Power Page

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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Index

To save writing exponents, we'll denote power sums like this: terms on each side of the equality will be separated by commas, and opposite sides of the equality will be separated by a semicolon. Thus, for Pythagorean triplets, a, b; c means a2 + b2= c2. For chains of squares,
21, 22, 23, 24; 25, 26, 27 means 212+ 222+ 232+ 242=252+ 262+ 272 and similarly for sums of higher powers. Where the exponent is not clear from context, it will be noted in parentheses. Thus: 95,800, 217,519, 414,560; 422,481 (n=4) means 95,8004 + 217,5194 + 414,5604 = 422,4814.

Pythagorean Triplets

The simplest triangle with integer sides that satisfies the Pythagorean Theorem is 3, 4, 5.
That is, 32 + 42 = 52. Such a triplet is called a Pythagorean Triplet. There are infinitely many, and they are easy to generate. A classic formula, known since ancient times, can generate them at will. If the numbers in the triplet are a, b, and c, then:
a = n2 -m2, b=2mn, c=m2+n2, where m and n are two integers and m is less than n.
(Verify that a2 + b2= c2.)

We can see several things of interest:

1. Looking at the b term, at least one number in every Pythagorean triple must be even.
2. Looking at the a and c terms, if m and n are both even or both odd, then a and c must be even. Thus, all members of the triple are even. Since we can divide all the members of the triple by 2 to get a smaller triple, we are not really interested in these cases. So we can restrict our attention to even-odd sets of m and n.
3. Obviously, if m and n have a common factor, so will the triple. We aren't interested in these cases either.
4. Thus, the only interesting (relatively prime) triples are generated when m and n are relatively prime and only one is even.
5. The c term (hypotenuse) will always be odd in any relatively prime triple.
6. The difference between c and b is always a square: c - b = (n - m)2. Since we only get relatively prime triples when only one of (m,n) is odd, c - b is always an odd square.
7. The sum of c and b is always a square: c + b = (n + m)2.
8. Since a2 = c2 - b2 = (c - b)(c + b), c + b always equals a2 times an odd square. We can use this relationship to group triples into related series:
9. Also, c - a = (n2 + m2) - (n2 - m2) = 2m2. Thus the difference between c and a is always two times a square.

We can use the relationships above to group Pythagorean triples in many ways. This set of series displays relationships 6-8 above:

 c - b = 1 m= n= c - b = 9 m= n= c - b = 25 m= n= 1, 0; 1 0 1 9, 0; 9 0 3 25, 0; 25 0 5 3, 4; 5 1 2 15, 8; 17 1 4 35, 12; 37 1 6 5, 12; 13 2 3 21, 20; 29 2 5 45, 28; 53 2 7 7, 24; 25 3 4 27, 36; 45 3 6 55, 48; 73 3 8 9, 40; 41 4 5 33, 56; 65 4 7 65, 72; 97 4 9 11, 60; 61 5 6 39, 80; 89 5 8 75, 100; 125 5 10

The set of series below emphasizes relationship 9 above:

 c - a = 2 m= n= c - a = 8 m= n= c - a = 18 m= n= 4, 3; 5 1 2 12, 5; 13 2 3 24, 7; 25 3 4 8, 15; 17 1 4 20, 21; 29 2 5 36, 27; 45 3 6 12, 35; 37 1 6 28, 45; 53 2 7 48, 55; 73 3 8 16, 63; 65 1 8 36, 77; 85 2 9 60, 91; 109 3 10 20, 99; 101 1 10 44, 117; 125 2 11 72, 135; 153 3 12

Almost-Isosceles Triples

The square root of two is irrational, therefore there cannot be any Pythagorean triplets a, a; c. But there are an infinite number of triplets a, a + 1; c. Here are the first ten:

 0 1 1 c= 3 4 5 6*1 -1 20 21 29 6*5 -1 119 120 169 6*29 - 5 696 697 985 6*169 -29 4059 4060 5741 6*985 - 169 23660 23661 33461 6*5741 -985 137903 137904 195025 6*33461 - 5741 803760 803761 1136689 6*195025 - 33461 4684659 4684660 6625109 6*1136689-195025

The recursion relations are:

 a0 b0=a0+1 c0 a1 = 3*a0 + 2*b0 + 1 b1 = 3*a0 + 2*b0 +2 c1 = 4*a0 + 3*c0 + 2 a2 = 6*a1 - a0 + 2 b2 = 6*a1 - a0 + 2 c2 = 6*c1 - c0

Almost 30-60 Triples

Like the square root of two, the square root of three is irrational, so there cannot be any Pythagorean triples that are 30-60 triangles (c = 2a). But we can come close; there are an infinite number of Pythagorean triples where c = 2a +/- 1. Here are the first 13.

 0 1 1 3 4 5 8 15 17 33 56 65 120 209 241 451 780 901 1680 2911 3361 6273 10864 12545 23408 40545 46817 87363 151316 174725 326040 564719 652081 1216801 2107560 2433601 4541160 7865521 9082321

The recursion relations are:

 a0 b0 c0 = 2*a0 +/- 1 a1 = b0 + c0 +/- 1 b1 = c0 + 2*b0 - a0 c1 = 2*b0 + 2*c0 +/- 1

Chains of Consecutive Squares

The Pythagorean triplet 3,4;5 is the first example of another interesting pattern. The first few examples are:
3, 4; 5
10, 11, 12; 13, 14
21, 22, 23, 24; 25, 26, 27

For any value of a, there is a chain of a squares (n-a), (n-a+1) ..... (n-1),n; (n+1), n+2)....(n+a)
The value of n is given by n=2a(a+1).

Proof: Pair up terms across the equality as shown:
...... (n-3)2 + (n-2)2 + (n-1)2 + n2 = (n+1)2 + (n+2)2 + (n+3)2 .... or,
.......n2 - 6n + 9 + n2 - 4n + 4 + n2 - 2n + 1 + n2 = n2 + 2n + 1 + n2 + 4n + 4 + n2 + 6n + 9 ....
All the squares of integers cancel, as do all the n2 terms except one. We get:
...... -2an - ..... - 6n - 4n - 2n + n2 = 2n + 4n + 6n + ..... 2an. Dividing by n and rearranging:
n = 4(1 + 2 + 3 + 4 + 5 ...... a). The sum of integers from 1 to a is a(a+1)/2 (see Sums of Powers of Consecutive Integers below). Thus n = 2a(a+1). There is a chain of a+1 squares left of the equality and a to the right. The examples above are for a = 1, 2, and 3. For a = 4, we find n = 40 and the resulting chain is: 36, 37, 38, 39, 40; 41, 42, 43, 44

Pythagorean Quartets

There are an infinite number of quartets of squares a, b, c; d. There are so many that they are not all that interesting, except for one point. The three-dimensional version of the Pythagorean Theorem is a2 + b2 + c2 = d2. In other words, in three-dimensional Cartesian Coordinates, point (a,b,c) is distance d from the origin. (Proof: Consider the x-y plane. Point (a,b,c) - call it Q - projects to (a,b,0). The distance from this point - call it P - to the origin is given by a2 + b2 = r2. Now look at the vertical plane through Q and P. The distance from Q to the origin is r2 + c2 = d2, and since a2 + b2 = r2, a2 + b2 + c2 = d2.) Thus Pythagorean quartets are the coordinates of points in three dimensions that lie an integer distance from the origin.

A formula that generates Pythagorean quartets is:
a = m2, b = 2mn, c = 2n2; d = (m2 + 2n2) = a + c. Also note that b2 = 2ac. When m =1 and n = 1, we get 1, 2, 2; 3, the simplest example.

Fermat's Last Theorem

There are an infinite number of Pythagorean triplets. Are there any triplets for higher powers? That is, are there any integers for which an + bn = cn, where n is a power higher than 2? The French mathematician Pierre Fermat claimed about 1637 that there were none. He wrote in the margin of a book that he had a marvelous proof too complex to fit in the margin. For centuries, mathematicians wished Fermat had been reading a bigger book. The only reasons for taking such an unsupported conjecture seriously are (a) the proofs Fermat did furnish were brilliant, (b) in every other case where Fermat claimed to have proven something profound without actually providing a proof, a proof was later found and (c) Fermat's Last Theorem survived every assault made on it. He has credibility.

The possibility that there might be a simple solution to a famous problem attracted cranks by the score. (Why exactly cranks think they will become famous by solving famous problems is a mystery. How many mathematicians have megabuck endorsement contracts?) Most mathematicians now think Fermat was probably mistaken in thinking he had a proof (even Brett Favre gets sacked occasionally.) Fermat's Last Theorem was proven for specific powers of n to such large values - in the millions - that nobody would ever actually write a counter-example on paper. Finally, in 1993, the theorem was proven by Andrew Weil of Princeton University. (Weil is now endorsing sneakers and married to a supermodel - not.)

Euler's Conjecture

There are no triplets of numbers for powers larger than 2, but there are longer sums. It is an interesting coincidence that we have 3, 4; 5 for squares and 3, 4, 5; 6 for cubes (The pattern does not hold for larger powers.) The mathematician Leonhard Euler conjectured that it always required n terms to sum to an nth power: two squares, three cubes, four fourth powers, and so on. Nice guess, but wrong. In 1966, L. J. Lander and T. R. Parkin found the first counterexample: four fifth powers that summed to a fifth power: 27, 84, 110, 133; 144.

In 1988 Noam Elkies of Harvard found a counterexample for fourth powers:
2,682,440, 15,365,639, 187,960; 20,615,673. Roger Frye of Thinking Machines Corporation did a computer search to find the smallest example:
95,800, 217,519, 414,560; 422,481

Sums of Cubes (Cubic Quartets)

Even though 3, 4, 5; 6 for cubes is as pretty a mathematical formula as anyone could hope to find, sums of cubes do not yield to formulas anything like as neat as those for Pythagorean triplets. One additional complication is that cubes can be negative. We could ignore negative numbers in dealing with Pythagorean triplets, but we cannot in dealing with cubes. Also, Pythagorean triplets generally have terms that are more or less the same size. The largest triplet with a term less than 10 is 9, 40; 41. Any term larger than 41 will differ from all other squares by more than 81. But there is much less limitation on cubic quartets. For example:

• 1, 6 , 8 ; 9
• 1 , 71 , 138 ; 144
• 1 , 135 , 138 ; 172
• 1 , 236 , 1207 ; 1210
• 1 , 242 , 720 ; 729
• 1 , 372 , 426 ; 505
• 1 , 426 , 486 ; 577
• 1 , 566 , 823 ; 904
• 1 , 575 , 2292 ; 2304
• 1 , 791 , 812 ; 1010
• 1 , 1124 , 5610 ; 5625
• 1 , 1851 , 8675 ; 8703
• 1 , 1938 , 2820 ; 3097
• 1 , 1943 , 6702 ; 6756
• 1 , 2196 , 5984 ; 6081
• 1 , 2676 , 3230 ; 3753

The list above resulted from a search where b and c ranged up to 10,000, so it appears the cutoff at b = 2676 is a real limit. These are "almost" exceptions to Fermat's Last Theorem. Since cubes can be negative, we also find this set of "almost" exceptions to Fermat's Last Theorem.

• -1 , 9,  10 ;12
• -1 , 73 ,144 ; 150
• -1 , 244 , 729 ; 738
• -1 , 368 , 1537 ;1544
• -1 , 577 , 2304 ;2316
• -1 , 1010 ,1897 ;1988
• -1 , 1033 ,1738 ;1852
• -1 , 1126 ,5625 ;5640
• -1 , 3097 ,3518 ;4184
• -1 , 3753 ,4528 ;5262

Again, since the search ran to b = c = 10,000, the cutoff at 3753 is probably the real limit.

Polynomial Generators

There are quite a few polynomials known that generate cubic quartets. Here are some:

• 3m2+5mn-5n2, 4m2-4mn+6n2, 5m2-5mn-3n2; 6m2-4mn+4m2
• m2+16m-21, 16m-m2+21, 2m2-4m+42; 2m2+4m+42
• (2m-1)(2m3-6m2-1), (m+1)(5m3-9m2+3m-1), 3m(m+1)(m2-m+1); 3m(2m-1)(m2-m+1)

Vieta, 1591

In 1591, Francois Vieta developed the formula below. We can generate numbers a, b, c; d for cubes as follows:

a = m(m3 - 2n3), b = n(2m3 - n3), c = n(m3 + n3), d = m(m3 + n3), where m and n are any two numbers. If m = 2 and n =1 we get 12, 15, 9; 18. If m = 3 and n = 2 we get 33, 70, 92; 105.

Unfortunately, the sum of two squares has no simple factors, but the sum of two cubes does. Vieta's formulas yield large numbers that are often not relatively prime. Elementary reasoning suggests that if it takes two variables to create a Pythagorean triple, it should take at least three to make the cubic analogue, so Vieta's solution is not complete.

Euler, 1760

We can write A3 + B3 = D3 - C3, and then factor both polynomials. Unfortunately, this seductively symmetrical approach tends to lead to solutions with four variables. Euler's solution (1760-63) is typical. He let A,B,C and D be as follows:

A = (m-n)p + q3     B = (m+n)p - q3     C = p2 - (m+n)q     D = p2 + (m-n)q

A3 + B3 = D3 - C3 factors to: (A+B)(A2-AB+B2) = (D-C)(D2+CD+C2)

Substituting and simplifying, we have m2 +3n2 = 3pq. Obviously m is divisible by 3. Call m = 3k. Also pq = n2 + 3k2

However, Euler also proved that every divisor of n2 + 3k2, if n and k are relatively prime, is of the same form. Thus we can write

p = x2 + 3y2     q = z2 + 3w2     m = 3(yz +- xw)     n =(xz -+ 3yw)

and plug these back into the formulas for A,B,C and D. This, unfortunately, is pretty typical of most of the solutions of this problem. We end up with a chain of substitutions leading back to extremely complex formulas with no attempt to condense or simplify them. And since D is wholly dependent on A, B, and C, there should be no more than three independent variables.

Korneck, 1873

G. Korneck in 1873 derived this solution in three variables for the formula
A3 + B3 = C3 + D3. Since cubes can be negative, this is merely a slight rearrangement of the formula above.

• A = 6m3tf + t(t+m)(m4 + m2t2 + t4) + 3t(t-m)f2
• B = 6m3tf - t(t+m)(m4 + m2t2 + t4) - 3t(t-m)f2
• C = -6t3mf +m(t+m)(m4 + m2t2 + t4) + 3m(m-t)f2
• D = 6t3mf +m(t+m)(m4 + m2t2 + t4) + 3m(m-t)f2

Table of Quartets with a and b < 100

Below is a table of cube sums for a and b less than 100. Click here for a complete listing of quartets with a, b and c up to 1000.

 a b c d 1 6 8 9 1 71 138 144 2 17 40 41 3 4 5 6 3 10 18 19 3 34 114 115 3 36 37 46 4 17 22 25 4 57 248 249 5 76 123 132 5 86 460 461 6 32 33 41 7 14 17 20 7 54 57 70 9 55 116 120 9 58 255 256 11 15 27 29 12 19 53 54 12 31 102 103 12 81 136 145 12 86 159 167 13 51 104 108 13 65 121 127 14 23 70 71 15 42 49 58 15 64 297 298 15 82 89 108 16 23 41 44 16 47 108 111 16 51 213 214 17 40 86 89 17 57 177 179 18 19 21 28 19 53 90 96 19 60 69 82 19 92 101 122 19 93 258 262 20 54 79 87 21 43 84 88 21 46 188 189 22 51 54 67 22 57 255 256 22 75 140 147 23 81 300 302 23 86 97 116 23 94 105 126 25 31 86 88 25 38 87 90 25 48 74 81 26 55 78 87 27 30 37 46 27 46 197 198 27 64 306 307 28 53 75 84 29 34 44 53 29 75 96 110 31 33 72 76 31 64 137 142 31 95 219 225 32 54 85 93 33 70 92 105 34 39 65 72 35 77 202 206 36 38 61 69 36 147 341 350 38 43 66 75 38 48 79 87 38 57 124 129 42 83 205 210 44 51 118 123 44 73 128 137 45 53 199 201 45 69 79 97 46 47 148 151 47 75 295 297 47 97 162 174 48 85 491 492 49 80 263 266 49 84 102 121 50 61 64 85 50 67 216 219 50 74 97 113 51 82 477 478 53 58 194 197 54 80 163 171 56 61 210 213 57 68 180 185 57 82 495 496 58 59 69 90 58 75 453 454 59 93 148 162 61 90 564 565 64 75 477 478 65 87 142 156 66 97 632 633 69 99 146 164 71 73 138 150 71 81 384 386 72 85 122 141 86 95 97 134 88 95 412 415 94 96 99 139

Sums of Higher Powers

Sums of Fourth Powers

If we look at the fourth powers of the digits 0-9, we see an interesting fact. The fourth powers are: 04=0, 14=1, 24=16, 34=81, 44=256, 54=625, 64=1296, 74=2401, 84=4096, 94=6561. Fourth powers only end in 0,1,5 and 6. To get fourth powers that sum to a fourth power, we have to find combinations that sum to something ending in those digits. Using sums of four terms, we find only (0,0,0,1=1), (0,0,0,5=5), (0,0,0,6=6), (0,0,1,5=6), (0,0,5,5=0), (0,0,5,6=1), (0,1,5,5=1), (0,5,5,5=5), (0,5,5,6=6), (1,5,5,5=6), (5,5,5,5=0), (5,5,5,6=1). Note here that we're talking about the powers, not the numbers themselves; a fourth power ending in 6 could belong to a number ending in any even digit. Nevertheless, this improves the odds a lot over the 10,000 combinations if any combination of four ending digits were possible. Noting that Euler's conjecture isn't true, but Fermat's Last Theorem is, one of the four terms at most can be zero (although other terms can end in zero.)

Kermit Rose and Simcha Brudno, (More about four biquadrates equal one biquadrate, Mathematics of computation, vol. 27, no. 123, July 1973, p. 491-494.) found the following sets, where A4+B4+C4+D4=E4

 A B C D E 30 120 315 272 353 240 340 430 599 651 2420 710 435 1384 2487 2365 1190 1130 1432 2501 2745 1010 850 1546 2829 2460 2345 2270 3152 3723 3395 3230 350 1652 3973 2650 1060 205 4094 4267 3670 3545 1750 1394 4333 4250 2840 700 699 4449 1880 1660 380 4907 4949 5080 1120 1000 3233 5281 5055 3910 410 1412 5463 5400 1770 955 2634 5491 5400 1680 30 3043 5543 5150 4355 1810 1354 5729 5695 4280 2770 542 6167 5000 885 50 5984 6609 6185 4790 1490 3468 6801 5365 2850 1390 6368 7101 2790 1345 160 7166 7209 6635 5440 800 3052 7339 6995 5620 2230 3196 7703 5670 5500 4450 7123 8373 7565 5230 4730 4806 8433 7630 5925 4910 524 8493 7815 6100 3440 1642 8517 8230 2905 1050 5236 8577 5780 3695 3450 8012 8637 8570 6180 3285 816 9137 6435 2870 680 8618 9243 7820 6935 5800 5192 9431 8760 6935 1490 1394 9519 8570 7050 305 5264 9639 8835 6800 5490 2922 9797 6485 5660 4840 8864 9877 8870 8635 1620 2294 10419 9145 8530 5300 5936 10939 10490 8635 5300 3556 11681 11455 6200 4490 1476 11757 8735 8170 1180 10144 12019 11720 7270 3710 2833 12167 9360 8655 7480 8862 12259 8925 4410 3450 11234 12287 11390 8045 320 7352 12439 12435 6190 5780 1616 12759 10310 6870 2935 10678 12771 12845 5950 2870 5934 13137 11210 7590 7025 9712 13209 13040 4975 1700 7896 13521 12035 3610 3440 10738 13637 13410 6420 1275 8278 14029 13740 7920 6660 3929 14297 14405 2630 210 34 14409 13355 8010 1530 9498 14489 13900 2040 1920 9219 14531 13760 10245 800 4682 14751 14815 8940 4250 2512 15309 11110 6800 3890 14579 15829 11815 5640 2880 14598 16027 15780 4790 4140 7701 16049 15940 6670 5430 137 16113 14320 13110 2275 1088 16359 14890 8830 1220 12107 16643 15160 11015 10850 412 16891 11810 2350 1845 15776 16893 15375 11050 6690 11658 17381 13060 8495 1220 15644 17519 16405 6500 950 11896 17521 16215 12850 5450 1802 17661 10660 3235 3220 17068 17693 17320 9860 1945 7256 17881 17510 8340 2760 9423 18077 16805 13660 5270 5898 18477 15365 12430 11410 12668 18701 16560 8355 610 15906 19483 13940 9305 4460 17726 19493 17595 13440 5370 12772 19871 19255 3090 780 12702 20111 11980 8975 1090 19244 20131 19670 10030 1880 9579 20253 19480 7550 1660 12969 20469 18100 13690 12140 11801 20699 13970 8855 8720 19142 20719 17740 16525 12070 3362 21013 13915 5950 5420 24802 25427 16260 12860 8545 34178 34803 1840 30690 41000 89929 91179

Sums of Fifth and Higher Powers

Amazingly enough, there is a formula for fifth powers that generates an infinite number of solutions:
(75y5-x5), (x5+25y5), (x5-25y5), (10x3y2), (50xy4); (x5+75y5)

Intuition suggests there have to be a lot more than two variables needed for a complete solution, so this formula, though interesting, is only a special case.

There are no examples or formulas known for powers above 5.

Sums of Powers of Consecutive Integers

What is the general formula for the sum 1n + 2n + 3n + 4n + ..... xn for any value of n? This turns out to be a tougher and deeper problem than one might guess. It will be useful to write the sum as a function Sn(x), that is, the sum of the nth powers of consecutive integers up to x. For n=0, the answer is trivial: S0(x) = x.

For n=1, a little trick helps. Pair up the first and last terms in the series, the second and next to last, and so on. We have 1 + x, 2 + (x-1), 3+ (x-2) and so on. If x is even, we have x/2 pairs each summing to x+1, so the total is (x/2)(x+1). Thus S1(x) = x(x+1)/2. If x is odd, leave off the last term x for a moment. Now there is one less pair plus the leftover term x, so S1(x) = (x-1)/2)(x) + x or S1(x) = x(x+1)/2, again.

n = 2:

For higher powers we have to get trickier. Consider n=2. We assume S2(x)=ax3 + bx2 + cx + d. Since S2(0) = 0, d=0. This will be true for any Sn(x); there is no constant term.

Now calculate S2(x+1)-s2(x): We have:

S2(x+1)-S2(x)=a(x+1)3 + b(x+1)2 + c(x+1) - (ax3 + bx2 + cx) =
ax3 + 3ax2 + 3ax + a - ax3
+ bx2 + 2bx + b - bx2
+ cx + c - cx
or
S2(x+1)-S2(x) = 3ax2 + 3ax + a + 2bx + b + c
but also
S2(x+1)-S2(x) = (x+1)2 = x2 + 2x + 1
Now, when two polynomials are equal for any x, their coefficients must be equal. Thus: 3a = 1 and a = 1/3; 3a + 2b = 2, thus b = 1/2; a + b + c = 1, thus c = 1/6. Thus:
S2(x)=(1/3)x3 + (1/2)x2 + (1/6)x = (2x3 + 3x2 + x)/6 = x(x+1)(2x+1)/6

n=3:

Now that we have a general algorithm, we can apply it to higher powers.
Assume S3(x)=ax4 + bx3 + cx2 + dx. Write:

S3(x+1)-S3(x)=a(x+1)4 + b(x+1)3 + c(x+1)2 + dx - (ax4 + bx3 + cx2 + d) =
ax4 + 4ax3 + 6ax2 + 4ax + x - ax4
+ bx3 + 3bx2 + 3bx + b - bx3
+ cx2 + 2cx + c - cx2
+ dx + d - dx
or
S3(x+1)-S3(x) = 4ax3 + 6ax2 + 4ax + a + 3bx2 + 3bx + b + 2cx + c + d =
(x+1)3 = x3 + 3x2 + 3x + 1
Hence:
4a = 1 and a = 1/4; 6a + 3b = 3 and b = 1/2; 4a + 3b + 2c = 3 and c = 1/4;
a + b + c + d =1 and d= 0
Combining terms and simplifying: S3(x) = x2(x + 1)2/4 or S3(x) = (S1(x))2.
It looks like we might be starting a nice pattern.

n=4:

Using the routine above, assume S4(x)=ax5 + bx4 + cx3 + dx2 + ex.

S4(x+1)-S4(x)=
ax5 + 5ax4 + 10ax3 + 10ax2 + 5ax + a - ax5 + bx4 + 4bx3 + 6bx2 + 4bx + b - bx4
+ cx3 + 3cx2 + 3cx + c - cx3
+ dx2 + 2dx + d - dx2
+ ex + e - ex
or
S4(x+1)-S4(x) = 5ax4 + (10a + 4b)x3 + (10a + 6b + 3c)x2 + (5a + 4b + 3c + 2d)x + (a + b + c + d + e)
= x4 + 4x3 + 6x2 + 4x + x. Thus:
5a =1 and a = 1/5; 10a + 4b = 4 and b = 1/2; 10a + 6b + 3c = 6 and c=1/3;
5a + 4b + 3c + 2d = 4 and d = 0; a + b + c + d + e = 1 and e = -1/30. Clearing fractions, we get:
S4(x) = (6x5 + 15x4 + 10x3 - x)/30 = x(x + 1)(2x + 1)(3x2 + 3x -1)/30.
So much for our elegant pattern.

The coefficients for powers up to 12 are shown below.

 x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 n=1 1/2 1/2 n=2 1/6 1/2 1/3 n=3 0 1/4 1/2 1/4 n=4 -1/30 0 1/3 1/2 1/5 n=5 0 -1/12 0 5/12 1/2 1/6 n=6 1/42 0 -1/6 0 1/2 1/2 1/7 n=7 0 1/12 0 -7/24 0 7/12 1/2 1/8 n=8 -1/30 0 2/9 0 -7/15 0 2/3 1/2 1/9 n=9 0 -3/20 0 1/2 0 -7/10 0 3/4 1/2 1/10 n=10 5/66 0 -1/2 0 1/1 0 -1/1 0 5/6 1/2 1/11 n=11 0 5/12 0 -11/8 0 11/16 0 -11/8 0 11/12 1/2 1/12

Readers with some advanced math background may recognize the first column as Bernoulli Numbers (with alternating signs) for the even powers. The patterns in the table are easiest to see along the diagonals. From right to left, the terms in each diagonal are as follows (n is the row number)

• 1/(n+1)
• 1/2
• n/12 (Each term is n/(n-1) times the preceding term - this will make sense when compared to the patterns below)
• 0
• Each term is n/(n-3) times the preceding term
• 0
• Each term is n/(n-5) times the preceding term
• 0
• Each term is n/(n-7) times the preceding term

Factored polynomials for each n are:
n=0: x
n=1: x(x + 1)/2
n=2: x(x + 1)(2x + 1)/6
n=3: x2(x + 1)2/4
n=4: x(x + 1)(2x + 1)(3x2 + 3x - 1)/30
n=5: x2(x + 1)2(2x2 + 2x - 1)/12

One might guess that it is very unlikely to have sums of powers that hold for several different exponents. In fact, there are many such equations, called multigrades. For example:

• 1 + 8 + 10 + 17 = 36 = 2 + 5 + 13 + 16
• 12 + 82 + 102 + 172 = 454 = 22 + 52 + 132 + 162
• 13 + 83 + 103 + 173 = 6426 = 23 + 53 + 133 + 163

It is a remarkable fact that we can add any integer to all the terms of the multigrade and it will still hold. Adding 1 to the example above, we get (2, 9, 11, 18); (3, 6, 14, 17). (n=1,2,3). For clarity we group the terms on each side of the equality in parentheses.

A few remarkable high-order multigrades are:

(1, 50, 57, 15, 22, 71); (2, 45, 61, 11, 27, 70); (5, 37, 66, 6, 35, 67) (n=1,2,3,4,5)

(1, 9, 25, 51, 75, 79, 107, 129, 131, 157, 159, 173);
(3, 15, 19, 43, 89, 93, 97, 137, 139, 141, 167, 171) (n=1, 3, 5, 7, 9, 11, 13)

References

Most of the historical results are from:

Dickson, Leonard E., History of the theory of numbers,  New York, Chelsea Pub. Co., 1952.

This is a rich but maddeningly disorganized book, with no attempt to use consistent notation or format.

Created 2 January 1997: Last Update 14 December 2009

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