For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the ``perfect cube'' equation a^3=b^3 c^3 d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 8^3 10^3is indeed true). This problem requires that you write a program to find all sets of numbers {a, b, c, d} which satisfy this equation for . The first part of the output is shown here: Cube = 6, Triple = (3,4,5) Cube = 12, Triple = (6,8,10) Cube = 18, Triple = (2,12,16) Cube = 18, Triple = (9,12,15) Cube = 19, Triple = (3,10,18) Cube = 20, Triple = (7,14,17) Cube = 24, Triple = (12,16,20)

這是歷史上很有名證明題嗎? 不是很難解嗎? 用到最後只能用理論方式證明.... 練習! 練習! 再練習!

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嘿嘿嘿

供參考 這個好像是作業....    

    //---------------------------------------------------------------------------
#include 
#pragma hdrstop    #include "Unit1.h"
#include "math.h"
//---------------------------------------------------------------------------
#pragma package(smart_init)
#pragma resource "*.dfm"    long Sum[1000];
TForm1 *Form1;
//---------------------------------------------------------------------------
__fastcall TForm1::TForm1(TComponent* Owner)
    : TForm(Owner)
{
    for (int i=0;i< 1000 ; i  )
    {
        Sum[i] = pow( (i 1) , 3 ) ;
    };
}
//---------------------------------------------------------------------------    void __fastcall TForm1::Button1Click(TObject *Sender)
{
    long KeyIn; //輸入數字
    KeyIn = Edit1->Text.ToInt()-1 ;        RichEdit1->Text ="" ;        for (int A = (KeyIn-1); A >2; A--)
    {
        if ( Sum[KeyIn] > ( Sum[A]   Sum[A-1]  Sum[A-2] ) )
        {
            break;
        } ;            for (int B = (A-1) ; B >1;B--)
        {
            if ( Sum[KeyIn] > ( Sum[A]   Sum[B]  Sum[B-1] ) )
            {
                break;
            }                for (int C = (B-1) ; C >0;C--)
            {
                if ( Sum[KeyIn] == ( Sum[A]   Sum[B]  Sum[C] ) )
                {   //OK
                    RichEdit1->Text =  RichEdit1->Text  
                    IntToStr(A 1)   "  ;  "  
                    IntToStr(B 1)   "  ;  "  
                    IntToStr(C 1)   "  :" ;
                    RichEdit1->Text = RichEdit1->Text   "\n";
                } ;
            } ;
        };
    } ;
} ;
//---------------------------------------------------------------------------    

練習! 練習! 再練習!

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嘿嘿嘿