Proof of Fermat's Last Theorem for specific exponents
is a theorem in number theory, originally stated by Pierre de Fermat in 1637 and proved by Andrew Wiles in 1995. The statement of the theorem involves an integer exponent n larger than 2. In the centuries following the initial statement of the result and before its general proof, various proofs were devised for particular values of the exponent n. Several of these proofs are described below, including Fermat's proof in the case n = 4, which is an early example of the method of infinite descent.
Mathematical preliminaries
Fermat's Last Theorem states that no three positive integers can satisfy the equation an + bn = cn for any integer value of n greater than two.Factors of exponents
A solution for a given n leads to a solution for all the factors of n: if h is a factor of n then there is an integer g such that n = gh. Then is a solution for the exponent h:Therefore, 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 ap + bp = cp corresponds to a general integer solution to the equation ap + bp + cp = 0. For example, if solves the first equation, then 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 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 equationIf 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 of a, b, and c. Then may be written as a = gx, b = gy, and c = gz where the three numbers are pairwise coprime. In other words, the greatest common divisor of each pair equals one
If is a solution of Fermat's equation, then so is, since the equation
implies the equation
A pairwise coprime solution 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 evenly 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 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 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 states that any natural number can be written in only one way 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 is built.One consequence of this unique factorization property is that if a pth power of a number equals a product such as
and if u and v are coprime, then u and v are themselves the pth power of two other numbers, u = rp and v = sp.
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, Fermat's Last Theorem has been separated into two cases that are proven separately. The first case is to show that there are no primitive solutions to the equation xp + yp = zp under the condition that p does not divide the product xyz. The second case 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.''n'' = 4
Only one mathematical proof by Fermat has survived, in which Fermat uses the technique of infinite descent to show that the area of a right triangle with integer sides can never equal the square of an integer. This result is known as Fermat's right triangle theorem. As shown below, his proof is equivalent to demonstrating that the equationhas no primitive solutions in integers. In turn, this is sufficient to prove Fermat's Last Theorem for the case n = 4, since the equation a4 + b4 = c4 can be written as c4 − b4 = 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 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, where the area equals and, by the Pythagorean theorem, u2 + v2 = w2. If the area were equal to the square of an integer sgiving
Adding u2 + v2 = w2 to these equations gives
which can be expressed as
Multiplying these equations together yields
But as Fermat proved, there can be no integer solution to the equation
of which this is a special case with z =, x = w and y = 2s.
The first step of Fermat's proof is to factor the left-hand side
Since x and y are coprime, the greatest common divisor of x2 + y2 and x2 − y2 is either 2 or 1. 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 form a primitive Pythagorean triple, they can be writtenwhere d and e are coprime and d > e > 0. Thus,
which produces another solution that is smaller. As before, there must be a lower bound on the size of solutions, while this argument always produces a smaller solution than any given one, and thus the original solution is 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 squareThe 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, so we define integers u and v as
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 form a primitive Pythagorean triple
they can be expressed in terms of smaller integers d and e using Euclid's formula
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
The solution is another solution to the original equation, but smaller. Applying the same procedure to would produce another solution, still smaller, and so on. But this is impossible, since natural numbers cannot be shrunk indefinitely. Therefore, the original solution was impossible.
''n'' = 3
Fermat sent the letters in which he mentioned the case in which n = 3 in 1636, 1640 and 1657.Euler sent a letter in which he gave a proof of the case in which n = 3 to Goldbach on 4 August 1753.
Euler had the complete and pure elemental proof in 1760.
The case n = 3 was proven by Euler in 1770. Independent proofs were published by several other mathematicians, including Kausler, Legendre, Calzolari, Lamé, Tait, Günther, Gambioli, Krey, Rychlik, Stockhaus, Carmichael, van der Corput, Thue, and Duarte.
| date | result/proof | published/not published | work | name |
| 1621 | none | published | Latin version of Diophantus's Arithmetica | Bachet |
| around 1630 | only result | not published | a marginal note in Arithmetica | Fermat |
| 1636, 1640, 1657 | only result | published | letters of n = 3 | Fermat |
| 1670 | only result | published | a marginal note in Arithmetica | Fermat's son Samuel published the Arithmetica with Fermat's note. |
| 4 August 1753 | only result | published | letter to Goldbach | Euler |
| 1760 | proof | not published | complete and pure elemental proof | Euler |
| 1770 | proof | published | incomplete but elegant proof in Elements of Algebra | Euler |
As Fermat did for the case n = 4, Euler used the technique of infinite descent. The proof assumes a solution 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
where the non-zero integers u and v are coprime and have different parity. Since x = u + v and y = u − v, it follows that
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 or 3.
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 sSince 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
so that
Since u is even and v odd, then e is even and f is odd. Since
The factors 2e,, and 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
which yields a smaller solution k3 + l3 + m3= 0. Therefore, by the argument of infinite descent, the original solution 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
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
By the lemma above, since s is odd and its cube is equal to a number of the form 3w2 + v2, it too can be expressed in terms of smaller coprime numbers, e and f.
A short calculation shows that
Thus, e is odd and f is even, because v is odd. The expression for 18w then becomes
Since 33 divides r3 we have that 3 divides r, so 3 is an integer that equals 2f . 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.
which yields a smaller solution k3 + l3 + m3= 0. Therefore, by the argument of infinite descent, the original solution was impossible.
''n'' = 5
Fermat's Last Theorem for n = 5 states that no three coprime integers x, y and z can satisfy the equationThis was proven neither independently nor collaboratively by Dirichlet and Legendre around 1825. Alternative proofs were developed by Gauss, Lebesgue, Lamé, Gambioli, Werebrusow, Rychlik, van der Corput, and Terjanian.
Dirichlet's proof for n = 5 is divided into the two cases defined by Sophie Germain. In case I, the exponent 5 does not divide the product xyz. In case II, 5 does divide xyz.
- Case I for n = 5 can be proven immediately by Sophie Germain's theorem if the auxiliary prime θ = 11.
- Case II is divided into the two cases and II) by Dirichlet in 1825. [|Case II] is the case which one of x, y, z is divided by either 5 and 2. Case II is the case which one of x, y, z is divided by 5 and another one of x, y, z is divided by 2. In July 1825, Dirichlet proved the case II for n = 5. In September 1825, Legendre proved the case II for n = 5. After Legendre's proof, Dirichlet completed the proof for the case II for n = 5 by the extended argument for the case II.
Proof for Case A
and therefore
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 indivisible 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.
Without loss of generality, x and y can be designated as the two equivalent numbers modulo 5. That equivalence implies that
However, the equation x ≡ y also implies that
Combining the two results and dividing both sides by x5 yields a contradiction
Thus, case A for n = 5 has been proven.