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In H. Eves Return to Mathematical Circles, Boston: Prindle, Weber and Schmidt, 1988.

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# Sums of Powers of Positive Integers

## Pythagoras (c. 570-500 BCE), Turkey-Greece-Italy

The Greek mathematician, philosopher, and mystic Pythagoras is said to have lived with his followers, the Pythagoreans, at Croton in what is now southern Italy.  The Pythagoreans took as their motto “All is Number” and were believed to have experimented with number properties by arranging pebbles on a flat surface.  As a result, they saw what we would describe as a sum of successive positive integers as a triangle or triangular number (Fig. 1).

Figure 1.  The first four triangular numbers

Their pebble experiments led them to see that two copies of the same triangular number could be fitted together to form an oblong number; hence, for example, twice the triangular number 15 = 1 + 2 + 3 + 4 + 5 could be viewed as the oblong number 5 x 6 = 30 (Fig. 2).

Figure 2. Twice a triangular number is an oblong number, or, in modern notation, $$2(1 + 2 + 3 + \cdots + n) = n(n + 1).$$  Here we see that $$2(1 + 2 + 3 + 4 + 5) = 5\cdot 6.$$

Equivalently, any triangular number was half an oblong number; for example,

$$1 + 2 + 3 + 4 + 5 = {{5 \cdot 6} \over 2},$$

and, in general,

$$1 + 2 + 3 + \cdots + n = {{n(n + 1)} \over 2}$$

for any positive integer n.

Exercise 1:  Use pebbles to illustrate that the sum of every two consecutive triangular numbers is a square number.  Hint:  Examples of pairs of consecutive triangular numbers include 6 and 10, 10 and 15, and 15 and 21.

The solution to this exercise is available here.

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Beery, Janet, "Sums of Powers of Positive Integers," Loci (February 2009), DOI: 10.4169/loci003284

### Karaji's solution

I just wonder if the Karaji Generalized his method of solutions? Is there any information about that. For instance did he found sum of fifth powers?

Al-Karaji and higher powers by Janet Beery (posted: 01/20/2010 )
My understanding is that there is no evidence that al-Karaji derived a "formula" for the sum of the fourth, fifth, or higher powers. His justification for his formula for the sum of the cubes was ingenious but al-Haytham's idea seems more readily generalizable.

### On the symmetry of the resulting polynomials

== Conjecture == It would be interesting to know that * sums of odd powers can be factored as a polynomial on (n Ã¢â€¹â€¦ (n + 1) / 2) and is therefore symmetric with the center at (Ã¢Ë†â€™Ã‚Â½); and that * sums of even powers can be factored to a function like above multiplied by (2 Ã¢â€¹â€¦ n + 1) and therefore antisymmetric with the center at (Ã¢Ë†â€™Ã‚Â½). It would be interesting to know whether any rule concerning the polynomials thus obtained.

#### Replies:

Re: On the symmetry of the resulting polynomials by Rick Mabry (posted: 12/30/2011 )
Christopher, see the following article, especially section 3 (Faulhaber Polynomials). Although my browsers don't let me view the code you pasted, I think the article confirms some of what you're getting at (some of which might be gleaned from page 8 of the article we're viewing). A. F. Beardon, Sums of Powers of Integers, American Mathematical Monthly, Vol. 103, No. 3 (Mar., 1996), pp. 201-213.

More about symmetry: by Janet Beery (posted: 01/03/2013 )
The most succinct, beautiful, and perhaps even historically plausible explanation I've seen for the symmetry of the Faulhaber polynomials about -1/2 was in a recent article by Reuben Hersh in the College Math Journal 43:4 (Sept. 2012), pp. 322-4. His approach? Extend the polynomials to negative values of n.