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m & n (Posted on 2014-12-18) Difficulty: 2 of 5
Let m and n be integers each less than 2014.
Determine the maximum value of m^3+n^3 fulfilling the equation
(n^2 - mn - m^2)^2 = 1

No Solution Yet Submitted by Ady TZIDON    
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Solution computer-aided solution | Comment 1 of 2
Here are all the cases:

m n m^3+n^3
1 1 2
1 -1 0
1 2 9
2 -1 7
2 3 35
3 -2 19
3 5 152
5 -3 98
5 8 637
8 -5 387
8 13 2709
13 -8 1685
13 21 11458
21 -13 7064
21 34 48565
34 -21 30043
34 55 205679
55 -34 127071
55 89 871344
89 -55 538594
89 144 3690953
144 -89 2281015
144 233 15635321
233 -144 9663353
233 377 66231970
377 -233 40933296
377 610 280563633
610 -377 173398367
610 987 1188485803
987 -610 734523803
987 1597 5034507976
1597 -987 3111498370

The max is

987 1597 ---> 5034507976

Some things are apparent: The numbers are the Fibonacci numbers. Only the smaller number, in absolute value, can be negative, so if numbers below -2014 were included, there would be a number in the pair that's larger than 2014, so we need not consider such negative numbers.

 For m = 1 To 2013
  For n = 1 To 2013
    v1 = n * n - m * n - m * m
    If Abs(v1) = 1 Then
      Text1.Text = Text1.Text & m & Str(n) & Str(m * m * m + n * n * n) & crlf
      If m * m * m + n * n * n > mx Then
        mx = m * m * m + n * n * n
        m1 = m: n1 = n
      End If
    End If
    v2 = n * n + m * n - m * m
    If Abs(v2) = 1 Then
      Text1.Text = Text1.Text & m & " " & Str(-n) & Str(m * m * m - n * n * n) & crlf
      If Abs(m * m * m - n * n * n) > mx Then
        mx = Abs(m * m * m - n * n * n)
        m1 = m: n1 = -n
      End If
    End If
  Next
 Next

 Text1.Text = Text1.Text & m1 & Str(n1) & Str(mx)


  Posted by Charlie on 2014-12-18 09:53:58
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