Pythagorean Triple!!!

[chaos]

Can someone help me with a program that shows all the possible values for a Pythagorean Triple (i.e. 3,4,5). By the way, im putting it in a list box,if that would say anything

Thanks,

-Keith

[phinds]

since there are an infinite number of such triples, I assume you don't REALLY mean you want them all.

[chaos]

Oh yeah, sorry about that. I meant only up to 100.

Sorry,

-Keith

[Spooner]

Loop from i =1 to 100, and each time through check if i^2 + (i+1)^2 = (i+2)^2.

If it does, keep it and carry on; if it doesn't, just carry on.

[Bruce Fox]

Private Sub Command1_Click()
   Dim a As Long
   Dim b As Long
   
   For a = 1 To 100
      For b = 1 To 100
         List1.AddItem "a= " & a & "  " & "b= " & b & "  " & "Hyp= " & Round(Sqr(a ^ 2 + b ^ 2), 3)
      Next
   Next
End Sub

[Spooner]

Bruce, a Pythagorean triple is 3 consecutive integers which fit the equation, like 3-4-5, not just all the combinations of a & b with c rounded. 4-5-6 isn't one, because the 6 should be 6.4, and 30-31-43 is way off.

[KingMoogle]

Originally posted by Spooner
Loop from i =1 to 100, and each time through check if i^2 + (i+1)^2 = (i+2)^2.

If it does, keep it and carry on; if it doesn't, just carry on.

A Pythagorean triple does not have to be x,x+1,x+2

For example: 5, 12, 13 is a Pythagorean triple, but follows the form x,x+7,x+8. That method would not catch this one.

This should do it, but it will produce some repeats (x,y,z & y,x,z), although it would be easy to fix this.

Dim i As Integer
Dim j As Integer
Dim k As Double
For i = 1 To 100
    For j = 1 To 100
        k = Sqr(i ^ 2 + j ^ 2)
        If k = Int(k) And k <= 100 Then
            List1.AddItem CStr(i) + ", " + CStr(j) + ", " + CStr(k)
        End If
    Next j
Next i

[Spooner]

I could have sworn they had to be consecutive, but I checked up at Mathworld and you're right, they just need to be all integers.

[Frans C]

A little modification of KingMoogle's code, and the double entries are gone.

Dim i As Integer
Dim j As Integer
Dim k As Double
For i = 1 To 100
    For j = i To 100 'changed 1 into i
        k = Sqr(i ^ 2 + j ^ 2)
        If k = Int(k) And k <= 100 Then
            List1.AddItem CStr(i) + ", " + CStr(j) + ", " + CStr(k)
        End If
    Next j
Next i

[MarkT]

Here is my version.

Private Sub Form_Load()
Dim a As Integer
Dim b As Integer
Dim c As Integer
 
    For a = 1 To 100
        For b = a + 1 To 100
            For c = b + 1 To 100
                If (a ^ 2) + (b ^ 2) = c ^ 2 Then
                    List1.AddItem a & ", " & b & ", " & c
                End If
            Next c
        Next b
    Next a
End Sub

[pi3-14159]

As all the Pythagorean triples are given by
a=m^2-n^2
b=2mn
c=m^2+n^2
such that m, n are natural numbers

So we can use the following code to save over 99% of CPU time

For m=2 to 100
    For n=1 to m-1
        If m^2+n^2<=100^2
            List1.AddItem Str(m^2-n^2) & ", " & 2*m*n & ", " & Str(m^2-n^2)
        Else
            Exit For
        End if
    Next n
Next m

[KingMoogle]

pi3-14159, some of the triples that your code gives are over 100. While the method you used for getting the triples works, it's not quite as easy to ensure that you're getting all values between any particular range. (not to mention that it doesn't save that much CPU time)

[pi3-14159]

Thank you very much for your comments.
I've look at my code closely and I've found out what the problem is. It is because I am not following the original mathematical formula rigorously. It should be
For all natural numbers k,m,n , such that m,n are coprime
a=k(m^2-n^2)
b=2kmn
c=k(m^2+n^2)
Then {a,b,c} denotes the set of all Pythagorean triples

But in that case, the program would be much more complicated.

[Bruce Fox]

Originally posted by Spooner
Bruce, a Pythagorean triple is 3 consecutive integers which fit the equation, like 3-4-5, not just all the combinations of a & b with c rounded. 4-5-6 isn't one, because the 6 should be 6.4, and 30-31-43 is way off.

Spooner,

I appreciate that only Int were required (consec sets) and
I didn't alow for that.
However, I would like to point out the code I posted would NOT
do as you stated as I rounded to 3 decimal places. So 4-5-6.403 is
a result, not 4-5-6!


Bruce.