Download - PELL ’ S EQUATION
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