Power Page

Steven Dutch, Professor Emeritus, Natural and Applied Sciences, Universityof Wisconsin - Green Bay

Interesting stuff about powers of numbers

Index

• 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

About Notation

To save writing exponents, we'll denote power sums like this: terms on each side of theequality will be separated by commas, and opposite sides of the equality will be separatedby 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 exponentis 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 PythagoreanTriplet. 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 aretwo 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 setof series displays relationships 6-8 above:

The set of series below emphasizes relationship 9 above:

Almost-Isosceles Triples

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

The recursion relations are:

Almost 30-60 Triples

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

The recursion relations are:

Chains of Consecutive Squares

The Pythagorean triplet 3,4;5 is the first example of another interesting pattern. Thefirst 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 byn 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). Thereis a chain of a+1 squares left of the equality and a to the right. The examples above arefor 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 thatthey are not all that interesting, except for one point. The three-dimensional version ofthe Pythagorean Theorem is a² + b² + c² = d². In other words, inthree-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 fromthis 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 quartetsare the coordinates of points in three dimensions that lie an integer distance from theorigin.

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, thesimplest example.

Fermat's Last Theorem

There are an infinite number of Pythagorean triplets. Are there any triplets for higherpowers? 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 about1637 that there were none. He wrote in the margin of a book that he had a marvelous prooftoo complex to fit in the margin. For centuries, mathematicians wished Fermat had beenreading a bigger book. The only reasons for taking such an unsupported conjectureseriously are (a) the proofs Fermat did furnish were brilliant, (b) in every othercase where Fermat claimed to have proven something profound without actually providing aproof, a proof was later found and (c) Fermat's Last Theorem survived every assault madeon it. He has credibility.

The possibility that there might be a simple solution to a famous problem attractedcranks by the score. (Why exactly cranks think they will become famous by solving famousproblems is a mystery. How many mathematicians have megabuck endorsement contracts?) Mostmathematicians now think Fermat was probably mistaken in thinking he had a proof (evenBrett Favre gets sacked occasionally.) Fermat's Last Theorem was proven for specificpowers of n to such large values - in the millions - that nobody would ever actually writea counter-example on paper. Finally, in 1993, the theorem was proven by Andrew Weil ofPrinceton 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 Eulerconjectured that it always required n terms to sum to an nth power: two squares, threecubes, 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 Corporationdid 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 couldhope to find, sums of cubes do not yield to formulas anything like as neat as those forPythagorean triplets. One additional complication is that cubes can be negative. We couldignore negative numbers in dealing with Pythagorean triplets, but we cannot in dealingwith cubes. Also, Pythagorean triplets generally have terms that are more or less the samesize. The largest triplet with a term less than 10 is 9, 40; 41. Any term larger than 41will differ from all other squares by more than 81. But there is much less limitation oncubic 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, soit 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 findthis 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 probablythe real limit.

polynomial Generators

There are quite a few polynomialsknown 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; dfor 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 cubesdoes. Vieta's formulas yield large numbers that are often not relatively prime. Elementaryreasoning suggests that if it takes two variables to create a Pythagorean triple, itshould take at least three to make the cubic analogue, so Vieta's solution isnot complete. 

Euler, 1760

We can write A³ + B³ = D³ - C³, and then factor bothpolynomials. Unfortunately, this seductively symmetrical approach tends to leadto 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 achain of substitutions leading back to extremely complex formulas with noattempt 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 theformula 
A³ + B³ = C³ + D³. Since cubescan 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. Clickhere for a complete listing of quartets with a, b and c up to 1000.

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. Thefourth 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, wehave to find combinations that sum to something ending in those digits. Using sums of fourterms, 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 powerending in 6 could belong to a number ending in any even digit. Nevertheless, this improvesthe odds a lot over the 10,000 combinations if any combination of four ending digits werepossible. 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.)

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

Sums of Fifth and Higher Powers

Amazingly enough, there is a formula for fifth powers that generates an infinitenumber 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 completesolution, 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 deeperproblem than one might guess. It will be useful to write the sum as a function Sn(x), thatis, the sum of the nth powers of consecutive integers up to x. For n=0, the answer istrivial: S0(x) = x.

For n=1, a little trick helps. Pair up the first and last terms in the series, thesecond and next to last, and so on. We have 1 + x, 2 + (x-1), 3+ (x-2) and so on. If x iseven, 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 pairplus 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 isno 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.

Readers with some advanced math background may recognize the first column asBernoulli Numbers (with alternating signs) for the even powers. The patterns inthe table are easiest to see along the diagonals. From right to left, the termsin 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

Multigrades

One might guess that it is very unlikely to have sums of powers that hold for severaldifferent 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 multigradeand 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 inparentheses.

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 useconsistent notation or format.

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Created 2 January 1997: Last Update 24 May 2020

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