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 Three ways - same sum (Posted on 2016-04-14)
What is the smallest number that can be written as the sum of of 3 distinct squares in 3 ways?

What is the 2nd smallest such number?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 computer solution | Comment 2 of 3 |
The first line of each group identifies the number, then the number of ways. After that are two more numbers: the number of ways if zero is excluded as one of the squares and the number of ways if both zero and one are disallowed.

65 3 1 1
64 1 0
49 16 0
36 25 4
74 3 2 1
64 9 1
49 25 0
49 16 9
89 3 2 2
64 25 0
64 16 9
49 36 4
90 3 2 1
81 9 0
64 25 1
49 25 16
101 4 3 2
100 1 0
81 16 4
64 36 1
49 36 16
110 3 3 2
100 9 1
81 25 4
49 36 25
117 3 2 1
100 16 1
81 36 0
64 49 4
122 3 2 2
121 1 0
81 25 16
64 49 9
125 4 2 2
121 4 0
100 25 0
100 16 9
64 36 25
126 3 3 1
121 4 1
100 25 1
81 36 9
134 3 3 3
121 9 4
100 25 9
81 49 4
145 3 1 1
144 1 0
100 36 9
81 64 0
146 4 3 2
121 25 0
121 16 9
81 64 1
81 49 16
149 4 3 2
144 4 1
100 49 0
81 64 4
64 49 36
161 4 4 3
144 16 1
121 36 4
100 36 25
81 64 16
170 4 2 1
169 1 0
144 25 1
121 49 0
81 64 25
173 4 3 3
169 4 0
144 25 4
121 36 16
100 64 9
174 3 3 2
169 4 1
121 49 4
100 49 25
181 3 2 1
144 36 1
100 81 0
81 64 36
182 3 3 2
169 9 4
121 36 25
100 81 1
185 5 3 3
169 16 0
144 25 16
121 64 0
100 81 4
100 49 36
186 3 3 1
169 16 1
121 64 1
121 49 16
189 4 4 4
169 16 4
144 36 9
121 64 4
100 64 25
194 5 4 3
169 25 0
169 16 9
144 49 1
121 64 9
81 64 49
197 3 2 2
196 1 0
144 49 4
100 81 16

To make them easier to find, repeated below are those where there must be at least three ways discounting zero as an addend:

(within a qualifying number, disqualified ways are still shown, for completeness)

101 4 3 2
100 1 0
81 16 4
64 36 1
49 36 16
110 3 3 2
100 9 1
81 25 4
49 36 25
126 3 3 1
121 4 1
100 25 1
81 36 9
134 3 3 3
121 9 4
100 25 9
81 49 4
146 4 3 2
121 25 0
121 16 9
81 64 1
81 49 16
149 4 3 2
144 4 1
100 49 0
81 64 4
64 49 36
161 4 4 3
144 16 1
121 36 4
100 36 25
81 64 16
173 4 3 3
169 4 0
144 25 4
121 36 16
100 64 9
174 3 3 2
169 4 1
121 49 4
100 49 25
182 3 3 2
169 9 4
121 36 25
100 81 1
185 5 3 3
169 16 0
144 25 16
121 64 0
100 81 4
100 49 36
186 3 3 1
169 16 1
121 64 1
121 49 16
189 4 4 4
169 16 4
144 36 9
121 64 4
100 64 25
194 5 4 3
169 25 0
169 16 9
144 49 1
121 64 9
81 64 49

And if anything less than four is disallowed, here's the list:

134 3 3 3
121 9 4
100 25 9
81 49 4
161 4 4 3
144 16 1
121 36 4
100 36 25
81 64 16
173 4 3 3
169 4 0
144 25 4
121 36 16
100 64 9
185 5 3 3
169 16 0
144 25 16
121 64 0
100 81 4
100 49 36
189 4 4 4
169 16 4
144 36 9
121 64 4
100 64 25
194 5 4 3
169 25 0
169 16 9
144 49 1
121 64 9
81 64 49

DefDbl A-Z
Dim crlf\$, w(20, 3)

Form1.Visible = True

Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)

For n = 5 To 200
ways = 0: ct2 = 0: ct3 = 0
sr0 = Int(Sqr(n))
For b = sr0 To lwrlim Step -1
DoEvents
b2 = b * b: brem = n - b2
For a = b - 1 To lwrlim Step -1
a2 = a * a: arem = brem - a2
If arem >= lwrlim And arem < a2 Then
sr = Int(Sqr(arem) + 0.5)
If sr * sr = arem Then
ways = ways + 1
If arem > 0 Then ct2 = ct2 + 1
If arem > 1 Then ct3 = ct3 + 1
w(ways, 1) = b2
w(ways, 2) = a2
w(ways, 3) = arem
End If
End If
Next a
Next b
If ct3 >= 3 Then
Text1.Text = Text1.Text & n & Str(ways) & Str(ct2) & Str(ct3) & crlf
For i = 1 To ways
Text1.Text = Text1.Text & "    " & w(i, 1) & Str(w(i, 2)) & Str(w(i, 3)) & crlf
Next
End If
Next

Text1.Text = Text1.Text & crlf & " done"

End Sub

The test ct3 >= 3  was at first ways >=3, then ct2 >= 3.

 Posted by Charlie on 2016-04-14 09:42:14

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