.^2));a=z(end)
x/a^2

Explanation

x=input('')       -- takes input
y=1:x             -- numbers from 1 to x
z=y(~rem(x,y.^2)) -- numbers such that their squares divide x
a=z(end)          -- biggest such number (first part of output)
x/a^2             -- remaining part

Answer 12

JavaScript (ECMAScript 2016), 40 bytes

n=>{for(k=n;n%k**2;k--);return[k,n/k/k]}

Basically a JavaScript port of Dennis's Python 2 answer.

Try it on JSBin.

Note: it doesn't work in strict mode, because k is not initialized anywhere. To make it work in strict mode, k=n in the loop should be changed to let k=n.

Answer 13

Haskell, 43> 42 bytes

Brute force solution.

Saved 1 byte thanks to Xnor

f n=[(x,y)|y<-[1..],x<-[1..n],x*x*y==n]!!0

Answer 14

APL, 25 chars

 {(⊢,⍵÷×⍨)1+⍵-0⍳⍨⌽⍵|⍨×⍨⍳⍵}

In English:

• 0⍳⍨⌽⍵|⍨×⍨⍳⍵: index of the largest of the squares up to n that divides completely n;
• 1+⍵-: the index is in the reversed array, so adjust the index
• (⊢,⍵÷×⍨): produce the result: the index itself (a) and the quotient b (that is, n÷a*a)

Test:

     ↑{(⊢,⍵÷×⍨)⊃z/⍨0=⍵|⍨×⍨z←⌽⍳⍵}¨⍳36
1  1
1  2
1  3
2  1
1  5
1  6
1  7
2  2
3  1
1 10
1 11
2  3
1 13
1 14
1 15
4  1
1 17
3  2
1 19
2  5
1 21
1 22
1 23
2  6
5  1
1 26
3  3
2  7
1 29
1 30
1 31
4  2
1 33
1 34
1 35
6  1

Answer 15

JavaScript (ECMAScript 6), 35 bytes

f=(n,k=n)=>n/k%k?f(n,--k):[k,n/k/k]

JavaScript 1+, 37 B

for(k=n=prompt();n/k%k;--k);[k,n/k/k]

Answer 16

J, 19 bytes

(0|:0 2#:])&.(_&q:)

Try it online!

Same as the Jelly solution.

Answer 17

Vyxal, 9 bytes

K'∆²;t₍/√

Try it Online!

Very cool and good. Outputs as [constant, surd]

Explained

K'∆²;t₍√/
K        # from the factors of the input, 
 '∆²;    # keep only square numbers
     t   # and keep the last.
      ₍√/ # Push [sqrt(^), input / ^]

Answer 18

Python 2.7 (ungolfed) - 181 Bytes

def e(n):   
 for x in range(1,n+1):
  r=(1,x)
  for i in range(1,x+1):
   l=i**.5
   for j in range(1,x+1): 
    if i*j==x and l%1==0 and j<r[1]:r=(int(l),j)                
  print x,r

Run as: e(number) eg. e(24)

Sample output:

>> e(24)
1 (1, 1)
2 (1, 2)
3 (1, 3)
4 (2, 1)
5 (1, 5)
6 (1, 6)
7 (1, 7)
8 (2, 2)
9 (3, 1)
10 (1, 10)
11 (1, 11)
12 (2, 3)
13 (1, 13)
14 (1, 14)
15 (1, 15)
16 (4, 1)
17 (1, 17)
18 (3, 2)
19 (1, 19)
20 (2, 5)
21 (1, 21)
22 (1, 22)
23 (1, 23)
24 (2, 6)