Pierre de Fermat, who lived most of his life in Toulouse, France, was a lawyer who spent his spare time doing mathematics. Like Harriot, he never published any of his work, but he did share hints of it in letters to Marin Mersenne, Gilles Persone de Roberval, and others. Fermat is famous for stating, but not proving, Fermat’s Last Theorem, which finally was proved by Andrew Wiles in 1994.

In letters to Mersenne and Roberval in 1636, Fermat stated the following results about the generalized triangular numbers (Mahoney, pp. 230-231):

The last side multiplied by the next greater makes twice the triangle. The last side multiplied by the triangle of the next greater side makes three times the pyramid. The last side multiplied by the pyramid of the next greater side makes four times the triangulotriangle. And so on by the same progression in infinitum.

We have seen the triangular numbers. The triangular numbers 1, 3, 6, 10, 15, 21, … are sums of successive positive integers; the pyramidal numbers 1, 4, 10, 20, 35, 56, … are sums of successive triangular numbers; the triangulotriangular numbers (as Fermat called them) 1, 5, 15, 35, 70, 126, … are sums of successive pyramidal numbers; and so on. For example, the fourth triangular number, 10, is given by 10 = 1 + 2 + 3 + 4, and the third pyramidal number, 10, is given by 10 = 1 + 3 + 6. If we denote the *n*th triangular number by \(T_n,\) the *n*th pyramidal number by \(P_n,\) and the *n*th triangulotriangular number by \(TT_n,\) then we can translate Fermat’s first three sentences to the following three equations in our notation: $$n(n + 1) = 2T_n,\ nT_{n + 1} = 3P_n, \ {\rm and}\ nP_{n + 1} = 4TT_n.$$

The first equation gives us the formula, $$T_n = {{n(n + 1)} \over 2}$$

for the *n*th triangular number or, equivalently, the sum of the first *n* positive integers. This result, substituted into the second equation, then yields $$n\cdot{(n + 1)(n + 2) \over 2} = 3P_n\ {\rm or}\ P_n = {n(n + 1)(n + 2)\over 2 \cdot 3}$$

for the *n*th pyramidal number. Finally, this last result, substituted into the third equation, gives $$TT_n = {{n(n + 1)(n + 2)(n + 3)} \over {2 \cdot 3 \cdot 4}}$$

for the *n*th triangulotriangular number.

In the same letters, Fermat gave two hints about sums of powers of integers. First, he wrote the following formula for the sum of the fourth powers of the first *n* positive integers (Mahoney, p. 231).

If you multiply four times the greatest number increased by two by the square of the triangle of numbers, and from the product you subtract the sum of the squares of the individual numbers, five times the sum of the fourth powers will result.

Translating Fermat’s words into our modern notation, we have

$$(4n + 2)\,T_n ^2 - (1^2 + 2^2 + \cdots + n^2 ) = 5(1^4 + 2^4 + \cdots + n^4 ),$$

or, substituting formulas for \(T_n\) and for $$1^2 + 2^2 + \cdots + n^2,$$

$$(4n + 2)\,\left( {{{n(n + 1)} \over 2}} \right)^2 - {{n(n + 1)(2n + 1)} \over 6} = 5(1^4 + 2^4 + \cdots + n^4 ).$$

Solving for $$1^4 + 2^4 + \cdots + n^4,$$

we get

$$1^4 + 2^4 + \cdots + n^4 = {{n(n + 1)(2n + 1)\left( {2\, \cdot {{n(n + 1)} \over 4} - {1 \over 6}} \right)} \over 5} = {{n(n + 1)(2n + 1)(3n^2 + 3n - 1)} \over {30}}$$

$$ = {{6n^5 + 15n^4 + 10n^3 - n} \over {30}} = {{n^5 } \over 5} + {{n^4 } \over 2} + {{n^3 } \over 3} - {n \over {30}}.$$

It would make sense that Fermat would report his formula for the sum of fourth powers, because it would be the first such formula he thought was unknown (Mahoney, p. 232). In his letter to Roberval, Fermat explained his use for the formulas described above (Mahoney, p. 231).

All these propositions, however pretty in themselves, have aided me in the quadrature that I’m pleased you value.

Fermat’s biographer, Michael Mahoney, has argued that, like Archimedes, Fermat’s results in “quadrature” would have required that he have verbal formulas for sums of powers of integers (Mahoney, pp. 231-233). More specifically, Fermat’s success in computing the definite integral of a function of the form *cx*^{k} would have required that he describe a formula for the sum of the *k*th powers of the first *n* positive integers. In reality, an inequality like that of Archimedes or a limit involving only terms with exponent *k* + 1 would have sufficed, so we cannot say with certainty that Fermat had formulas beyond the one he described for the sum of fourth powers.

Based on the evidence above, Mahoney surmised that Fermat used the following method to derive sums of powers (Mahoney, pp. 231-232). We will assume that Fermat already knew or had derived formulas for the sums of the first *n* positive integers and of their squares and cubes, and now wished to obtain a formula for the sum of the fourth powers of the first *n* positive integers. The generalized triangular numbers after the triangulotriangular numbers are sometimes called triangulopyramidal numbers. If \(TP_n\) is the *n*th triangulopyramidal number, then, by definition, \(TP_n\) is the sum of the first *n* triangulotriangular numbers; that is, $$TP_n = \sum_{k = 1}^n {TT_k }.$$

According to Fermat’s first result above and our reasoning there, $$TP_n = {{n(n + 1)(n + 2)(n + 3)(n + 4)} \over {2 \cdot 3 \cdot 4 \cdot 5}}.$$

Using our formula for \(TT_n,\) the defining equation then becomes $${{n(n + 1)(n + 2)(n + 3)(n + 4)} \over {2 \cdot 3 \cdot 4 \cdot 5}} = \sum_{k = 1}^n {{{k(k + 1)(k + 2)(k + 3)} \over {2 \cdot 3 \cdot 4}}}, $$

or $${{n(n + 1)(n + 2)(n + 3)(n + 4)} \over {2 \cdot 3 \cdot 4 \cdot 5}} = \sum_{k = 1}^n {{{k^4 + 6k^3 + 11k^2 + 6k} \over {24}}}.$$

Solving for the sum of the fourth powers, we obtain $${1 \over {24}}\sum_{k = 1}^n {k^4 } = {{n(n + 1)(n + 2)(n + 3)(n + 4)} \over {120}} - {1 \over 4}\sum_{k = 1}^n {k^3 } - {{11} \over {24}}\sum_{k = 1}^n {k^2 } - {1 \over 4}\sum_{k = 1}^n k .$$

If we now substitute our formulas for the sums of the first *n* positive integers and of their squares and cubes, multiply both sides of the equation by 24, and simplify, we will obtain one (or all) of the formulas for the sums of fourth powers given above.

Harriot actually wrote symbolic formulas like those labeled \(T_n,\) \(P_n,\) \(TT_n,\) and \(TP_n\) (Harriot, folios 108-109; or Beery and Stedall, pp. 58-61), but, as we have seen, he used another method to derive his formulas for sums of powers of integers. This method, explained in detail above, was described more briefly in the book by Beery and Stedall (pp. 19-20).

**Exercise 15:** Show how to write Fermat’s formula for the sum of the first *n* fourth powers as a (rational) multiple of *n*(*n* + 1)(2*n* + 1)/6.

**Exercise 16:** Write Fermat’s formula for the sixth order triangular numbers, 1, 7, 28, 84, … in words and in symbols, perhaps denoting the *n*th sixth order triangular number as *T*_{n}^{6}. *Note:* The second order triangular numbers are the triangular numbers 1, 3, 6, 10, …; the third order triangular numbers the pyramidal numbers 1, 4, 10, 20, …; the fourth order triangular numbers the triangulotriangular numbers 1, 5, 15, 35, …; and the fifth order triangular numbers the triangulopyramidal numbers 1, 6, 21, 56, ….

**Exercise 17:** Use your formula for the *n*th sixth order triangular number *T*_{n}^{6} from Exercise 16, along with the formula for the *n*th triangulopyramidal number *TP*_{n} (or *T*_{n}^{5}) and the four formulas already derived for the sums of the first *n* positive integers and their squares, cubes, and fourth powers, to derive a formula for the sum of the first *n* fifth powers.

For solutions to these exercises, click here.