ON FIFTH-POWER NUMBERS WHOSE SUM IS
A FIFTH POWER.
BY
ARTEMAS MARTIN OF WASHINGTON.
IT is known that the sum of two fifth-power numbers cannot be
a rational fifth power, but, so far as the present writer knows, it
has not been proved that the sum of three, of four, and of five
fifth-power numbers cannot be a fifth power. The writer, however,
has not been able to discover fewer than sisc fifth-power numbers
whose sum is a fifth power, and thus far has succeeded in finding
only two such sets, although probably many exist.
In the present state of algebraic science, fifth-power numbers
whose sum is a fifth power can be most easily found by resorting
to some artifice or some tentative process, two of which methods
it is the object of this paper to present.
1. Take any two numbers p and q and put p5 — q5 = d; then,
by transposition
q5 + d=p*
(A).
Now if d can in any way be separated' into fifth-power
numbers, all different and p5 not among them, we shall obviously
have
q5 + (these fifth-power numbers) = p5
(B).
Examples.
1. In (A), take p = 12, g = 11; then we have
d = 87781 = 9 5 H-7 B + 66+ 55 + 45;
therefore by (B), 45 + 55 + 65 + 75 + 95 + 11B = 125,
six fifth-power numbers whose sum is a fifth power.
2. Take p = 30, q = 29; then
d = 3788851 = 195 +165 + II5 + 105 + 55;
therefore
55 + 105 + II5 + 165 +195 + 295 = SO5,
another set of six fifth-power numbers whose sum is a fifth power.
ON FIFTH-POWER NUMBERS.
3.
169
Take p = 32, q = 31, then
d = 4925281 = 18B -f!6 B + 15B -f 14B + IS5 + 11B
+ 10B + 8 B -|-7 B H-6 B + 3B;
therefore
35 + fj5 + 76 + s* + 1QB + II5 + 13B + 148 + 15B + 165 + 18B + 31B = 32B,
twelve fifth-power numbers whose sum is a fifth power.
4.
Take p = 20, g = 19 ; then
d = 723901 = 13B + 12B + 9B + 8B + 6B + 5B + 4B + 2B + 1B + 1B ;
therefore
1B + 1B + 2B + 4B + 5B + 6B + 8B + 95 + 12B + 13B + 1 95 = 20B,
in which 1B appears twice, a remarkable set.
5.
Take p = 22, q = 21 ; then
d = 1069531 = 16B H- 7e + 5B -f 4B - I6 ;
therefore
6.
4B + 5B + 7B + 16B + 21B = I6 + 22B.
Take p = 36, q = 33 ; then
d = 21330783 = 27B + 21B + IS6 + 15B + 11B + 9B + 7B + 65 + 5B + 4B ;
therefore
5B + 6B -f 75 + 9B + 11B+ 15B + 185 + 21B -f 27B + 33B= 36B.
7.
Take p = 40, q = 39 ; then
d = 12175801 = 25B + 176 + IS5 + 12B + IP -f 10B
+ 9B + 8B -f 7B + 3B + 3B + 2B + I6 ;
therefore
B
+ 3B + 3B + 7B + 8B -f 9B + 10B + II6 + 12B
in which 35 appears twice.
8. Take p = 51, q = 50 ; then
d = 32525251 = 30B + 23B + 16B + 13B + 12B + 10B + 7B + 5B + ^B + 2B ;
... 2B + 3B + 5B + 7B -1- 10B + 12B + 13B + 16B + 23B + 30B + 50B = 516.
Also, 32525251 = 30B + 20B + 19B 4- 17B + 15B-f 11B+ 10B + 9 B -i-8 B
+ 7B + 3B + 3 B +1 B +1 B ;
+ 20B + 30 B +50 B =51 B ,
B
another set in which 1 appears twice.
170
II.
ARTEMAS MARTIN.
Put j8f (tf5) = T + 2B 4- 3B + 4B + 5B + . ..
-1) ......... (G).
Assume b less than S (of) and put r for their difference, and we
have
S(fl*)-&« = r,
or by transposition of b5 and r,
r:=6 B ........................ (D).
If r can be separated into fifth-power numbers, all different
and none of them greater than ac5, we shall evidently have
8 (of) — (these fifth-power numbers) = 6B ......... (E).
I devised this formula in 1887, and have used it in finding
square numbers whose sum is a square ; cube numbers whose sum
is a cube ; biquadrate numbers whose sum is a biquadrate ; fifthpower numbers whose sum is a fifth power, and sixth-power
numbers whose sum is a sixth power. See tl^Q Mathematical
Magazine, Vol. u., No. 6^ pp. 89 — 96 ; Quarterly Journal of
Mathematics, No. 103, pp. 225—227.
5
Examples. 9. In (D), take x = 1 1 ; then S (of) = 381876.
Take 6 = 12, then
r = 133044 = 105 + 8B + 3B + 2B + 1B ;
therefore, by (E),
I5 + 25 + 35 + . . . + 1 15 - (I5 + 2B + 35 + 86 + 105) = 12B,
or
4 B +5 B + 6B + 7 B +9 B + 115 = 12BJ
the same as found in Ex. 1 by the first method.
10. Take x = 22, then S (x5) = 2157103. Take b = 24, then
r = 13608409 = 215 + 20B -h 19B + 17B + 16B + 15B + 13B
therefore, by (E),
4B + 5B + 6B + 7B + 8B + 9B + 10B + II5 -M45 + 18B + 22B = 24B.
11. Take x = 35, then S 0B) = 333263700. Take b = 50, then
r = 20763700 = 26B + 24B + 14B + II5 + 10B + 9B -f . . . + 1B ;
.-. 12B + 13B + 15B 4- 165 + 17B + . . . + 23B + 25B + 27B + 28B
+ 29 e + ...+35 B =50 5 .
ON FIFTH-POWER NUMBERS.
171
Also, 20763700 = 26B + 24B + 14B + 12G + 10B + 85 + 35 4- 25 + 1B ;
... 4,' + G « + 6 B + 7 B + 9 B + llB + 13B + 15B + 16 B +l7 B + ... + 23"
+ 255 + 275 + 285 + 29B + . . . + 35B = 50B.
Again,
20763700 = 26B + 22B + 1 9B + 16B+ 105 + 9B + 8B+ 6B+ 5B+ 4B+ 35+ 2B;
. •.
I5 + 76 + 11B + 12B -1- 13B + 14e + 155 + 17B + 18B + 20B + 215
+ 23B + 24B + 25B + 27B + 28B + 29B + . . . + 355 = 50B,
12. Take a? = 46, then S(x*) = 1683896401. Take 6 = 70,
then
r = 3196401 = 17B + 15B + 14B + 13B + 10B + 6B + 35 4- 2B + I5 ;
It may be well to remind those who would object to these
methods on the ground that they are tentative and not rigorous
because d and r have to be separated into fifth-power numbers
by trial, that all inverse methods in arithmetic and the higher
branches of mathematics are tentative and depend upon trial —
division, extraction of the square and cube roots are tentative
processes and depend upon trial.
"The process of Integration is of a tentative nature, depending on a previous knowledge of differentiation, as explained
in Chapter I.; just as Division in Arithmetic is a tentative
process, depending on a knowledge of Multiplication and the
Multiplication Table." GREENHILL'S Differential and Integral
Galculm, Second Edition, page 84.
In finding these sets of fifth-power numbers whose sum is a
fifth power I have used Barlow's Tables, edition of 1814, which
contains on pp. 170-173 a table of the first ten powers of all
numbers from 1 to 100, and on pp. 176-194 a table of the fourth
and fifth powers of all numbers from 100 to 1000. The use of
these tables very materially facilitates the work.
To farther facilitate the work I have formed the appended
table of the values of S (#5) for all values of x from 1 to 60 by
means of the formula
£[(* + !)']=£ (O + OB + 1)",
checking the work at intervals by the formula
(as + I)2 (2a2 + 20-1).
172
ARTEMAS MARTIN.
X
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
*(*»)
X
1
21
33
22
276
23
1300
24
4425
25
26
12201
29008
27
61776
28
120825
29
220825
30
381876
31
630708
32
1002001 • 33
1539825
34
2299200
35
3347776
36
4767633
37
38
6657201
9133300'
39
12333300
40
8 (x*}
X
8(afi)
16417401
21571033
28007376
35970000
45735625
57617001
71965908
89176276
109687425
133987425
162616576
196171008
235306401
280741825
333263700
393729876
463073833
542309001
632533200
734933200
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
850789401
981480633
1128489076
12934-05300
1477933425
1683896401
1913241408
2168045376
2450520625
2763020625
3108045876
3488249908
3906445401
4365610425
4868894800
5419626576
6021318633
6677675401
7392599700
8170199700
III. When we have found one set of numbers the sum of
whose fifth powers is a fifth power "other sets may be deduced
from it.
If
e*+f5 + g5+hG+ ..................... =W...., ..... (F),
we have obviously
(me)5 + (mff + (mg)5 + (mh)5 + ......... = (mw)s ...... (G).
Now if m be so taken that w ;= me, mf, mg} mh, or some other
one of the numbers in (G), we can substitute e5 +f5 + gs + h5 + ...
for the fifth power of that number and thus obtain another set of
fifth powers whose sum is a fifth power, if all the numbers in the
left-hand member of (F) are different from those in the left-hand
member of (G) — except the one substituted for.
Examples. 13. Multiply the set in Ex. 1 by 25, and substitute
the value of 125, and we have
45 + 55 4. g5 + fa + 35 + 95 + 1 Q5+ iIB
which is the set found in Ex. 10.
ON FIFTH-POWER NUMBERS.
173
14. In Ex, 1, take m = 5 and substitute the value of 30B from
Ex. 2, and we have
5B + 10B 4- 11B 4- 16B + 19B + 206 + 25B + 29B + 356 + 45B + 55B = 60B.
15. In Ex. 3, take m = 4 and substitute the value of 12B from
Ex. 1 and we have
4B + 5B + 6B + 7B + 9B + HB + 24B + 285 + 32B + 40B
+ 44B + 52B + 565 + 60B + 64B + 72B + 124B = 128B.
16. In Ex. 1, take w= 5, and in Ex. 3, take m= 4; substitute
the value of 605 thus obtained from Ex. 1, in the set obtained
from Ex. 3 and we have
12B + 20B + 24B + 25B + 28B + 30B + 32B + 35B + 40B + 44B
+ 45B + 52B + 55B + 56B + 64B + 72B + 124B = 128B.
17.
have
In Ex. 16, substitute the value of 30B from Ex. 2, and we
5B + 10B + HB + 12B + 16B + 19B + 20B + 24B + 25B + 28B + 29B + 32B
+ 35B + 40B + 44B + 4<5B + 52B + 55" + 56B + 64B + 72B + 124B = 128B.
18. In Ex. 10, take m = 2 and substitute the value of 12B from
Ex. 1 and we have
45 + 55 + 65 + 75^8B + 95 + 1()6 + 11B+ 14s+ 18^+ 20B
19. In Ex. 10, take m = 3 and substitute the value of 30B from
Ex. 2 and we have
5B + 10B + 11B + 12B + 15B + 16B + 18B + 19B + 21B
+ 24B + 27B + 296 + 33B + 42B + 54B + 66B = 72B.
20. In the second set of Ex. 11, take m — 2, and substitute
the value of 12B from Ex. 1 and we have
4" + 6B + 6B + 7B + 8B + 9B + 10B + 11B + 14" + 18B + 22B + 26B
+ 30B + 32B + 345 + 36B + 38B + 40B + 42B + 44B + 46B + 50B + 54B
+ 56B + 58B + 60B + 62B + 646 + 66B + 68B + 70B s= 100B.
In this way may be found an infinite number of sets of fifthpower numbers whose sum is a fifth power.
I am not aware that any other person than myself has ever
174
ARTEMAS MAETIN.
discovered any sets of fifth-power numbers whose sum is a fifth
power.
In my search for fifth-power numbers whose sum is a fifth
power I have discovered the following equalities:
I5 + 65 + 98 + 11B + 22B = 125 + 16B + 215,
IB + 55 + 105+135 + 14B = 8B + 95+ II5 + lo5,
105 + 11B +125 +135 + 17B + 255 = 1B + 5B + 75 + 215 + 24B,
' 98 + II5 + 12B + 13B + 16B + 18B = I5 + 68 + 8B + 14B + 20B,
105 + 22B + 325 -I- 38B + 585 = 256 + 30B + 35B + 455 + 55B,