(IV.D) PELL S EQUATION AND RELATED PROBLEMS

Transcription

1 (IV.D) PELL S EQUATION AND RELATED PROBLEMS Let d Z be non-square, K = Q( d). As usual, we take S := Z[ [ ] d] (for any d) or Z 1+ d (only if d 1). We have roved that (4) S has a least ( fundamental ) unit u > 1, and S = {±u r r Z}, u = a + b d (res. a+b d ) where (a, b) is the ositive-integral solution of x y d = ±1 (res. ±4) with b as small as ossible, and romised some examles. Here they are: Examle 1 (S = Z[ ]). (a, b) = (1, 1) yields the minimal-b solution (in N N) to a b = ±1; so u = 1 + is the fundamental unit. Examle (S = Z[ 0]). Notice that a 0b = ±1 is equivalent to 0b ± 1 is a square. So we make a table b = 1 0b + 1 = b 1 = in which (from left to right) the first square to aear is 81 = a, a = 9. So a minimal-b solution is (a, b) = (9, ), and the fundamental unit is u = Examle 3 (S = Z[ 14]). To solve a 14b = ±1, we again make a table: b = b + 1 = = 15 14b 1 = and conclude that u =

2 MATH (IV.D) Examle 4 (S = Z[ ]). Look at a 17b = ±4, and = u = = b = 1 17b + 4 = b 4 = Of course, you won t be able to do them all by hand: Examle 5 (S = Z[ 46]). u = But obviously the algorithm we have been using could be set u vey easily in PARI. Aarently the integral solutions of Pell s equation (1) x dy = ±1 (d N nonsquare) had been studied with artial success in medieval India, and before that by Diohantus himself. We know that they are in 1-1 corresondence with the units Z[ d] (via x + y d), and that the same goes (via x+y d ) for solutions of () x dy = ±4 (d N nonsquare, d (4) 1) and Z[ 1+ d ]. We can therefore find all solutions to (1)-() by comuting a fundamental unit u and taking ±u r (r Z). If u = x 1 + y 1 d, then setting u r =: x r + y r d, we have ũ r = x r y r d and (3) x r = ur + ũ r, y r = ur ũ r d. Now suose for simlicity that (x 1, y 1 ) satisfies (4) X dy = +1.

3 MATH (IV.D) 3 We would like to geometrically interret multilication in Z[ d] on the hyerbola X y X (1,0) x comrising solutions of (4), so as to view our corresondence between units and solutions as a homomorhism (in fact, isomorhism) of grous. That is, we want a grou law 1 on the oints of X (and for the integer oints to be closed under this binary oeration). Given oints 1 and on X, draw the line L = L 1 : y X 1 x L 1 This is another standard term for binary oeration satisfying the grou axioms.

4 4 MATH (IV.D) (If 1 =, then L is the line tangent to X at this oint.) Next, aralleltranslate this line until it goes through (1, 0) and one other oint 3, y 3 1 (1,0) x L L X and call the result L. Finally, set 1 := 3 ; (1, 0) is clearly the identity element in this (evidently abelian) grou. But wait: we have not checked existence of inverses, or closure, or associativity! So how do we know the integer oints of X, written X(Z), constitute a grou? The following will take care of that: Theorem 6. The ma ϕ : Z[ d] X(Z) sending x + y d (x, y), identifies (multilication) on the lefthand side with on the right-hand side, and is 1-to-1 and onto. Hence (X(Z), ) is a grou, and ϕ is an isomorhism of grous. Proof. It is easy to see that ϕ is a bijection: if x + y d, x + y d Z[ d] have (x, y) = (x, y ), then obviously they re equal. So ϕ is 1-1. Moreover, if (x, y) X(Z), then (x, y) Z and N(x + y d) = x dy = +1 = x + y d Z[ d], and ϕ is onto. Now comes the work: to comute 1 = (x, y) (z, w), observe that L has sloe w y z x, so that L = {(X, Y) X = 1 + w y z x Y}. To find For those of you with some exosure to algebraic geometry, this is the grou law on the singular cubic obtained by adding the line at infinity in P to the rojective closure of the conic X.

5 3 L X: has nonzero solution Y = MATH (IV.D) 5 ( 1 + z x w y Y ) = dy + 1 z x w y d ( z x w y ) = (z x)(w y) d(w y) (z x), but this is not yet in a useful form. We need to use the fact that 1 and lie on X, i.e. that (5) z = dw + 1, x = dy + 1. Alying (5) reeatedly to the exression for Y gives = Y = (z x)(w y) zx (dwy + 1) (zx + dwy + 1)(z x)(w y) d(w y) = = (zx + dwy + 1)(z x)(w y) z x (dwy + 1) (zx + dwy + 1)(z x) d(w y) d(w y)(wx + zy) = = wx + zy, d(w y) where I have left some of the work for you. Similarly, using (5), one shows that hence (w y) + (z x)(wx + zy) = (w y)(xz + dyw) X = 1 + z x w y Y = 1 + z x (wx + zy) = xz + dyw. w y (Note that this already roves directly that X(Z) is closed under!) The verification we are after is now simle: ϕ{(x + y d) (z + w d)} = ϕ{(xz + dyw) + (wx + zy) d} = (xz + dyw, wx + zy) = (x, y) (z, w) = ϕ(x + y d) ϕ(z + w d),

6 6 MATH (IV.D) which (together with ϕ being bijective) identifies with. Hence (Z[ d], ) is a grou = (X(Z), ) is a grou. With this established, we also see that ϕ is a homomorhism, hence an isomorhism. Now to the business of solving equations. We will use the notation ϕ(x + y d) := (x, y) more broadly than in the secific (grouhomomorhism) context above. Examle 7. Suose we wish to find all integer solutions of x y = 1. The fundamental unit of S = Z[ ] is u = 1 +, so S = {±u m m Z} gives all solutions of N( ) = ±1. Since N(u) = 1 1 = 1, {±u m m Z} gives all solutions to N( ) = 1; that is, ϕ(±u m ) =: ±(x m, y m ), m Z yields all solutions to the equation. As in (3), x m = um + ũ m comuting u = 3 + gives x m = (3 + ) m + (3 ) m and y m = um ũ m ; and y m = (3 + ) m (3 ) m. Examle 8. Consider the equation x 5y = 1. We take S = Z[ 5] (even though 5 (4) 1), for which u = + 5, with N(u) = 5 1 = 1. Noting that u = , the comlete list of integral solutions is {ϕ(±u m )} = { ( ± (9+4 5) m +(9 4 5) m, (9+4 5) m (9 4 )} 5) m. 5