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Challenges

Solve Goldbach's Conjecture

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Goldbach's Conjecture states that every even whole number greater than 2 is the sum of 2 prime numbers. Your task is to return those 2 prime numbers, given an even whole number as input. There are often multiple solutions - any solution will do.

Input/Output Examples

These examples only show one of potentially many possible outputs.

4 -> 2, 2
6 -> 3, 3
24 -> 5, 19
120 -> 7, 113
1000 -> 3, 997

This is code-golf, so shortest answer in each language wins.

P.S. If no one finds a test case that has no solution, I'll consider the problem solved by engineer's induction.

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10 answers

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J, 46 char

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:

Sample runs

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:4

2 2

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:6

3 3

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:24

5 19

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:120

7 113

({.(+/m)#a),{:(+/"1 m=:x=+/~a)#a=:p:i._1 p:x=:1000

3 997

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JavaScript (Node.js), 87 bytes

f=(a,b=2,c=a-b,d=(a,b=2)=>b<a?a%b&&d(a,b+1):1,e=a=>d(++a)?a:e(a))=>d(c)?[b,c]:f(a,e(b))

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Husk, 9 bytes

ḟo=⁰Σπ2İp

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Explanation

ḟo=⁰Σπ2İp
       İp take the infinite list of primes
     π2   cartesian power 2 (all possible pairs)
ḟo        first pair which satisfies:
    Σ     sum
  =       equals
   ⁰      input?
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APL (Dyalog Unicode), 32 bytes (SBCS)

{⊃(⊢(/⍨)⍵=+/¨),∘.,⍨(⊢~∘.×)⍨1↓⍳⍵}

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{⊃(⊢(/⍨)⍵=+/¨),∘.,⍨(⊢~∘.×)⍨1↓⍳⍵}
                               ⍳⍵ ⍝ Range from 1 to n
                             1↓   ⍝ Drop the 1st number (make 2 to n)
                    (⊢~∘.×)⍨     ⍝ Filter prime numbers
                           ⍨     ⍝ Use range as both arguments for train
                       ∘.×       ⍝ Outer product - multiply all numbers 
                                 ⍝ in [2..n] by every other number in [2..n],
                                 ⍝ giving all composite numbers up to n
                      ~          ⍝ Remove those composite numbers from
                    ⊢           ⍝ that same range (2 to n)
               ∘.,⍨             ⍝ Cartesian product with itself
              ,                 ⍝ Flatten into vector of prime pairs
  (⊢(/⍨)⍵=+/¨)                 ⍝ Filter the ones that sum to n
           +/¨                  ⍝ Map each pair to its sum
        ⍵=                      ⍝ Check if it equals n
     (/⍨)                       ⍝ Keep elements where the pair equals n in
  ⊢                            ⍝ that same vector of prime pairs
 ⊃                             ⍝ Get the first pair that works

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Japt, 13 bytes

o ï æ@¶Xx*Xej

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o ï æ@¶Xx*Xej     :Implicit input of integer U
o                 :Range [0,U)
  ï               :Cartesian product
    æ             :First pair X that returns true
     @            :When passed through the following function
      ¶           :  Test U for equality with
       Xx         :  X Reduced by addition
         *        :  Multiplied by
          Xe      :  Is every element in X
            j     :    Prime?
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J, 50 bytes

{{({~[:y&(i.&1@:=)[:>+/&.>),{@(,&<)~p:i.p:^:_1 y}}

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A nasty DD solution.

{{({~[:y&(i.&1@:=)[:>+/&.>),{@(,&<)~p:i.p:^:_1 y}}
                                    p:i.p:^:_1 NB. primes less than y
                            {@(,&<)            NB. cartesian product, 2d array of boxed pairs
                           ,                   NB. flatten boxes
  ({~[:y&(i.&1@:=)[:>+/&.>)                    NB. monadic hook
                  [:>+/&.>                     NB. sum each pair and unbox
     [:y&(i.&1@:=)                             NB. first occurrence of y in list of sums
   {~                                          NB. flip arguments to index boxed pairs
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Python 3.8, 112 bytes

def f(x):y=[i for i in range(2,x)if not[j for j in range(2,i)if i%j<1]];return[(q,x-q)for q in y if x-q in y][0]

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Ruby, 69 bytes

require'prime';->n{Prime.first(n).then{_1.product _1}.find{_1+_2==n}}

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MATL, 19 bytes

XH:YqtZ*!tsH=f1Z)Z)

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I'm still a novice at MATL, so I am all ears to any improvements.

XH:YqtZ*!tsH=f1Z)Z)
XH                  - copy implicit input to H clipboard
  :                 - range 1..n
   Yq               - vectorized n-th prime
     t              - duplicate
      Z*            - cartesian product
        !           - transpose
         t          - duplicate
          s         - sum columns
           H=       - paste from clipboard H and vectorized equal
             f      - find indices of nonzeros
              1Z)   - first index returned from f
                 Z) - use that index to find first pair from Z*!
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Ruby, 62 bytes

require'prime';f=->n,k=0{n.prime?&&k.prime?? [n,k]:f[n-1,k+1]}

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