Find a solution to:
x
1^4 + x
2^4 + x
3^4 + ... + x
n^4 = 1999
where each xy is a distinct integer.
(Or prove that it is impossible).
If 1999 is to be made the sum of 4th powers of integers, but we limit the restriction that the bases be distinct to only one occurrence of -1 and one occurrence of 1, but as many occurrences as necessary for the other bases, then the following table shows all the ways of doing that. The distinction of positive or negative of the base is not made in the table, so 1 is allowed to occur twice -- once as positive and once as negative:
1 2 3 4 5 6
2 59 13 0 0 0
1 54 14 0 0 0
0 49 15 0 0 0
2 43 13 1 0 0
1 38 14 1 0 0
0 33 15 1 0 0
2 27 13 2 0 0
1 22 14 2 0 0
0 17 15 2 0 0
2 11 13 3 0 0
1 6 14 3 0 0
0 1 15 3 0 0
2 25 12 0 1 0
1 20 13 0 1 0
0 15 14 0 1 0
2 9 12 1 1 0
1 4 13 1 1 0
where this bottom line for example indicates 1^4 + 4*2^4 + 13*3^4 + 4^4 + 5^4. As mentioned, of course, 4*2^4, for example, could actually be 2*(-2)^4 + 2*2^4.
These are the only ways that still restrict -1 and +1 to one occurrence each as the base.
DECLARE SUB brkDwn (lvl#)
DEFDBL A-Z
DIM SHARED fp(6)
FOR i = 1 TO 6
fp(i) = INT(i ^ 4 + .5)
PRINT fp(i)
NEXT
DIM SHARED used(6), remain
remain = 1999
brkDwn 6
SUB brkDwn (lvl)
FOR i = 0 TO remain / fp(lvl)
remain = remain - i * fp(lvl)
used(lvl) = i
IF remain = 0 THEN
FOR j = 1 TO 6
PRINT USING "####"; used(j);
NEXT
PRINT
ELSE
IF lvl > 2 OR lvl = 2 AND remain < 3 THEN
brkDwn (lvl - 1)
END IF
END IF
remain = remain + i * fp(lvl)
NEXT i
used(lvl) = 0
END SUB
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Posted by Charlie
on 2004-09-12 14:32:23 |
Observation
