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

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- About Notation
- Pythagorean Triplets
- Almost-Isoceles Triples
- Almost 30-60 Triples
- Chains of Consecutive Squares
- Pythagorean Quartets
- Fermat's Last Theorem
- Euler's Conjecture
- Sums of Cubes (Cubic Quartets)
- Sums of Fourth Powers
- Sums of Fifth and Higher Powers
- Sums of Powers of Consecutive Integers
- Multigrades

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 a^{2} + b^{2}=
c^{2}. For chains of squares,

21, 22, 23, 24; 25, 26, 27 means 21^{2}+ 22^{2}+ 23^{2}+ 24^{2}=25^{2}+
26^{2}+ 27^{2} 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 95,800^{4} + 217,519^{4} + 414,560^{4}
= 422,481^{4}.

The simplest triangle with integer sides that satisfies the Pythagorean Theorem is 3,
4, 5.

That is, 3^{2} + 4^{2} = 5^{2}. 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 = n^{2} -m^{2}, b=2mn, c=m^{2}+n^{2}, where m and n are
two integers and m is less than n.

(Verify that a^{2} + b^{2}= c^{2}.)

We can see several things of interest:

- Looking at the b term, at least one number in every Pythagorean triple must be even.
- 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.
- Obviously, if m and n have a common factor, so will the triple. We aren't interested in these cases either.
- Thus, the only interesting (relatively prime) triples are generated when m and n are relatively prime and only one is even.
- The c term (hypotenuse) will always be odd in any relatively prime triple.
- 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. - The sum of c and b is always a square: c + b = (n + m)
^{2}. - Since a
^{2}= c^{2}- b^{2}= (c - b)(c + b), c + b always equals a^{2}times an odd square. We can use this relationship to group triples into related series: - Also, c - a = (n
^{2}+ m^{2}) - (n^{2}- m^{2}) = 2m^{2}. 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 |

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 |

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 |

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} + n^{2} = (n+1)^{2}
+ (n+2)^{2} + (n+3)^{2} .... or,

.......n^{2} - 6n + 9 + n^{2} - 4n + 4 + n^{2} - 2n + 1 + n^{2}
= n^{2} + 2n + 1 + n^{2} + 4n + 4 + n^{2} + 6n + 9 ....

All the squares of integers cancel, as do all the n^{2} terms except one. We get:

...... -2an - ..... - 6n - 4n - 2n + n^{2} = 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

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
a^{2} + b^{2} + c^{2} = d^{2}. 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 a^{2} + b^{2} = r^{2}.
Now look at the vertical plane through Q and P. The distance from Q to the origin is r^{2}
+ c^{2} = d^{2}, and since a^{2} + b^{2} = r^{2},
a^{2} + b^{2} + c^{2} = d^{2}.) 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 = m^{2}, b = 2mn, c = 2n^{2}; d = (m^{2} + 2n^{2}) = a +
c. Also note that b^{2} = 2ac. When m =1 and n = 1, we get 1, 2, 2; 3, the
simplest example.

There are an infinite number of Pythagorean triplets. Are there any triplets for higher
powers? That is, are there any integers for which a^{n} + b^{n} = c^{n},
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*.)

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

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.

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

- 3m
^{2}+5mn-5n^{2}, 4m^{2}-4mn+6n^{2}, 5m^{2}-5mn-3n^{2}; 6m^{2}-4mn+4m^{2} - m
^{2}+16m-21, 16m-m^{2}+21, 2m^{2}-4m+42; 2m^{2}+4m+42 - (2m-1)(2m
^{3}-6m^{2}-1), (m+1)(5m^{3}-9m^{2}+3m-1), 3m(m+1)(m^{2}-m+1); 3m(2m-1)(m^{2}-m+1)

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

a = m(m^{3} - 2n^{3}), b = n(2m^{3} - n^{3}), c = n(m^{3}
+ n^{3}), d = m(m^{3} + n^{3}), 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.

We can write A^{3} + B^{3} = D^{3} - C^{3}, 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 + q^{3} B = (m+n)p - q^{3}
C = p^{2} - (m+n)q D = p^{2} + (m-n)q

A^{3} + B^{3} = D^{3} - C^{3 }factors to: (A+B)(A^{2}-AB+B^{2})
= (D-C)(D^{2}+CD+C^{2})

Substituting and simplifying, we have m^{2} +3n^{2} = 3pq.
Obviously m is divisible by 3. Call m = 3k. Also pq = n^{2} + 3k^{2}

However, Euler also proved that every divisor of n^{2} + 3k^{2},
if n and k are relatively prime, is of the same form. Thus we can write

p = x^{2} + 3y^{2} q = z^{2}
+ 3w^{2} 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.

G. Korneck in 1873 derived this solution in three variables for the
formula

A^{3} + B^{3} = C^{3} + D^{3}. Since cubes
can be negative, this is merely a slight rearrangement of the formula above.

- A = 6m
^{3}tf + t(t+m)(m^{4}+ m^{2}t^{2}+ t^{4}) + 3t(t-m)f^{2} - B = 6m
^{3}tf - t(t+m)(m^{4}+ m^{2}t^{2}+ t^{4}) - 3t(t-m)f^{2} - C = -6t
^{3}mf +m(t+m)(m^{4}+ m^{2}t^{2}+ t^{4}) + 3m(m-t)f^{2} - D = 6t
^{3}mf +m(t+m)(m^{4}+ m^{2}t^{2}+ t^{4}) + 3m(m-t)f^{2}

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 |

If we look at the fourth powers of the digits 0-9, we see an interesting fact. The
fourth powers are: 0^{4}=0, 1^{4}=1, 2^{4}=16, 3^{4}=81, 4^{4}=256,
5^{4}=625, 6^{4}=1296, 7^{4}=2401, 8^{4}=4096, 9^{4}=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 A^{4}+B^{4}+C^{4}+D^{4}=E^{4}

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 |

Amazingly enough, there is a formula for *fifth* powers that generates an infinite
number of solutions:

(75y^{5}-x^{5}), (x^{5}+25y^{5}), (x^{5}-25y^{5}),
(10x^{3}y^{2}), (50xy^{4}); (x^{5}+75y^{5})

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.

What is the general formula for the sum 1^{n} + 2^{n} + 3^{n} + 4^{n}
+ ..... x^{n} 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.

For higher powers we have to get trickier. Consider n=2. We assume S2(x)=ax^{3}
+ bx^{2} + 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) - (ax^{3} + bx^{2}
+ cx) =

ax^{3} + 3ax^{2} + 3ax + a - ax^{3}

+ bx^{2} + 2bx + b - bx^{2}

+ cx + c - cx

or

S2(x+1)-S2(x) = 3ax^{2} + 3ax + a + 2bx + b + c

but also

S2(x+1)-S2(x) = (x+1)^{2} = x^{2} + 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)x^{3} + (1/2)x^{2} + (1/6)x = (2x^{3} + 3x^{2}
+ x)/6 = x(x+1)(2x+1)/6

Now that we have a general algorithm, we can apply it to higher powers.

Assume S3(x)=ax^{4} + bx^{3} + cx^{2} + dx. Write:

S3(x+1)-S3(x)=a(x+1)^{4} + b(x+1)^{3} + c(x+1)^{2} + dx - (ax^{4}
+ bx^{3} + cx^{2} + d) =

ax^{4} + 4ax^{3} + 6ax^{2} + 4ax + x - ax^{4}

+ bx^{3} + 3bx^{2} + 3bx + b - bx^{3}

+ cx^{2} + 2cx + c - cx^{2}

+ dx + d - dx

or

S3(x+1)-S3(x) = 4ax^{3} + 6ax^{2} + 4ax + a + 3bx^{2} + 3bx + b +
2cx + c + d =

(x+1)^{3} = x^{3} + 3x^{2} + 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) = x^{2}(x + 1)^{2}/4 or S3(x) =
(S1(x))^{2}.

It looks like we might be starting a nice pattern.

Using the routine above, assume S4(x)=ax^{5} + bx^{4} + cx^{3}
+ dx^{2} + ex.

S4(x+1)-S4(x)=

ax^{5} + 5ax^{4} + 10ax^{3} + 10ax^{2} + 5ax + a - ax^{5}
+ bx^{4} + 4bx^{3} + 6bx^{2} + 4bx + b - bx^{4}

+ cx^{3} + 3cx^{2} + 3cx + c - cx^{3}

+ dx^{2} + 2dx + d - dx^{2}

+ ex + e - ex

or

S4(x+1)-S4(x) = 5ax^{4} + (10a + 4b)x^{3} + (10a + 6b + 3c)x^{2} +
(5a + 4b + 3c + 2d)x + (a + b + c + d + e)

= x^{4} + 4x^{3} + 6x^{2} + 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) = (6x^{5} + 15x^{4} + 10x^{3} - x)/30 = x(x + 1)(2x + 1)(3x^{2}
+ 3x -1)/30.

So much for our elegant pattern.

The coefficients for powers up to 12 are shown below.

x | x^{2} |
x^{3} |
x^{4} |
x^{5} |
x^{6} |
x^{7} |
x^{8} |
x^{9} |
x^{10} |
x^{11} |
x^{12} |
||

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: x^{2}(x + 1)^{2}/4

n=4: x(x + 1)(2x + 1)(3x^{2} + 3x - 1)/30

n=5: x^{2}(x + 1)^{2}(2x^{2 }+ 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
- 1
^{2}+ 8^{2}+ 10^{2}+ 17^{2}= 454 = 2^{2}+ 5^{2}+ 13^{2}+ 16^{2} - 1
^{3}+ 8^{3}+ 10^{3}+ 17^{3}= 6426 = 2^{3}+ 5^{3}+ 13^{3}+ 16^{3}

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)

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.

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*Created 2 January 1997: Last Update
14 December 2009
*

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