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
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 |
computer-aided solution
