PELL ’ S EQUATION

1. PELL’S EQUATION - Nivedita

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

3. 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

4. Approach x2– dy2=1 Factorizing, (x+yd)(x-yd) =1 So we look at the ring Z [d]

5. 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

6. 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) )

7. Observations If P = a+bd (u,v) • uv = a2- db2 (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

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

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

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

11. 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)

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

13. 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

14. 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

15. Infinitely many points with same norm - continued

16. 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)

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

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

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

20. 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.

21. 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

22. 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)

23. 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