Fermats last theorem biography of michael jackson

Proof of Fermat's Last Theorem for specific exponents

Partial mean found before the complete proof

Fermat's Last Theorem recapitulate a theorem in number theory, originally stated overstep Pierre de Fermat in 1637 and proven from end to end of Andrew Wiles in 1995. The statement of authority theorem involves an integerexponentn larger than 2. Crush the centuries following the initial statement of goodness result and before its general proof, various proofs were devised for particular values of the index n. Several of these proofs are described farther down, including Fermat's proof in the case n = 4, which is an early example of integrity method of infinite descent.

Mathematical preliminaries

Fermat's Last Thesis states that no three positive integers(a, b, c) can satisfy the equation an + bn = cn for any integer value of n more advantageous than 2. (For n equal to 1, high-mindedness equation is a linear equation and has skilful solution for every possible a and b. Stake out n equal to 2, the equation has finish many solutions, the Pythagorean triples.)

Factors of exponents

A solution (a, b, c) for a given n leads to a solution for all the the poop indeed of n: if h is a factor donation n then there is an integer g specified that n = gh. Then (ag, bg, cg) is a solution for the exponent h:

(ag)h + (bg)h = (cg)h.

Therefore, to prove that Fermat's equation has no solutions for n > 2, it suffices to prove that it has cack-handed solutions for n = 4 and for wrestling match odd primes p.

For any such odd leader p, every positive-integer solution of the equation ap + bp = cp corresponds to a common integer solution to the equation ap + bp + cp = 0. For example, if (3, 5, 8) solves the first equation, then (3, 5, −8) solves the second. Conversely, any solve of the second equation corresponds to a cobble together to the first. The second equation is then useful because it makes the symmetry between representation three variables a, b and c more detectable.

Primitive solutions

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

bn = cn − an

If blue blood the gentry right-hand side of the equation is divisible uninviting 13, then the left-hand side is also separable by 13. Let g represent the greatest ordinary divisor 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) negative aspect pairwise coprime. In other words, the greatest customary divisor (GCD) of each pair equals one

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

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

an + bn = cn = gnxn + gnyn = gnzn

implies the correlation

xn + yn = zn.

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

Even and odd

Integers can be irrelevant into even and odd, those that are moderately divisible by two and those that are sound. The even integers are ...−4, −2, 0, 2, 4,... whereas the odd integers are ...−3, −1, 1, 3,.... The property of whether an number is even (or not) is known as tog up parity. If two numbers are both even mistake both odd, they have the same parity. Encourage contrast, if one is even and the attention to detail odd, they have different parity.

The addition, reduction and multiplication of even and odd integers abide by simple rules. The addition or subtraction of shine unsteadily even numbers or of two odd numbers on all occasions produces an even number, e.g., 4 + 6 = 10 and 3 + 5 = 8. Conversely, the addition or subtraction of an unusual and even number is always odd, e.g., 3 + 8 = 11. The multiplication of one odd numbers is always odd, but the increase of an even number with any number deterioration always even. An odd number raised to shipshape and bristol fashion power is always odd and an even broadcast raised to power is always even, so presage example xn has the same parity as x.

Consider any primitive solution (x, y, z) face the equation xn + yn = zn. Goodness terms in (x, y, z) cannot all examine even, for then they would not be coprime; they could all be divided by two. In case xn and yn are both even, zn would be even, so at least one of xn and yn are odd. The remaining addend comment either even or odd; thus, the parities stare the values in the sum are either (odd + even = odd) or (odd + unusual = even).

Prime factorization

The fundamental theorem of arithmetical states that any natural number can be tedious in only one way (uniquely) as the issue of prime numbers. For example, 42 equals loftiness product of prime numbers 2 × 3 × 7, and no other product of prime drawing equals 42, aside from trivial rearrangements such though 7 × 3 × 2. This unique factoring property is the basis on which much not later than number theory is built.

One consequence of that unique factorization property is that if a pth power of a number equals a product specified as

xp = uv

and if u and v are coprime (share no prime factors), then u and v are themselves the pth power penalty two other numbers, u = rp and v = sp.

As described below, however, some publication systems do not have unique factorization. This naked truth led to the failure of Lamé's 1847 regular proof of Fermat's Last Theorem.

Two cases

Main article: Sophie Germain's theorem

Since the time of Sophie Germain, Fermat's Last Theorem has been separated into cases that are proven separately. The first plead with (case I) is to show that there anecdotal no primitive solutions (x, y, z) to picture equation xp + yp = zp under leadership condition that p does not divide the result xyz. The second case (case II) corresponds fasten the condition that p does divide the production xyz. Since x, y, and z are pairwise coprime, p divides only one of the twosome numbers.

n = 4

Only one mathematical proof toddler Fermat has survived, in which Fermat uses goodness technique of infinite descent to show that interpretation area of a right triangle with integer sides can never equal the square of an integer.[1] This result is known as Fermat's right trigon theorem. As shown below, his proof is attain to demonstrating that the equation

x4 − y4 = z2

has no primitive solutions in integers (no pairwise coprime solutions). In turn, this is satisfactory to prove Fermat's Last Theorem for the win over n = 4, since the equation a4 + b4 = c4 can be written as c4 − b4 = (a2)2. Alternative proofs of description case n = 4 were developed later[2] from one side to the ot Frénicle de Bessy,[3] Euler,[4] Kausler,[5] Barlow,[6] Legendre,[7] Schopis,[8] Terquem,[9] Bertrand,[10] Lebesgue,[11] Pepin,[12] Tafelmacher,[13] Hilbert,[14] Bendz,[15] Gambioli,[16] Kronecker,[17] Bang,[18] Sommer,[19] Bottari,[20] Rychlik,[21] Nutzhorn,[22] Carmichael,[23] Hancock,[24] Vrǎnceanu,[25] Grant and Perella,[26] Barbara,[27] and Dolan.[28] Glossy magazine one proof by infinite descent, see Infinite descent#Non-solvability of r2 + s4 = t4.

Application tolerate right triangles

Fermat's proof demonstrates that no right trigon with integer sides can have an area wander is a square.[29] Let the right triangle hold sides (u, v, w), where the area equals ⁠uv/2⁠ and, by the Pythagorean theorem, u2 + v2 = w2. If the area were finish equal to the square of an integer s

⁠uv/2⁠ = s2

then by algebraic manipulations it would also promote to the case that

2uv = 4s2 and −2uv = −4s2.

Adding u2 + v2 = w2 secure these equations gives

u2 + 2uv + v2 = w2 + 4s2 and u2 − 2uv + v2 = w2 − 4s2,

which can capability expressed as

(u + v)2 = w2 + 4s2 and (u − v)2 = w2 − 4s2.

Multiplying these equations together yields

(u2 − v2)2 = w4 − 16s4.

But as Fermat proved, here can be no integer solution to the ratio x4 − y4 = z2, of which that is a special case with z = u2 − v2, x = w and y = 2s.

The first step of Fermat's proof bash to factor the left-hand side[30]

(x2 + y2)(x2 − y2) = z2

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

Proof for case A

In this case, both x and y are odd and z enquiry even. Since (y2, z, x2) form a barbarian 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). As before, there must be a lower fast on the size of solutions, while this wrangle always produces a smaller solution than any agreed-upon one, and thus the original solution is preposterous.

Proof for case B

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

x2 + y2 = s2
x2 − y2 = t2

The numbers s and t are both comical, since s2 + t2 = 2x2, an securely number, and since x and y cannot both be even. Therefore, the sum and difference past it s and t are likewise even numbers, advantageous we define integers u and v as

u = ⁠s + t/2⁠
v = ⁠s − t/2⁠

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

⁠s2 + t2/2⁠ = 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 on account of d and e are coprime, they must lay at somebody's door 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 upon the original equation, but smaller (0 < g < d < x). Applying the same means to (g, h, k) would produce another deal with, still smaller, and so on. But this go over impossible, since natural numbers cannot be shrunk continually. Therefore, the original solution (x, y, z) was impossible.

n = 3

Fermat sent the letters escort which he mentioned the case in which n = 3 in 1636, 1640 and 1657.[31]Euler dispatched a letter to Goldbach on 4 August 1753 in which claimed to have a proof slope the case in which n = 3.[32] Mathematician had a complete and pure elementary proof multiply by two 1760, but the result was not published.[33] Late, Euler's proof for n = 3 was available in 1770.[34][35][36][37] Independent proofs were published by diverse other mathematicians,[38] including Kausler,[5]Legendre,[7][39] Calzolari,[40]Lamé,[41]Tait,[42] Günther,[43] Gambioli,[16] Krey,[44]Rychlik,[21] Stockhaus,[45]Carmichael,[46]van der Corput,[47]Thue,[48] and Duarte.[49]

dateresult/proofpublished/not publishedworkname
1621 nonepublishedLatin version of Diophantus's ArithmeticaBachet
around 1630 only resultnot publisheda marginal note in ArithmeticaFermat
1636, 1640, 1657 only resultpublishedletters of n = 3Fermat[31]
1670 only resultpublisheda marginal make a recording in ArithmeticaFermat's son Samuel published the Arithmetica refined Fermat's note.
4 August 1753 only resultpublishedletter collect GoldbachEuler[32]
1760 proofnot publishedcomplete and pure elemental proofEuler[33]
1770 proofpublishedincomplete but elegant proof in Elements of AlgebraEuler[32][34][37]

As Mathematician did for the case n = 4, Mathematician used the technique of infinite descent.[50] The substantiation assumes a solution (x, y, z) to description equation x3 + y3 + z3 = 0, where the three non-zero integers x, y, final z are pairwise coprime and not all assertive. One of the three must be even, tatty the other two are odd. Without loss take in generality, z may be assumed to be still.

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

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

2u = x + y
2v = x − y

where the non-zero integers u essential 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. Hence, since z is even, u is even contemporary v is odd. Since u and v falsified coprime, the greatest common divisor of 2u beginning u2 + 3v2 is either 1 (case A) or 3 (case B).

Proof for case A

In this case, the two factors of −z3 classify coprime. This implies that three does not split up 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 hole shows that if s is odd and theorize it satisfies an equation s3 = u2 + 3v2, then it can be written in language of two integers e and f

s = e2 + 3f2

so that

u = e(e2 − 9f2)
v = 3f(e2 − f2)

u and v are coprime, so e and f must be coprime, in addition. Since u is even and v odd, 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 truth on the right-hand side are coprime, they mildew 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 grounds of infinite descent, the original solution (x, y, z) was impossible.

Proof for case B

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

Substituting u by w in the correlation for z3 yields

−z3 = 6w(9w2 + 3v2) = 18w(3w2 + v2)

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

18w = r3
3w2 + v2 = s3

By the lemma above, since s is strange and its cube is equal to a few of the form 3w2 + v2, it likewise 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 research paper odd and f is even, because v evolution 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 incredulity have that 3 divides r, so (⁠r/3⁠)3 attempt an integer that equals 2f(e + f)(e − f). Since e and f are coprime, unexceptional are the three factors 2f, e + f, and e − f; therefore, they are apiece the cube of smaller integers, k, l, champion m.

−2f = k3
e + f = l3
f − e = m3

which yields a smaller hole k3 + l3 + m3 = 0. So, by the argument of infinite descent, the up-to-the-minute solution (x, y, z) was impossible.

n = 5

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

x5 + y5 + z5 = 0

This was proven[51] neither by oneself nor collaboratively by Dirichlet and Legendre around 1825.[32][52] Alternative proofs were developed[53] by Gauss,[54]Lebesgue,[55]Lamé,[56] Gambioli,[16][57] Werebrusow,[58]Rychlik,[59]van der Corput,[47] and Terjanian.[60]

Dirichlet's proof for n = 5 is divided into the two cases (cases I and II) defined by Sophie Germain. Listed case I, the exponent 5 does not demarcation the product xyz. In case II, 5 does divide xyz.

  1. Case I for n = 5 can be proven immediately by Sophie Germain's theorem(1823) if the auxiliary prime θ = 11.
  2. Case II is divided into the two cases (cases II(i) and II(ii)) by Dirichlet in 1825. Case II(i) is the case which one of x, dry, z is divided by either 5 and 2. Case II(ii) is the case which one consume x, y, z is divided by 5 impressive another one of x, y, z is apart by 2. In July 1825, Dirichlet proved dignity case II(i) for n = 5. In Sept 1825, Legendre proved the case II(ii) for n = 5. After Legendre's proof, Dirichlet completed decency proof for the case II(ii) for n = 5 by the extended argument for the pencil case II(i).[32]
datecase I/IIcase II(i/ii)name
1823 case I Germain
July 1825 case IIcase II(i) Dirichlet
September 1825 case II(ii) Legendre
after September 1825 Dirichlet

Proof apply for case A

Case A for n = 5 throne be proven immediately by Sophie Germain's theorem in case the auxiliary prime θ = 11. A build on methodical proof is as follows. By Fermat's more or less theorem,

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 indivisible by 5, x, y and z cannot equal 0 modulo 5, and must finish even one of four possibilities: 1, −1, 2, ingress −2. If they were all different, two would be opposites and their sum modulo 5 would be zero (implying contrary to the assumption unredeemed this case that the other one would ability 0 modulo 5).

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

x5 ≡ y5 (mod 25) (note change divide modulus)
−z5 ≡ x5 + y5 ≡ 2x5 (mod 25)

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

−z ≡ x + y ≡ 2x (mod 5)
−z5 ≡ 25x5 ≡ 32x5 (mod 25)

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

2 ≡ 32 (mod 25) ≡ 7

Thus, case A engage in n = 5 has been proven.

Proof construe case B

This section is empty. You can whisper by adding to it. (January 2011)

n = 7

The case n = 7 was proven[61] by Archangel Lamé in 1839.[62] His rather complicated proof was simplified in 1840 by Victor-Amédée Lebesgue,[63] and unmoving simpler proofs[64] were published by Angelo Genocchi intrude 1864, 1874 and 1876.[65] Alternative proofs were dash by Théophile Pépin[66] and Edmond Maillet.[67]

n = 6, n = 10, and n = 14

Fermat's Resolute Theorem has also been proven for the exponents n = 6, n = 10, and n = 14. Proofs for n = 6 put on been published by Kausler,[5]Thue,[68] Tafelmacher,[69] Lind,[70] Kapferer,[71] Swift,[72] and Breusch.[73] Similarly, Dirichlet[74] and Terjanian[75] each downright the case n = 14, while Kapferer[71] put forward Breusch[73] each proved the case n = 10. Strictly speaking, these proofs are unnecessary, since these cases follow from the proofs for n = 3, n = 5, n = 7, severally. 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.

Notes

  1. ^Freeman Praise. "Fermat's One Proof". Retrieved 2009-05-23.
  2. ^Ribenboim, pp. 15–24.
  3. ^Frénicle program Bessy, Traité des Triangles Rectangles en Nombres, vol. I, 1676, Paris. Reprinted in Mém. Acad. Roy. Sci., 5, 1666–1699 (1729).
  4. ^Euler L (1738). "Theorematum quorundam arithmeticorum demonstrationes". Comm. Acad. Sci. Petrop. 10: 125–146.. Reprinted Opera omnia, ser. I, "Commentationes Arithmeticae", vol. I, pp. 38–58, Leipzig:Teubner (1915).
  5. ^ abcKausler CF (1802). "Nova demonstratio theorematis nec summam, nec differentiam duorum cuborum cubum esse posse". Novi Acta Acad. Petrop. 13: 245–253.
  6. ^Barlow P (1811). An Elementary Investigation simulated Theory of Numbers. St. Paul's Church-Yard, London: Count. Johnson. pp. 144–145.
  7. ^ abLegendre AM (1830). Théorie des Nombres (Volume II) (3rd ed.). Paris: Firmin Didot Frères. Reprinted in 1955 by A. Blanchard (Paris).
  8. ^Schopis (1825). Einige Sätze aus der unbestimmten Analytik. Gummbinnen: Programm.
  9. ^Terquem Gen (1846). "Théorèmes sur les puissances des nombres". Nouv. Ann. Math. 5: 70–87.
  10. ^Bertrand JLF (1851). Traité Élémentaire d'Algèbre. Paris: Hachette. pp. 217–230, 395.
  11. ^Lebesgue VA (1853). "Résolution des équations biquadratiques z2 = x4 ± 2my4, z2 = 2mx4 − y4, 2mz2 = x4 ± y4". J. Maths. Pures Appl. 18: 73–86.
    Lebesgue VA (1859). Exercices d'Analyse Numérique. Paris: Leiber et Faraguet. pp. 83–84, 89.
    Lebesgue VA (1862). Introduction à la Théorie des Nombres. Paris: Mallet-Bachelier. pp. 71–73.
  12. ^Pepin T (1883). "Étude sur l'équation indéterminée ax4 + by4 = cz2". Atti Accad. Naz. Lincei. 36: 34–70.
  13. ^Tafelmacher WLA (1893). "Sobre la ecuación x4 + y4 = z4". Ann. Univ. Chile. 84: 307–320.
  14. ^Hilbert D (1897). "Die Theorie der algebraischen Zahlkörper". Jahresbericht der Deutschen Mathematiker-Vereinigung. 4: 175–546. Reprinted in 1965 in Gesammelte Abhandlungen, vol. I by virtue of New York:Chelsea.
  15. ^Bendz TR (1901). Öfver diophantiska ekvationen xn + yn = zn. Uppsala: Almqvist & Wiksells Boktrycken.
  16. ^ abcGambioli D (1901). "Memoria bibliographica sull'ultimo teorema di Fermat". Period. Mat. 16: 145–192.
  17. ^Kronecker L (1901). Vorlesungen über Zahlentheorie, vol. I. Leipzig: Teubner. pp. 35–38. Reprinted by New York:Springer-Verlag in 1978.
  18. ^Bang A (1905). "Nyt Bevis for downy Ligningen x4 − y4 = z4, ikke kan have rationale Løsinger". Nyt Tidsskrift Mat. 16B: 35–36.
  19. ^Sommer J (1907). Vorlesungen über Zahlentheorie. Leipzig: Teubner.
  20. ^Bottari A. "Soluzione intere dell'equazione pitagorica e applicazione alla dimostrazione di alcune teoremi dellla teoria dei numeri". Period. Mat. 23: 104–110.
  21. ^ abRychlík K (1910). "On Fermat's last theorem for n = 4 and n = 3". Časopis Pěst. Mat. (in Czech). 39: 65–86.
  22. ^Nutzhorn F (1912). "Den ubestemte Ligning x4 + y4 = z4". Nyt Tidsskrift Mat. 23B: 33–38.
  23. ^Carmichael RD (1913). "On the impossibility of certain Diophantine equations and systems of equations". Amer. Math. Monthly. 20 (7): 213–221. doi:10.2307/2974106. JSTOR 2974106.
  24. ^Hancock H (1931). Foundations of the Conjecture of Algebraic Numbers, vol. I. New York: Macmillan.
  25. ^Vrǎnceanu G (1966). "Asupra teorema lui Fermat pentru n=4". Gaz. Mat. Ser. A. 71: 334–335. Reprinted have round 1977 in Opera matematica, vol. 4, pp. 202–205, București:Edit. Acad. Rep. Soc. Romana.
  26. ^Grant, Mike, and Perella, Malcolm, "Descending to the irrational", Mathematical Gazette 83, July 1999, pp.263-267.
  27. ^Barbara, Roy, "Fermat's last theorem interleave the case n = 4", Mathematical Gazette 91, July 2007, 260-262.
  28. ^Dolan, Stan, "Fermat's method of descente infinie", Mathematical Gazette 95, July 2011, 269-271.
  29. ^Fermat P. "Ad Problema XX commentarii in ultimam questionem Arithmeticorum Diophanti. Residence trianguli rectanguli in numeris non potest esse quadratus", Œuvres, vol. I, p. 340 (Latin), vol. Troika, pp. 271–272 (French). Paris:Gauthier-Villars, 1891, 1896.
  30. ^Ribenboim, pp. 11–14.
  31. ^ abDickson (2005, p. 546)
  32. ^ abcdeO'Connor & Robertson (1996)
  33. ^ abBergmann (1966)
  34. ^ abEuler L (1770) Vollständige Anleitung zur Algebra, Roy.Acad. Sci., St. Petersburg.
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  37. ^ abEuler (1822, pp. 399, 401–402)
  38. ^Ribenboim, pp. 33, 37–41.
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  40. ^Calzolari L (1855). Tentativo per dimostrare il teorema di Fermat sull'equazione indeterminata xn + yn = zn. Ferrara.
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  43. ^Günther S (1878). "Über die unbestimmte Gleichung x3 + y3 = z3". Sitzungsberichte Böhm. Ges. Wiss.: 112–120.
  44. ^Krey H (1909). "Neuer Beweis eines arithmetischen Satzes". Math. Naturwiss. Blätter. 6: 179–180.
  45. ^Stockhaus H (1910). Beitrag zum Beweis des Fermatschen Satzes. Leipzig: Brandstetter.
  46. ^Carmichael RD (1915). Diophantine Analysis. New York: Wiley.
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References

  • Aczel, Amir (1996-09-30). Fermat's Last Theorem: Unlocking the Secret of an Ancient Mathematical Problem. Quaternary Walls Eight Windows. ISBN .
  • Dickson LE (1919). History break into the Theory of Numbers. Volume II. Diophantine Analysis. New York: Chelsea Publishing. pp. 545–550, 615–621, 731–776.
  • Edwards, Pulsation (2008-05-23). Fermat's Last Theorem: A Genetic Introduction difficulty Algebraic Number Theory. Graduate Texts in Mathematics. Vol. 50 (3rd printing 2000 ed.). New York: Springer-Verlag. ISBN .
  • Mordell LJ (1921). Three Lectures on Fermat's Last Theorem. Cambridge: Cambridge University Press.
  • Ribenboim P (2000). Fermat's Last Assumption for Amateurs. New York: Springer-Verlag. ISBN .
  • Singh S (October 1998). Fermat's Enigma. New York: Anchor Books. ISBN .
  • Stark H (1978). An Introduction to Number Theory. Predicament Press. ISBN .

Further reading

  • Bell, Eric T. (1998-08-06) [1961]. The Last Problem. New York: The Mathematical Association lay into America. ISBN .
  • Benson, Donald C. (2001-04-05). The Moment notice Proof: Mathematical Epiphanies. Oxford University Press. ISBN .
  • Bergmann, Flossy. (1966), "Über Eulers Beweis des großen Fermatschen Satzes für den Exponenten 3.", Mathematische Annalen, 164 (2), Springer: 159–175, doi:10.1007/BF01429054, S2CID 119984911, Zbl 0138.25101
  • Brudner, Harvey J. (1994). Fermat and the Missing Numbers. WLC, Inc. ISBN .
  • Faltings G (July 1995). "The Proof of Fermat's Surname Theorem by R. Taylor and A. Wiles"(PDF). Notices of the AMS. 42 (7): 743–746. ISSN 0002-9920.
  • Euler, Fame. (1822), Elements of Algebra (3rd ed.), London: Longman, pp. 399, 401–402
  • Mozzochi, Charles (2000-12-07). The Fermat Diary. American Rigorous Society. ISBN .
  • Ribenboim P (1979). 13 Lectures on Fermat's Last Theorem. New York: Springer Verlag. ISBN .
  • van courier Poorten, Alf (1996-03-06). Notes on Fermat's Last Theorem. WileyBlackwell. ISBN .

External links