# Fermat's Last Theorem for Amateurs by Paulo Ribenboim

By Paulo Ribenboim

In 1995, Andrew Wiles accomplished an explanation of Fermat's final Theorem. even if this was once definitely an excellent mathematical feat, one shouldn't brush aside prior makes an attempt made by way of mathematicians and smart amateurs to resolve the matter. during this ebook, geared toward amateurs interested in the heritage of the topic, the writer restricts his consciousness completely to simple equipment that experience produced wealthy effects.

Similar number theory books

Number Theory 1: Fermat's Dream

This is often the English translation of the unique jap booklet. during this quantity, "Fermat's Dream", middle theories in sleek quantity thought are brought. advancements are given in elliptic curves, $p$-adic numbers, the $\zeta$-function, and the quantity fields. This paintings offers a sublime point of view at the ask yourself of numbers.

Initial-Boundary Value Problems and the Navier-Stokes Equations

This publication offers an advent to the monstrous topic of preliminary and initial-boundary price difficulties for PDEs, with an emphasis on functions to parabolic and hyperbolic structures. The Navier-Stokes equations for compressible and incompressible flows are taken to illustrate to demonstrate the consequences.

Extra resources for Fermat's Last Theorem for Amateurs

Example text

Special Cases (4) If the discriminant of a cubic polynomial with rational coefﬁcients is the sixth power of a√nonzero rational √ number, then 3 sin(2π/9), its roots are of the form r + s 3 sin(π/9), r + s √ r − s 3 sin(4π/9). Proof. (1) If x, y are integers and (x, y) satisﬁes 4x3 −3my 2 = m3 then letting u = −2x, v = y + m then u3 + v 3 = −8x3 + y 3 + 3y 2 m + 3ym2 + y 3 = −2m3 − 6my 2 + y 3 + 3y 2 m + 3ym2 + y 3 = (y − m)3 . Thus either x = 0 (which would imply −3y 2 = m2 , an absurdity) or y = ±m; in this case we have necessarily x = m.

For the equivalence of (1) and (2) we compute the Legendre symbol, using Gauss’ reciprocity law: −3 p = −1 p 3 p = (−1)(p−1)/2 (−1)(p−1)/2 p 3 = p . 3 So (−3/p) = +1 if and only if p ≡ 1 (mod 3), that is, p ≡ 1 (mod 6). For the equivalence of (2) and (3), we write X2 + X + 1 = X + 1 2 2 + 34 . 28 I. Special Cases 2 If there exists α ∈ Fp such that α2 + α + 1 = 0 then −3 = 4 α − 12 and conversely, if −3 = β 2 with β ∈ Fp , we take α = − 12 + β/2 so α2 + α + 1 = 0. 2. If k is a nonzero integer, if p is a prime, and p = c2 + 3d2 ∈ S, pk = a2 + 3b2 ∈ S then p divides ac ± 3bd and ad ∓ bc (with corresponding signs ) and k= ac ± 3bd p 2 +3 ad ∓ bc p ∈ S.

Paris, 61 (1865), 921–924 and 961–965. , Sur la d´ecomposition d’un nombre entier en une somme de deux cubes rationnels, J. Math. , (2), 15 (1870), 217–236. , Mathematical Notes, Proc. Roy. Soc. Edinburgh, 7 (1872), 144. , Sur√certains nombres complexes compris dans la formule a + b −c, J. Math. Pures Appl. (3), 1 (1875), 317–372. , Uber die unbestimmte Gleichung x3 + y 3 = z 3 , Sitzungsber. B¨ohm Ges. , 1878, pp. 112–120. , Sur l’´equation ind´etermin´ee X 3 + Y 3 = AZ 3 , Nouv. Ann. , (2), 17 (1879), 425–426.