The equation
a3+b3=c4
has infinite number of positive integer solutions.
Find the two lowest values of S (S=a+b+c).
Strictly speaking the below list is in order by sum of a+b rather than a+b+c, as evidenced by a+b+c=2016 preceding a+b+c=2010, to take the first such example. However, it does include all a+b under 10,000 and thus all a+b+c under 10,000. Only one a+b+c listed is above that value and thus there might exist unlisted entries above 10,000 and below 10,290, the only a+b+c listed that's over 10,000.
The two lowest values sought are 6 and 36.
a b c a+b+c
2 2 2 6
9 18 9 36
32 32 16 80
28 84 28 140
70 105 35 210
162 162 54 378
65 260 65 390
144 288 72 504
273 364 91 728
126 630 126 882
266 665 133 1064
512 512 128 1152
469 603 134 1206
456 760 152 1368
455 1001 182 1638
217 1302 217 1736
793 854 183 1830
756 945 189 1890
448 1344 224 2016
737 1072 201 2010
730 1241 219 2190
729 1458 243 2430
417 1807 278 2502
728 1729 273 2730
1250 1250 250 2750
344 2408 344 3096
721 2060 309 3090
1120 1680 280 3080
702 2457 351 3510
211 3165 422 3798
665 2926 399 3990
1110 2590 370 4070
1705 2046 341 4092
604 3473 453 4530
1628 2849 407 4884
513 4104 513 5130
1136 3905 497 5538
2592 2592 432 5616
1040 4160 520 5720
386 4825 579 5790
2340 3276 468 6084
217 5642 651 6510
1617 4312 539 6468
2304 4608 576 7488
3354 3913 559 7826
730 6570 730 8030
1474 6633 737 8844
3185 5096 637 8918
3666 5170 658 9494
2268 6804 756 9828
4802 4802 686 10290
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
For tot = 2 To 10000
For a = 1 To tot / 2
DoEvents
b = tot - a
c4 = a * a * a + b * b * b
c = Int(Sqr(Sqr(c4)) + 0.5)
If c * c * c * c = c4 Then
Text1.Text = Text1.Text & a & Str(b) & Str(c) & " " & a + b + c & crlf
End If
Next a
Next tot
Text1.Text = Text1.Text & crlf & " done"
End Sub
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Posted by Charlie
on 2015-11-17 18:57:42 |