PELL ’ S EQUATION - Nivedita

Notation d = positive square root of x Z = ring of integers Z[d] = {a+bd |a,b in Z}

Why? x 2 – dy 2 = 0 => x/y = d If d is not a perfect square, we cant find integer solutions to this equation The next best thing : x 2 – dy 2 = 1 which gives us good rational approximations to d

Approach x 2 – dy 2 =1 Factorizing, (x+yd)(x-yd) =1 So we look at the ring Z [d]

Representation of Z[rt(d)] Trivial : a+bd  (a,b) Other : a+bd  (u,v) u = a + bd v = a - bd

Lattice in R x R  L = { mx + ny | m,n in Z} with x,y two R independent vectors  {x,y} is a basis for L  Fundamental parallelogram = FP(L) = parallelogram formed by x and y  Z[d] in u-v plane is a lattice ( basis (1,1), (d,-d) )

Observations If P = a+bd  (u,v)  uv = a 2 - db 2 (Norm of P)  Norm is multiplicative  v = Conjugate(P) = a – bd Note: The same definitions of norm, conjugate go through for P = a+b d with a, b in Q

Solutions to Pell Latticepoints of Z[d] with norm 1 (or) Lattice points on hyperbola uv = 1

Idea and a glitch If Norm(P) = Norm (Q) then Norm (P/Q) =1 ! But … P/Q need ’ nt be in Z[d]

Any lattice point in the shaded region has absolute value of norm < B

Implementation of the idea If we can infinitely many lattice points inside the region, (note, they ’ ll all have |norm| < B ), then we can find infinitely many points which have the same norm r, |r| < B ( norms are integers and finitely many bet – B and B)

So, identifying lattice points in nice sets of R x R seems to be useful. Here follows a lemma

But what ’ s a nice set in RxR  Convex : S is convex if p in S, q in S => line-segment joining p and q is in S  Centrally symmetric : p in S => -p in S  Bounded : S is bounded if it lies inside a circle of radius R for big enough R

Minkowski ’ s lemma Let L be a lattice in R x R with fundamental parallelogram FP. If S is a bounded, convex, centrally symmetric set such that area(S) > 4* area(FP) then S contains a non-zero lattice point

Infinitely many points with same norm - continued

R(u) = the rectangle in the pic satisfies minkowski lemma conditions for all u >0. So each R(u) has a non zero lattice point No lattice point can be of form (x,0) And R(u) becomes narrower as u increases So, infinitely many lattice points P with |norm(P)| < B So infinitely many points with the same norm (as said before)

Go away glitch Pick infinitely many points P k =a k +b k d with same norm r Out of this pick infinitely many points such that a k =a j mod |r|, b k =b j mod |r| for all k, j Now evaluate P k /P j. (by rationalizing denominator) It belongs to Z[d]

Visual proof of Minkowski S = given set. U = {p | 2p in S }

Area of U Area (U) = ¼ Area(S) > area(FP) No. of red squares in U = no. of blue squares in S

Divide U into parallelograms Purple lines = L(lattice) Blue parallelogram = FP FP+a ={p+a| p in FP} Only finitely many a in L such that FP+a intersects U.

Translations Put U a = (FP+a)  U (example: red figure) V a = U a – a (the green one) So V a lies in FP ! Area of U =  U a =  V a

Area of U > area of FP   V a > area of FP But all V a lie in FP! So some two should overlap  V a  V b is not empty for some a b in L  u* + a = v* + b ( u*, v* in U)  u* - c = v* ( c in L, c = b-a  0)

Voila! u*- c in U  c – u* in U (Note U is also convex, centrally symmetric!) u* in U  Midpoint of c-u*, u* in U  c/2 in U  c in S