Trois entiers positifs A, B, C sont triples ABC s'ils sont premiers, avec A <B et satisfaisant la relation: A + B = C

Exemples :

• 1, 8, 9 est un ABC-triple car ils sont premiers, 1 <8 et 1 + 8 = 9
• 6, 8, 14 n'est pas parce qu'ils ne sont pas coprime
• 7, 5, 12 n'est pas parce que 7> 5

Vous pouvez voir cette présentation de Frits Beukers 2005 pour plus de détails sur les triplets ABC.

Entrée sortie

Trois entiers, décimaux écrits. Peut être des valeurs séparées ou une liste. La sortie devait être une valeur véridique / fausse, que les trois entiers soient un triple ABC.

Remarque: Il est important de respecter l'ordre des entiers dans la liste, par exemple: 1, 8, 9n'est pas considéré comme la même liste 9, 1, 8ou toute autre combinaison. Le premier est donc un triple ABC et le second ne l'est pas.

Ainsi, A est le premier élément de la liste, B le deuxième et C le troisième.

Cas de test

Chacune des listes suivantes doit produire une valeur véridique

[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]

Chacune des listes suivantes doit afficher une valeur de falsey

[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]

Gelée , 10 9 octets

Ṫ=S×</=g/

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Comment ça fonctionne

Ṫ=S×</=g/  Main link. Argument: [a, b, c] (positive integers)

Ṫ          Tail; pop and yield c.
  S        Take the sum of [a, b], yielding (a + b).
 =         Yield t := (c == a + b).
    </     Reduce by less than, yielding (a < b).
   ×       Multiply, yielding t(a < b).
       g/  Reduce by GCD, yielding gcd(a, b).
      =    Check if t(a < b) == gcd(a, b).

Haskell , 48 38 29 octets

^{-10 octets en raison de TFeld de » gcdtruc!}

^{-7 octets grâce à HPWiz pour avoir amélioré le test de co-primalité et repéré un espace superflu!}

^{-2 octets grâce à nimi avoir proposé un infix-opérateur!}

(a!b)c=a<b&&a+b==c&&gcd a b<2

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Explication

Les deux premières conditions a < bet a + b == csont assez évidentes, le troisième utilisations qui $\gcd(a,b) = \gcd(a,c) = \gcd(b,c)$ :

Écrire $\gcd(a,c) = U \cdot a + V \cdot c$ utilisant l'identité de Bézout et en substituant $c = a + b$ donne:

Puisque le $\gcd$ est la solution positive minimale à cette identité, il s'ensuit que $\gcd(a,b) = \gcd(a,c)$ . L'autre cas est symétrique.

Perl 6 , 33 32 octets

-1 octet grâce à nwellnhof

{(.sum/.[2]/2*[<] $_)==[gcd] $_}

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Bloc de code anonyme qui prend une liste de trois nombres et renvoie Vrai ou Faux.

Explication

{                              }  # Anonymous code block
                       [gcd] $_   # Is the gcd of all the numbers
 (                  )==           # Equal to
  .sum        # Whether the sum of numbes
      /       # Is equal to
       .[2]/2 # The last element doubled
             *[<] $_   # And elements are in ascending order

Excel, 33 octets

=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)

bash, 61 octets

factor $@|grep -vzP '( .+\b).*\n.*\1\b'&&(($1<$2&&$1+$2==$3))

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Entrée comme arguments de ligne de commande, sortie dans le code de sortie (produit également une sortie sur stdout comme effet secondaire, mais cela peut être ignoré).

The second part (starting from &&(() is pretty standard, but the interesting bit is the coprime test:

factor $@      # produces output of the form "6: 2 3\n8: 2 2 2\n14: 2 7\n"
|grep -        # regex search on the result
v              # invert the match (return truthy for strings that don't match)
z              # zero-terminated, allowing us to match newlines
P              # perl (extended) regex
'( .+\b)'      # match one or more full factors
'.*\n.*'       # and somewhere on the next line...
'\1\b'         # find the same full factors

Java 10, 65 64 bytes

(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}

-1 byte thank to @Shaggy.

Try it online.

Explanation:

(a,b,c)->{        // Method with three integer parameters and boolean return-type
  var r=          //  Result-boolean, starting at:
        a<b       //   Check if `a` is smaller than `b`
        &a+b==c;  //   And if `a+b` is equal to `c`
  for(;b>0        //  Then loop as long as `b` is not 0 yet
      ;           //    After every iteration:
       a=b,       //     Set `a` to the current `b`
       b=c)       //     And set `b` to the temp value `c`
    c=a%b;        //   Set the temp value `c` to `a` modulo-`b`
                  //   (we no longer need `c` at this point)
  return r        //  Return if the boolean-result is true
         &a<2;}   //  And `a` is now smaller than 2

05AB1E, 12 11 10 bytes

Saved 1 byte thanks to Kevin Cruijssen

ÂÆ_*`\‹*¿Θ

Try it online! or as a Test Suite

Explanation

ÂÆ           # reduce a reversed copy of the input by subtraction
  _          # logically negate
   *         # multiply with input
    `        # push the values of the resulting list separately to stack
     \       # remove the top (last) value
      ‹      # is a < b ?
       *     # multiply by the input list
        ¿    # calculate the gcd of the result
         Θ   # is it true ?

Python 2, 69 67 63 62 55 bytes

lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*

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Python 3, 58 51 bytes

lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*

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_{-7 bytes, thanks to H.PWiz}

Japt, 16 14 13 11 bytes

<V¥yU «NÔr-

Try it

                :Implicit input of integers U=A, V=B & W=C
<V              :Is U less than V?
  ¥             :Test that for equality with
   yU           :The GCD of V & U
      «         :Logical AND with the negation of
       N        :The array of inputs
        Ô       :Reversed
         r-     :Reduced by subtraction

JavaScript (ES6),  54 43 42  40 bytes

Thanks to @Shaggy for pointing out that we don't need to compute $\gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.

Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.

f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2

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Wolfram Language 24 30 28 26 bytes

With 2 bytes shaved by Doorknob. A further 2 bytes shaved off by @jaeyong sung

#<#2&&GCD@##==1&&#+#2==#3&

C# (Visual C# Interactive Compiler), 90 bytes

n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]

Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.

Try it online!

C# (Visual C# Interactive Compiler), 59 bytes

(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c

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ECMAScript Regex, 34 bytes

Input is in unary, in the domain ^x*,x*,x*$ (repeated xs delimited by ,).

^(?!(xx+)\1*,\1+,)(x*)(,\2x+)\3\2$

Try it online! (.NET regex engine)
Try it online! (SpiderMonkey regex engine)

# see /codegolf/178303/find-if-a-list-is-an-abc-triple
^
(?!                # Verify that A and B are coprime. We don't need to include C in the
                   # test, because the requirement that A+B=C implies that A,B,C are
                   # mutually comprime if and only if A and B are coprime.
    (xx+)\1*,\1+,  # If this matches, A and B have a common factor \1 and aren't coprime.
)
(x*)(,\2x+)\3\2$   # Verify that A<B and A+B=C. The first comma is captured in \3 and
                   # reused to match the second comma, saving one byte.

The question does say "Three integers, decimal written", so this might not qualify (as it takes input in unary), but it makes for such an elegant pure regex that I hope it will at least be appreciated.

However, note that if the phrasing is to be literally interpreted, lambda and function submissions that take integer arguments are to be disqualified too, as to strictly adhere to the question's specification they would need to take the input in the form of a string.

J, 27 bytes

(+/=2*{:)*({.<1{])*1=+./ .*

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Inspired by Jo King's Perl solution

C# (.NET Core), 68 bytes

Without Linq.

(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};

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Stax, 12 bytes

ü╡v╕7+Pü°╔|g

Run and debug it

Clean, 43 bytes

import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2

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Similar to basically everything else because the direct test is the same.

Pari/GP, 30 bytes

Saved 2 bytes thanks to @Shaggy.

(a,b,c)->a<b==gcd(a,b)&&a+b==c

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Befunge-98 (FBBI), 83 bytes

&:&:03p&:04pw>03g04g\:v_1w03g04g+w1.@
00:    7j@.0[^j7      _^;>0.@;j7;>0.@;:%g00\p

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The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.

Mathematica 35 bytes

CoprimeQ @@ # && #[[1]] + #[[2]] == #[[3]] & 

if order is important:

CoprimeQ @@ # && Sort[#]==# && #[[1]] + #[[2]] == #[[3]] & 

or...

And[CoprimeQ @@ #, Sort@# == #, #[[1]] + #[[2]] == #[[3]]] &

Retina 0.8.2, 42 41 bytes

\d+
$*
A`^(11+)\1*,\1+,
^(1+)(,1+\1)\2\1$

Try it online! Link includes test cases. Edit: Saved 1 byte thanks to @Deadcode. Explanation:

\d+
$*

Convert to unary.

A`^(11+)\1*,\1+,

Check that A and B have no common factor.

^(1+)(,1+\1)\2\1$

Check that A < B and A + B = C.

Common Lisp, 51 bytes

(lambda(a b c)(and(< a b)(=(+ a b)c)(=(gcd a c)1)))

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