3 added 181 characters in body
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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)))              

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, 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)))              
2 deleted 66 characters in body
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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++)))

Pari/GP-interpreter, 107 chars findsFinds 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

Oneliner 107 chars: as function F(n)

 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++)))

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-interpreter, 107 chars 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

Oneliner 107 chars: as function F(n)

 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++)))

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, 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++);
          );
    );}
1
source | link

Pari/GP-interpreter, 107 chars 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

Oneliner 107 chars: as function F(n)

 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++)))

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++);
          );
    );}