Pierre de Fermat (1630 - 1665)






David Wasserman
ma163sar
Written 6/8/95
Presented 6/9/95
Pierre de Fermat was born in 1601 and died in 1665. By profession he was a lawyer and jurist, and he studied mathematics in his spare time. He is mostly remembered for his work in number theory. He also worked in geometry. Fermat was a contemporary of Descartes, and they both independently discovered analytic geometry.

Fermat's most famous result, his so-called "Last Theorem," was found written in the margin of his copy of Diophantus' "Arithmetica." He wrote that he had a proof of the theorem, but the proof was not found in his effects. Also, in letters to other mathematicians, he frequently claimed to have proved results, but did not provide the proofs. He wanted to let others have the pleasure of discovering the proof, but this practice alienated some of his correspondents. Sometimes he provided proofs that were incomplete. This also annoyed his correspondents, who pressed him for more details.

Some of the gaps in these proofs were due to the fact that Fermat expected his readers to be familiar with certain works of François Viete (1540-1603) and the ancient Greek geometers, and to be able to apply the correct theorems from these works. Also, Fermat was strongly influenced by Viete, who revived interest in Greek analysis. The ancient Greeks divided their geometric arguments into two categories, analysis and synthesis. Synthesis was what we now call proof. Analysis meant assuming the proposition in question, and deducing from it something already known. This is not a logically valid way to verify the proposition. However, many theorems of geometry have converses that are also theorems, so analysis often produced correct results, and it worked often enough that it became common practice. Fermat recognized the need for synthesis, but he would often give an analysis of a theorem, and then state that it could easily be converted to a synthesis.

Fermat enjoyed producing his results, but was not willing to do the clean-up work required to make them suitable for publication. When others published his results, he would not allow them to put his name on them, because he wanted to avoid controversy. The following is a sample of theorems that Fermat did leave proofs of, written in modern notation and terminology.

Theorem: If a right triangle has integral sides, its area is not the square of an integer.

The proof is to long to include here. It can easily be modified to a proof that no fourth power is a sum of two fourth powers, a special case of the Last Theorem.

Theorem: If points A and B and a real number k are given, then the set of all points D, such that

(length of AD)2 - (length of BD)2 = k, is a line.

Proof: Let C be the point on line AB such that

(length of AC)2 - (length of BC)2 = k.
Let m be the line through A and B, and Let D be any point on the perpendicular to m through C. Then
(length of AD)2 - (length of BD)2 = (length of AC)2 - (length of BC)2 = k.
(Here Fermat has appplied the Pythagorean theorem to triangles ACD and BCD, and subtracted (length of CD)2 from both sides.)

Theorem: Let a spiral be described by the polar equation

R/(R - r) = (alpha/theta)2,
where theta ranges from 0 to alpha. Then the area enclosed by this spiral and the x-axis is 8/15 of the area of the sector of angle alpha in the circle of radius R.

Proof: By solving for (R - r), we see that the section of the radius of the circle that lies outside the spiral, i.e., R - r, grows as the square of theta. Let the sector of angle alpha be split into N equal sectors. For each sector there is a circular arc inscribed in the spiral, and also a circumscribed circular arc. The segment between the circle and the spiral at the end of the ith sector is to that at the end of the (i - 1)th as

i2:(i - 1)2.
Hence, in the ith subsector, the radius of the inscribed arc is to the radius of the circle (both now measured from the center) as
N2 - i2:N2
and the radius of the circumscribed arc to the radius of the circle as
N2 - (i - 1)2:N2.
Now, since the areas of similar sectors of circles are to each other as the squares of the radii of the circles, the area of the inscribed sector is to that of the sector of the referent circle as
(N2 - i2)2:N4,
and for the circumscribed sector the ratio is
(N2 - (i - 1)2)2:N4.
Forany value of N, the ratio of the area of the ith sector of the spiral to the area of the ith sector of the referent circle lies between these two limits. Summing from 1 to N, we get for the ratio of the spiral to the circle, an upper bound of 1/(N times N4) times Sum 1:N of
[(N2 - (i - 1)2)2]
and a lower bound of 1/(N times N4) times Sum 1:N of
[(N2 - i2)2] = 1/(N5) times [N5 - 2N2 (sum 1: N of i2) + Sum 1:N of i4].
These two bounds differ by 1/N, and the ratio is between them for all N, so the last expression must converge to the ratio as N increases. We can simplify it using the formula
5(Sum 1:N of i4) = (4N + 2)N2(N + 1)2/4 - Sum 1:N of i2
to eliminate the sum of fourth powers, and using the formula
Sum 1:N of i2 = N(N + 1)(2N + 1)/6,
the result follows.

Theorem: No number of the form 4k - 1, where k is an integer, is a sum of two squares.

Proof: A square of an even number is divisible by 4. A square of an odd number 2h + 1 is 4h2 + 4h + 1 = 4k + 1. Therefore 4k - 1 is not a square. Suppose

4k - 1 = x2 + y2.
Since 4k - 1 is odd, one of x2 and y2 must be odd, but not both. Assume without loss of generality that y2 is odd. Then 4k = x2 + (y2 + 1), so k = (x/2)2 + (y2 + 1)/4. Since x2 is even, x is even, so (x2)/2 is an integer, so (y2 + 1)/4 is an integer. Call this number m; then y2 = 4m - 1. But we already showed that a number of this form can not be a square, a contradiction.

Problem: Find the maximum value of bx2 - x3 in the first quadrant, where b is any positive number.

Solution: Let c be any value between 0 and the maximum value. The equation bx2 - x3 = c has two positive roots. Call them x and y. Then

bx2 - x3 = by2 - y3.
Rearranging,
b(x2 - y2) = x3 - y3.
Dividing by x - y,
b(x + y) = x2 + xy + y2.
This holds for any value of c. At the maximum value, the two roots become equal, so setting x = y, 2bx = 3x2, so x = 2b/3.

This solution is correct but there is an flaw in the derivation.

Exercise: Find the flaw in the derivation.





Answer to exercise: Fermat divides by (x - y) and then sets x equal to y; this constitutes division by 0. This problem can be circumvented by taking the limit of x as c approaches the maximum, which suggests why we now find maxima and minima using analysis instead of algebra.

Source: Mahoney, Michael Sean, "The Mathematical Career of Pierre de Fermat."

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