Định lý cuối cùng của Fermat nói rằng không có giải pháp tích cực, tích phân a^n + b^n = c^nnào cho phương trình cho bất kỳ n>2. Điều này đã được chứng minh là đúng bởi Andrew Wiles vào năm 1994.

Tuy nhiên, có rất nhiều "gần bỏ lỡ" gần như thỏa mãn phương trình diophantine nhưng lại bỏ lỡ nó. Chính xác, tất cả chúng đều lớn hơn 1 và là các giải pháp tích phân của a^3 + b^3 = c^3 + 1(chuỗi là giá trị của mỗi bên của phương trình, theo thứ tự tăng dần).

Nhiệm vụ của bạn được đưa ra n, để in ra các ngiá trị đầu tiên của chuỗi này.

Dưới đây là một vài giá trị đầu tiên của chuỗi:

1729, 1092728, 3375001, 15438250, 121287376, 401947273, 3680797185, 6352182209, 7856862273, 12422690497, 73244501505, 145697644729, 179406144001, 648787169394, 938601300672, 985966166178, 1594232306569, 2898516861513, 9635042700640, 10119744747001, 31599452533376, 49108313528001, 50194406979073, 57507986235800, 58515008947768, 65753372717929, 71395901759126, 107741456072705, 194890060205353, 206173690790977, 251072400480057, 404682117722064, 498168062719418, 586607471154432, 588522607645609, 639746322022297, 729729243027001

Đây là mã golf , vì vậy mã ngắn nhất tính bằng byte sẽ thắng!

Thạch , 16 byte

*3‘
ḊŒc*3S€ċǵ#Ç

Giải pháp vũ lực. Hãy thử trực tuyến!

*3‘           Helper link. Maps r to r³+1.

ḊŒc*3S€ċǵ#Ç  Main link. No arguments.

         µ    Combine the links to the left into a chain.
          #   Read an integer n from STDIN and execute the chain to the left for
              k = 0, 1, 2, ... until n matches were found. Yield the matches.
Ḋ             Dequeue; yield [2, ..., k].
 Œc           Yield all 2-combinations of elements of that range.
   *3         Elevate the integers in each pair to the third power.
     S€       Compute the sum of each pair.
        Ç     Call the helper link, yielding k³+1.
       ċ      Count how many times k³+1 appears in the sums. This yields a truthy 
              (i.e., non-zero) integer if and only if k is a match.
           Ç  Map the helper link over the array of matches.

Brachylog , 31 byte

:{#T#>>:{:3^}aLhH,Lb+.-H,#T=,}y

Hãy thử trực tuyến!

Đây không phải là hoàn toàn vũ phu vì điều này sử dụng các ràng buộc. Đây là một chút chậm trên TIO (khoảng 20 giây cho N = 5). Mất khoảng 5 giây cho N = 5và 13 giây cho N = 6máy của tôi.

Giải trình

:{                           }y    Return the first Input outputs of that predicate
  #T                               #T is a built-in list of 3 variables
    #>                             #T must contain strictly positive values
      >                            #T must be a strictly decreasing list of integers
       :{:3^}aL                    L is the list of cubes of the integers in #T
              LhH,                 H is the first element of L (the biggest)
                  Lb+.             Output is the sum of the last two elements of L
                     .-H,          Output - 1 = H
                         #T=,      Find values for #T that satisfy those constaints

Perl, 78 byte

#!perl -nl
grep$_<(($_+2)**(1/3)|0)**3,map$i**3-$_**3,2..$i++and$_-=print$i**3+1while$_

Phương pháp vũ phu. Giới thiệu shebang là hai, đầu vào được lấy từ stdin.

Sử dụng mẫu

$ echo 10 | perl fermat-near-miss.pl
1729
1092728
3375001
15438250
121287376
401947273
3680797185
6352182209
7856862273
12422690497

Hãy thử trực tuyến!

Mathematica, 95 bytes

(b=9;While[Length[a=Select[Union@@Array[#^3+#2^3&,{b,b},2],IntegerQ[(#-1)^3^-1]&,#]]<#,b++];a)&

Unnamed function taking a single positive integer argument # and returning a list of the desired # integers. Spaced for human readability:

1  (b = 9; While[
2    Length[ a =
3      Select[
4        Union @@ Array[#^3 + #2^3 &, {b, b}, 2],
5        IntegerQ[(# - 1)^3^-1] &
6      , #]
7    ] < #, b++
8  ]; a) &

Line 4 calculates all possible sums of cubes of integers between 2 and b+1 (with the initialization b=9 in line 1) in sorted order. Lines 3-5 select from those sums only those that are also one more than a perfect cube; line 6 limits that list to at most # values, which are stored in a. But if this list has in fact fewer than # values, the While loop in lines 1-7 increments b and tries again. Finally, line 8 outputs a once it's the right length.

Holy hell this version is slow! For one extra byte, we can change b++ in line 7 to b*=9 and make the code actually run in reasonable time (indeed, that's how I tested it).

Racket 166 bytes

(let((c 0)(g(λ(x)(* x x x))))(for*((i(in-naturals))(j(range 1 i))(k(range j i))#:final(= c n))
(when(=(+(g j)(g k))(+ 1(g i)))(displayln(+ 1(g i)))(set! c(+ 1 c)))))

Ungolfed:

(define (f n)
  (let ((c 0)
        (g (λ (x) (* x x x))))
    (for* ((i (in-naturals))
           (j (range 1 i))
           (k (range j i))
           #:final (= c n))
      (when (= (+ (g j) (g k))
               (+ 1 (g i)))
        (displayln (+ 1(g i)))
        (set! c (add1 c))))))

Testing:

(f 5)

Output:

1729
1092728
3375001
15438250
121287376

Python 2, 102 98 bytes

def f(n,c=2):z=c**3+1;t=z in[(k/c)**3+(k%c)**3for k in range(c*c)];return[z]*n and[z]*t+f(n-t,c+1)

Try it online!

Pari/GP, 107 bytes

F(n)=c=2;while(n>0,c++;C=c^3+1;a=2;b=c-1;while(a<b,K=a^3+b^3;if(K==C,print(C);n--;break);if(K>C, b--,a++)))

Finds the first 10 solutions in 10 sec.

Goal: a^3+b^3 = c^3+1

1. Gets number of required solutions by function-argument n

2. Increases c from 3 and for each c^3+1 searches a and b with 1 < a <= b < c such that a^3+b^3=c^3+1 . If found, decrease required number of further soulutions n by 1 and repeat

3. Finishes, when number of furtherly required solutions (in n) equals 0

Call it to get the first ten solutions:

F(10)

Readable code (requires leading and trailing braces as indicators for block-notation of the function. Also for convenience prints all variables of a solution):

{F(m) = c=2;
   while(m>0,        
     c++;C=c^3+1;             
     a=2;b=c-1;                
     while(a<b,                
           K=a^3+b^3;               
            if(K==C,print([a,b,c,C]);m--;break);
            if(K>C, b--,a++);
          );
    );}

Pari/GP, 93 bytes

(Improvement by Dennis)

F(n)=c=2;while(n,C=c^3+1;a=2;b=c++;while(a<b,if(K=a^3+b^3-C,b-=K>0;a+=K<0,print(C);n--;b=a)))              

Python 2, 122 119 Bytes

^{why are you still upvoting? Dennis crushed this answer ;)}

Welcome to the longest solution to this question :/ I managed to shave off a whole byte by making longer conditions and removing as much indentation as possible.

x,y,z=2,3,4
n=input()
while n:
 if y**3+x**3-z**3==1and x<y<z:print z**3+1;n-=1
 x+=1
 if y<x:y+=1;x=2
 if z<y:z+=1;y=3

Output for n = 5:

1729
1092728
3375001
15438250
121287376

TI-Basic, 90 bytes

There must be a shorter way...

Prompt N
2->X
3->Y
4->Z
While N
If 1=X³+Y³-Z³ and X<Y and Y<Z
Then
DS<(N,0
X+1->X
If Y<X
Then
2->X
Y+1->Y
End
If Z<Y
Then
3->Y
Z+1->Z
End
End

MATLAB, 94 bytes

Another brute-force solution:

for z=4:inf,for y=3:z,for x=2:y,c=z^3+1;if x^3+y^3==c,n=n-1;c,if~n,return,end,end,end,end,end

Output for n=4:

>> n=4; fermat_near_misses    
c =
        1729
c =
     1092728
c =
     3375001
c =
    15438250

Suppressing the c= part of the display increases the code to 100 bytes

for z=4:inf,for y=3:z,for x=2:y,c=z^3+1;if x^3+y^3==c,n=n-1;disp(c),if~n,return,end,end,end,end,end

>> n=4; fermat_near_misses_cleandisp    
        1729
     1092728
     3375001
    15438250

C#, 188 174 187 136 bytes

Golfed version thanks to TheLethalCoder for his great code golfing and tips (Try online!):

n=>{for(long a,b,c=3;n>0;c++)for(a=2;a<c;a++)for(b=a;b<c;b++)if(a*a*a+b‌​*b*b==c*c*c+1)System‌​.Console.WriteLin‌e(‌​c*c*(a=c)+n/n--);};

Execution time to find the first 10 numbers: 33,370842 seconds on my i7 laptop (original version below was 9,618127 seconds for the same task).

Ungolfed version:

using System;

public class Program
{
    public static void Main()
    {
        Action<int> action = n =>
        {
            for (long a, b, d, c = 3; n > 0; c++)
                for (a = 2; a < c; a++)
                    for (b = a; b < c; b++)
                        if (a * a * a + b‌ * b * b == c * c * c + 1)
                            System‌.Console.WriteLin‌e( c * c * (a = c) + n / n--);
        };

        //Called like
        action(5);
    }
}

Previous golfed 187 bytes version including using System;

using System;static void Main(){for(long a,b,c=3,n=int.Parse(Console.ReadLine());n>0;c++)for(a=2;a<c;a++)for(b=a;b<c;b++)if(a*a*a+b*b*b==c*c*c+1)Console.WriteLin‌​e(c*c*(a=c)+n/n--);}

Previous golfed 174 bytes version (thanks to Peter Taylor):

static void Main(){for(long a,b,c=3,n=int.Parse(Console.ReadLine());n>0;c++)for(a=2;a<c;a++)for(b=a;b<c;b++)if(a*a*a+b*b*b==c*c*c+1)Console.WriteLin‌​e(c*c*(a=c)+n/n--);}

Previous (original) golfed 188 bytes version (Try online!):

static void Main(){double a,b,c,d;int t=0,n=Convert.ToInt32(Console.ReadLine());for(c=3;t<n;c++)for(a=2;a<c;a++)for(b=a;b<c;b++){d=(c*c*c)+1;if(a*a*a+b*b*b==d){Console.WriteLine(d);t++;}}}

Execution time to find the first 10 numbers: 9,618127 seconds on my i7 laptop.

This is my first ever attempt in C# coding... A bit verbose compared to other languages...

GameMaker Language, 119 bytes

Why is show_message() so long :(

x=2y=3z=4w=argument0 while n>0{if x*x*x+y*y*y-z*z*z=1&x<y&y<z{show_message(z*z*z+1)n--}x++if y<x{x=2y++}if z<y{y=3z++}}

x,y,z=2,3,4 n=input() while n: if y3+x3-z3==1and x3+1;n-=1 x+=1 if y

Axiom, 246 bytes

h(x:PI):List INT==(r:List INT:=[];i:=0;a:=1;repeat(a:=a+1;b:=1;t:=a^3;repeat(b:=b+1;b>=a=>break;q:=t+b^3;l:=gcd(q-1,223092870);l~=1 and(q-1)rem(l^3)~=0=>0;c:=round((q-1)^(1./3))::INT;if c^3=q-1 then(r:=cons(q,r);i:=i+1;i>=x=>return reverse(r)))))

ungof and result

-- hh returns x in 1.. numbers in a INT list [y_1,...y_x] such that 
-- for every y_k exist a,b,c in N with y_k=a^3+b^3=c^3+1 
hh(x:PI):List INT==
   r:List INT:=[]
   i:=0;a:=1
   repeat
      a:=a+1
      b:=1
      t:=a^3
      repeat
          b:=b+1
          b>=a=>break
          q:=t+b^3
          l:=gcd(q-1,223092870);l~=1 and (q-1)rem(l^3)~=0 =>0 -- if l|(q-1)=> l^3|(q-1)
          c:=round((q-1.)^(1./3.))::INT
          if c^3=q-1 then(r:=cons(q,r);i:=i+1;output[i,a,b,c];i>=x=>return reverse(r))

(3) -> h 12
   (3)
   [1729, 1092728, 3375001, 15438250, 121287376, 401947273, 3680797185,
    6352182209, 7856862273, 12422690497, 73244501505, 145697644729]
                                                       Type: List Integer