难题，高级人员请进！

[后八十生人]

编一个程序，对给定的自然数N，找出满足下述关系的最小S：

S=P^N + Q^N = R^N + T^N

其中P,Q,R,T都是自然数，且P,Q和R,T不全同。有解时输出解，否则给出无解信.

高手请用QQ回复我，同时给出数学算法。

[efsnake]

能说的明白点吗，看的糊涂！

[goodname008]

' 算法大概应该就是这样，但数字稍大就不行了。

Option Explicit

Private Sub Command1_Click()
Dim p As Long, q As Long, r As Long, t As Long, n As Long, s As Long
p = 1: q = 2: r = 1: t = 2
n = 1
Do Until p ^ n + q ^ n = r ^ n + t ^ n
n = n + 1
Loop
s = p ^ n + q ^ n
Debug.Print s
End Sub

[后八十生人]

无法满足题目的要求，还有没有其他算法？C语言的也可以。

[northwolves]

'add a listbox to form1,sorted=true
'add a commandbutton
'add these codes for the form

Function s(ByVal n As Integer) As Long
List1.Clear
For i = 1 To 20
For j = i To 20
List1.AddItem Format(i ^ n + j ^ n, "000000000")
Next
Next
For i = 0 To List1.ListCount - 1
s = List1.List(i)
If List1.List(i + 1) = List1.List(i) Then
Exit For
End If
Next
List1.Clear
End Function

Private Sub Command1_Click()
MsgBox "s(1)=" & s(1) & vbCrLf & "s(2)=" & s(2) & vbCrLf & "s(3)=" & s(3) & vbCrLf & "s(4)=" & s(4) & vbCrLf & "s(5)=" & s(5) & vbCrLf & "s(6)=" & s(6)
End Sub

[BraveHorse]

我有一个算法，但是我无法确定是否有解，也就是说会无穷无尽的计算下去。算法如下：
Option Explicit

Private Sub Command1_Click()

Dim S As Long
Dim i As Long
Dim j As Long
Dim m As Long
Dim P As Long
Dim Q As Long
Dim R As Long
Dim T As Long
Dim GeShu As Integer
Dim N As Long
N = CInt(txtN)
GeShu = 0
i = 2

Do While (True)
i = i + 1
GeShu = 0
For j = 1 To CInt(i / 2)
For m = i - 1 To CInt(i / 2) Step -1
If i ^ N = m ^ N + j ^ N Then
GeShu = GeShu + 1
If GeShu = 1 Then
P = m
Q = j
Exit For
ElseIf GeShu >= 2 Then
R = m
T = j
S = i
Exit Do
End If
End If
Next
Next
DoEvents
Loop
MsgBox S
End Sub

[programfish]

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