PELL’S EQUATION

- Nivedita

Notation

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

Why?

x2 – dy2 = 0 => x/y = d

If d is not a perfect square, we cant find integer solutions to this equation

The next best thing :x2 – dy2 = 1 which gives us good

rational approximations to d

Approach

x2 – dy2=1

Factorizing,(x+yd)(x-yd) =1

So we look at the ring Z [d]

Representation of Z[rt(d)]

Trivial : a+bd (a,b)

Other :a+bd (u,v)

u = a + bdv = a - bd

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+bd (u,v)

uv = a2- db2 (Norm of P) Norm is multiplicative v = Conjugate(P) = a – bd

Note: The same definitions of norm, conjugate go through for

P = a+bd 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 pointin the shadedregion has absolute value ofnorm < B

Implementation of the ideaIf we can infinitely many lattice points inside the region, (note, they’ll all have |norm| < B ),

then we can find infinitely manypoints 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 lemmaconditions 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 increasesSo, infinitely many lattice points P with |norm(P)| < B

So infinitely many points with the same norm (as said before)

Go away glitchPick infinitely many points Pk=ak+bkd with same norm r

Out of this pick infinitely many points suchthat ak=aj mod |r|, bk=bj mod |r| for all k, j

Now evaluate Pk/Pj .(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 Lsuch that FP+a intersects U.

Translations

Put Ua= (FP+a) U(example: red figure)

Va = Ua – a(the green one)

So Va lies in FP !

Area of U = Ua

= Va

Area of U > area of FP

Va > area of FP

But all Va lie in FP! So some two should overlap

Va Vb 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