Proof of Fermat's Last Theorem for specific exponents

Mathematical preliminaries

Fermat's Last Theorem states that no three positive integers (a, b, c) can satisfy the equation aⁿ + bⁿ = cⁿ for any integer value of n greater than two. If n equals two, the equation has infinitely many solutions, the Pythagorean triple

Pythagorean triple

A Pythagorean triple consists of three positive integers a, b, and c, such that . Such a triple is commonly written , and a well-known example is . If is a Pythagorean triple, then so is for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are pairwise coprime...

s.

A solution (a, b, c) for a given n is equivalent to a solution for all the factors of n. For illustration, let n be factored into g and h, n = gh. Then (a^g, b^g, c^g) is a solution for the exponent h

(a^g)^h + (b^g)^h = (c^g)^h

Conversely, to prove that Fermat's equation has no solutions for n > 2, it suffices to prove that it has no solutions for n = 4 and for all odd primes p.

For any such odd exponent p, every positive-integer solution of the equation a^p + b^p = c^p corresponds to a general integer solution to the equation a^p + b^p + c^p = 0. For example, if (3, 5, 8) solves the first equation, then (3, 5, −8) solves the second. Conversely, any solution of the second equation corresponds to a solution to the first. The second equation is sometimes useful because it makes the symmetry between the three variables a, b and c more apparent.

Primitive solutions

If two of the three numbers (a, b, c) can be divided by a fourth number d, then all three numbers are divisible by d. For example, if a and c are divisible by d = 13, then b is also divisible by 13. This follows from the equation

bⁿ = cⁿ − aⁿ

If the right-hand side of the equation is divisible by 13, then the left-hand side is also divisible by 13. Let g represent the greatest common divisor

Greatest common divisor

In mathematics, the greatest common divisor , also known as the greatest common factor , or highest common factor , of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.For example, the GCD of 8 and 12 is 4.This notion can be extended to...

of a, b, and c. Then (a, b, c) may be written as a = gx, b = gy, and c = gz where the three numbers (x, y, z) are pairwise coprime

Coprime

In number theory, a branch of mathematics, two integers a and b are said to be coprime or relatively prime if the only positive integer that evenly divides both of them is 1. This is the same thing as their greatest common divisor being 1...

. In other words, the greatest common divisor (GCD) of each pair equals one

GCD(x, y) = GCD(x, z) = GCD(y, z) = 1

If (a, b, c) is a solution of Fermat's equation, then so is (x, y, z), since the equation

aⁿ + bⁿ = cⁿ = gⁿxⁿ + gⁿyⁿ = gⁿzⁿ

implies the equation

xⁿ + yⁿ = zⁿ.

A pairwise coprime solution (x, y, z) is called a primitive solution. Since every solution to Fermat's equation can be reduced to a primitive solution by dividing by their greatest common divisor g, Fermat's Last Theorem can be proven by demonstrating that no primitive solutions exist.

Even and odd

Integers can be divided into even and odd, those that are divisible by two and those that are not. The even integers are ...−4, −2, 0, 2, 4, whereas the odd integers are −3, −1, 1, 3,... The property of whether an integer is even (or not) is known as its parity. If two numbers are both even or both odd, they have the same parity. By contrast, if one is even and the other odd, they have different parity.

The addition, subtraction and multiplication of even and odd integers obey simple rules. The addition or subtraction of two even numbers or of two odd numbers always produces an even number, e.g., 4 + 6 = 10 and 3 + 5 = 8. Conversely, the addition or subtraction of an odd and even number is always odd, e.g., 3 + 8 = 11. The multiplication of two odd numbers is always odd, but the multiplication of an even number with any number is always even. An odd number raised to a power is always odd and an even number raised to power is always even.

In any primitive solution (x, y, z) to the equation xⁿ + yⁿ = zⁿ, one number is even and the other two numbers are odd. They cannot all be even, for then they would not be coprime; they could all be divided by two. However, they cannot be all odd, since the sum of two odd numbers xⁿ + yⁿ is never an odd number zⁿ. Therefore, at least one number must be even and at least one number must be odd. It follows that the third number is also odd, because the sum of an even and an odd number is itself odd.

Prime factorization

The fundamental theorem of arithmetic

Fundamental theorem of arithmetic

In number theory, the fundamental theorem of arithmetic states that any integer greater than 1 can be written as a unique product of prime numbers...

states that any natural number can be written in only one way (uniquely) as the product of prime numbers. For example, 42 equals the product of prime numbers 2×3×7, and no other product of prime numbers equals 42, aside from trivial re-arrangements such as 7×3×2. This unique factorization property is the basis on which much of number theory

Number theory

Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...

is built.

One consequence of this unique factorization property is that if a p^th power of a number equals a product such as

x^p = uv

and if u and v are coprime (share no prime factors), then u and v are themselves the p^th power of two other numbers, u = r^p and v = s^p.

As described below, however, some number systems do not have unique factorization. This fact led to the failure of Lamé's 1847 general proof of Fermat's Last Theorem.

Two cases

Since the time of Sophie Germain

Sophie Germain

Marie-Sophie Germain was a French mathematician, physicist, and philosopher. Despite initial opposition from her parents and difficulties presented by a gender-biased society, she gained education from books in her father's library and from correspondence with famous mathematicians such as...

, Fermat's Last Theorem has been separated into two cases that are proven separately. The first case (case I) is to show that there are no primitive solutions (x,y,z) to the equation x^p + y^p = z^p under the condition that p does not divide the product xyz. The second case (case II) corresponds to the condition that p does divide the product xyz. Since x, y, and z are pairwise coprime, p divides only one of the three numbers. In general, it is easier to prove case I than case II.

n = 4

Only one mathematical proof by Fermat has survived, in which Fermat uses the technique of infinite descent

Infinite descent

In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the fact that the natural numbers are well ordered. One typical application is to show that a given equation has no solutions. Assuming a solution exists, one shows that another exists, that...

to show that the area of a right triangle with integer sides can never equal the square of an integer. As shown below, his proof is equivalent to demonstrating that the equation

x⁴ − y⁴ = z²

has no primitive solutions in integers (no pairwise coprime solutions). In turn, this is equivalent to proving Fermat's Last Theorem for the case n=4, since the equation a⁴ + b⁴ = c⁴ can be written as c⁴ − b⁴ = (a²)². Alternative proofs of the case n = 4 were developed later by Frénicle de Bessy, Euler, Kausler, Barlow, Legendre, Schopis, Terquem, Bertrand, Lebesgue, Pepin, Tafelmacher, Hilbert, Bendz, Gambioli, Kronecker, Bang, Sommer, Bottari, Rychlik, Nutzhorn, Carmichael, Hancock, Vrǎnceanu, Grant and Perella, Barbara, and Dolan. For one version of Fermat's proof by infinite descent, see Infinite descent#Non-solvability of r2 + s4 = t4.

Application to right triangles

Fermat's proof demonstrates that no right triangle with integer sides can have an area that is a square. Let the right triangle have sides (u, v, w), where the area equals uv/2 and, by the Pythagorean theorem

Pythagorean theorem

In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...

, u² + v² = w². If the area were equal to the square of an integer s

uv/2 = s²

then

(u + v)² = w² + 4s²

(u − v)² = w² − 4s²

Multiplying these equations together yields

(u² − v²)² = w⁴ − 2⁴s⁴

But as Fermat proved, there can be no integer solution to the equation

x⁴ - y⁴ = z²

of which this is a special case with z = (u² - v²), x = w and y = 2s.

The first step of Fermat's proof is to factor the left-hand side

(x² + y²)(x² − y²) = z²

Since x and y are coprime (this can be assumed because otherwise the factors could be cancelled), the greatest common divisor of x² + y² and x² − y² is either 2 (case A) or 1 (case B). The theorem is proven separately for these two cases.

Proof for Case A

In this case, both x and y are odd and z is even. Since (y², z, x²) form a primitive Pythagorean triple, they can be written

z = 2de

y² = d² − e²

x² = d² + e²

where d and e are coprime and d > e > 0. Thus,

x²y² = d⁴ − e⁴

which produces another solution (d, e, xy) that is smaller (0 < d < x). By the above argument that solutions cannot be shrunk indefinitely, this proves that the original solution was impossible.

Proof for Case B

In this case, the two factors are coprime. Since their product is a square z², they must each be a square

x² + y² = s²

x² − y² = t²

The numbers s and t are both odd, since s² + t² = 2 x², an even number, and since x and y cannot both be even. Therefore, the sum and difference of s and t are likewise even numbers

u = (s + t)/2

v = (s − t)/2

Since s and t are coprime, so are u and v; only one of them can be even. Since y² = 2uv, exactly one of them is even. For illustration, let u be even; then the numbers may be written as u=2m² and v=k². Since (u, v, x) form a primitive Pythagorean triple

Pythagorean triple

/2 = u² + v² = x²

they can be expressed in terms of smaller integers d and e using Euclid's formula

u = 2de

v = d² − e²

x = d² + e²

Since u = 2m² = 2de, and since d and e are coprime, they must be squares themselves, d = g² and e = h². This gives the equation

v = d² − e² = g⁴ − h⁴ = k²

The solution (g, h, k) is another solution to the original equation, but smaller (0 < g < d < x). Applying the same procedure to (g, h, k) would produce another solution, still smaller, and so on. But this is impossible, since natural numbers cannot be shrunk indefinitely. Therefore, the original solution (x, y, z) was impossible.

n = 3

The case n = 3 was proven by Euler

Leonhard Euler

Leonhard Euler was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion...

in 1770. Independent proofs were published by several other mathematicians, including Kausler, Legendre

Adrien-Marie Legendre

Adrien-Marie Legendre was a French mathematician.The Moon crater Legendre is named after him.- Life :...

, Calzolari, Lamé

Gabriel Lamé

Gabriel Léon Jean Baptiste Lamé was a French mathematician.-Biography:Lamé was born in Tours, in today's département of Indre-et-Loire....

, Tait

Peter Guthrie Tait

Peter Guthrie Tait FRSE was a Scottish mathematical physicist, best known for the seminal energy physics textbook Treatise on Natural Philosophy, which he co-wrote with Kelvin, and his early investigations into knot theory, which contributed to the eventual formation of topology as a mathematical...

, Günther, Gambioli, Krey, Rychlik

Karel Rychlík

Karel Rychlík was a Czech mathematician who contributed significantly to the fields of algebra, number theory, mathematical analysis, and the history of mathematics.-External links:**...

, Stockhaus, Carmichael

Robert Daniel Carmichael

Robert Daniel Carmichael was a leading American mathematician. Carmichael was born in Goodwater, Alabama. He attended Lineville College, briefly, and he earned his bachelor's degree in 1898, while he was studying towards his Ph.D. degree at Princeton University. Carmichael completed the...

, van der Corput

Johannes van der Corput

Johannes Gualtherus van der Corput was a Dutch mathematician, working in the field of analytic number theory....

, Thue

Axel Thue

Axel Thue was a Norwegian mathematician, known for highly original work in diophantine approximation, and combinatorics....

, and Duarte.

As Fermat did for the case n = 4, Euler used the technique of infinite descent

Infinite descent

. The proof assumes a solution (x, y, z) to the equation x³ + y³ + z³ = 0, where the three non-zero integers x, y, and z are pairwise coprime and not all positive. One of the three must be even, whereas the other two are odd. Without loss of generality, z may be assumed to be even.

Since x and y are both odd, they cannot be equal. If x = y, then 2x³ = −z³, which implies that x is even, a contradiction.

Since x and y are both odd, their sum and difference are both even numbers

2u = x + y

2v = x − y

where the non-zero integers u and v are coprime and have different parity (one is even, the other odd). Since x = u + v and y = u − v, it follows that

−z³ = (u + v)³ + (u − v)³ = 2u(u² + 3v²)

Since u and v have opposite parity, u² + 3v² is always an odd number. Therefore, since z is even, u is even and v is odd. Since u and v are coprime, the greatest common divisor of 2u and u² + 3v² is either 1 (case A) or 3 (case B).

Proof for Case A

In this case, the two factors of −z³ are coprime. This implies that three does not divide u and that the two factors are cubes of two smaller numbers, r and s

2u = r³

u² + 3v² = s³

Since u² + 3v² is odd, so is s. A crucial lemma shows that if s is odd and if it satisfies an equation s³ = u² + 3v², then it can be written in terms of two coprime integers e and f

s = e² + 3f²

so that

u = e ( e² − 9f²)

v = 3f ( e² − f²)

Since u is even and v odd, then e is even and f is odd. Since

r³ = 2u = 2e (e − 3f)(e + 3f)

The factors 2e, (e–3f ), and (e+3f ) are coprime since 3 cannot divide e: If e were divisible by 3, then 3 would divide u, violating the designation of u and v as coprime. Since the three factors on the right-hand side are coprime, they must individually equal cubes of smaller integers

−2e = k³

e − 3f = l³

e + 3f = m³

which yields a smaller solution k³ + l³ + m³= 0. Therefore, by the argument of infinite descent

Infinite descent

, the original solution (x, y, z) was impossible.

Proof for Case B

In this case, the greatest common divisor of 2u and u² + 3v² is 3. That implies that 3 divides u, and one may express u = 3w in terms of a smaller integer, w. Since u is divisible by 4, so is w; hence, w is also even. Since u and v are coprime, so are v and w. Therefore, neither 3 nor 4 divide v.

Substituting u by w in the equation for z³ yields

−z³ = 6w(9w² + 3v²) = = 18w(3w² + v²)

Because v and w are coprime, and because 3 does not divide v, then 18w and 3w² + v² are also coprime. Therefore, since their product is a cube, they are each the cube of smaller integers, r and s

18w = r³

3w² + v² = s³

By the lemma above, since s is odd and equal to a number of the form 3w² + v², it too can be expressed in terms of smaller coprime numbers, e and f.

s = e² + 3f²

A short calculation shows that

v = e (e² − 9f²)

w = 3f (e² − f²)

Thus, e is odd and f is even, because v is odd. The expression for 18w then becomes

r³ = 18w = 54f (e² − f²) = 54f (e + f) (e − f) = 3³×2f (e + f) (e − f).

Since 3³ divides r³ we have that 3 divides r, so (r /3)³ is an integer that equals 2f (e + f) (e − f). Since e and f are coprime, so are the three factors 2e, e+f, and e−f; therefore, they are each the cube of smaller integers, k, l, and m.

−2e = k³

e + f = l³

e − f = m³

which yields a smaller solution k³ + l³ + m³= 0. Therefore, by the argument of infinite descent

Infinite descent

, the original solution (x, y, z) was impossible.

n = 5

Fermat's Last Theorem for n = 5 states that no three coprime integers x, y and z can satisfy the equation

x⁵ + y⁵ + z⁵ = 0

This was proven independently by Legendre

Adrien-Marie Legendre

Adrien-Marie Legendre was a French mathematician.The Moon crater Legendre is named after him.- Life :...

and Dirichlet

Johann Peter Gustav Lejeune Dirichlet

Johann Peter Gustav Lejeune Dirichlet was a German mathematician with deep contributions to number theory , as well as to the theory of Fourier series and other topics in mathematical analysis; he is credited with being one of the first mathematicians to give the modern formal definition of a...

around 1825. Alternative proofs were developed by Gauss

Carl Friedrich Gauss

Johann Carl Friedrich Gauss was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.Sometimes referred to as the Princeps mathematicorum...

, Lebesgue, Lamé

Gabriel Lamé

Gabriel Léon Jean Baptiste Lamé was a French mathematician.-Biography:Lamé was born in Tours, in today's département of Indre-et-Loire....

, Gambioli, Werebrusow, Rychlik

Karel Rychlík

Karel Rychlík was a Czech mathematician who contributed significantly to the fields of algebra, number theory, mathematical analysis, and the history of mathematics.-External links:**...

, van der Corput

Johannes van der Corput

Johannes Gualtherus van der Corput was a Dutch mathematician, working in the field of analytic number theory....

, and Terjanian

Guy Terjanian

Guy Terjanian is a French-Armenian mathematician who has worked on algebraic number theory. He achieved his Ph.D under Claude Chevalley in 1966, and at that time published a counterexample to the original form of a conjecture of Emil Artin, which suitably modified had just been proved as the...

.

Dirichlet's proof for n = 5 is divided into the two cases (cases I and II) defined by Sophie Germain

Sophie Germain

. In case I, the exponent 5 does not divide the product xyz. In case II, 5 does divide xyz. Since the three integers are coprime, they share no prime factors; hence, in case II only one of x, y and z is divisible by 5. Since the integers x, y and z are interchangeable in the equation, z can be designated without loss of generality as the integer divisible by 5 in case II.

Proof for Case A

Case A for n = 5 can be proven immediately by Sophie Germain's theorem

Sophie Germain's theorem

In number theory, Sophie Germain's theorem is a statement about the divisibility of solutions to the equation xp + yp = zp of Fermat's Last Theorem...

if the auxiliary prime θ = 11. A more methodical proof is as follows. By Fermat's little theorem

Fermat's little theorem

Fermat's little theorem states that if p is a prime number, then for any integer a, a p − a will be evenly divisible by p...

x⁵ ≡ x mod 5

y⁵ ≡ y mod 5

z⁵ ≡ z mod 5

and therefore

x + y + z ≡ 0 mod 5

This equation forces two of the three numbers x, y, and z to be equivalent modulo 5, which can be seen as follows: Since they are indivisble by 5, x, y and z cannot equal 0 modulo 5, and must equal one of four possibilities: ±1 or ±2. If they were all different, two would be opposites and their sum modulo 5 would be zero (implying contrary to the assumption of this case that the other one would be 0 modulo 5).

Without loss of generality, x and y can be designated as the two equivalent numbers modulo 5. That equivalence implies that

x⁵ ≡ y⁵ mod 25 (note change in modulo)

−z⁵ ≡ x⁵ + y⁵ ≡ 2 x⁵ mod 25

However, the equation x ≡ y mod 5 also implies that

−z ≡ x + y ≡ 2 x mod 5

−z⁵ ≡ 2⁵ x⁵ ≡ 32 x⁵ mod 25

Combining the two results and dividing both sides by x⁵ yields a contradiction

2 ≡ 32 mod 25

Thus, case A for n = 5 has been proven.

n = 7

The case n = 7 was proven by Gabriel Lamé

Gabriel Lamé

Gabriel Léon Jean Baptiste Lamé was a French mathematician.-Biography:Lamé was born in Tours, in today's département of Indre-et-Loire....

in 1839. His rather complicated proof was simplified in 1840 by Victor Lebesgue, and still simpler proofs were published by Angelo Genocchi

Angelo Genocchi

Angelo Genocchi was an Italian mathematician who specialized in number theory. He worked with Giuseppe Peano. The Genocchi numbers are named after him.-References:...

in 1864, 1874 and 1876. Alternative proofs were developed by Théophile Pépin and Edmond Maillet.

n = 6, 10, and 14

Fermat's Last Theorem has also been proven for the exponents n = 6, 10, and 14. Proofs for n = 6 have been published by Kausler, Thue

Axel Thue

Axel Thue was a Norwegian mathematician, known for highly original work in diophantine approximation, and combinatorics....

, Tafelmacher, Lind, Kapferer, Swift, and Breusch. Similarly, Dirichlet

Johann Peter Gustav Lejeune Dirichlet

and Terjanian

Guy Terjanian

each proved the case n = 14, while Kapferer and Breusch each proved the case n = 10. Strictly speaking, these proofs are unnecessary, since these cases follow from the proofs for n = 3, 5, and 7, respectively. Nevertheless, the reasoning of these even-exponent proofs differs from their odd-exponent counterparts. Dirichlet's proof for n = 14 was published in 1832, before Lamé's 1839 proof for n = 7.

External links

A blog that covers the history of Fermat's Last Theorem from Pierre Fermat to Andrew Wiles. Discusses various material which is related to the proof of Fermat's Last Theorem: elliptic curves, modular forms, Galois representations and their deformations, Frey's construction, and the conjectures of Serre and of Taniyama–Shimura. The story, the history and the mystery. The title of one edition of the PBS television series NOVA, discusses Andrew Wiles's effort to prove Fermat's Last Theorem. Edited version of ~2,000-word essay published in Prometheus magazine, describing Andrew Wiles's successful journey. Simon Singh and John Lynch's film tells the enthralling and emotional story of Andrew Wiles.

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