INDEFINITE QUADRATIC FORMS AND PELL EQUATIONS INVOLVING QUADRATIC IDEALS

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1 INDEFINITE QUADRATIC FORMS AND PELL EQUATIONS INVOLVING QUADRATIC IDEALS AHMET TEKCAN Communicated by Alexandru Zaharescu Let p 1(mod 4) be a prime number, let γ P + p Q be a quadratic irrational, let I γ [Q, P + p be a quadratic ideal and let F γ (Q, P, Q) be an indefinite quadratic form of discriminant 4p, where P and Q are positive integers depending on p In this work, we first determined the cycle of I γ and then proved that the right and left neighbors of F γ can be obtained from the cycle of I γ Later we determined the continued fraction expansion of γ, and then we showed that the continued fraction expansion of p, the set of proper automorphisms of F γ, the fundamental solution of the Pell equation x py ±1 and the set of all positive integer solutions of the equation x py ±p can be obtained from the continued fraction expansion of γ AMS 010 Subject Classification: 11E08, 11E10, 11E16, 11D09, 11D41, 11D45 Key words: quadratic irrationals, quadratic ideals, quadratic forms, cycles, right and left neighbors, proper automorphisms, Pell equation 1 INTRODUCTION A real binary quadratic form (or just a form) F is a polynomial in two variables x and y of the type F F (x, y) ax + bxy + cy with real coefficients a, b, c We denote F briefly by F (a, b, c) The discriminant of F is defined by the formula b 4ac and is denoted by (F ) F is an integral form if and only if a, b, c Z and is indefinite if and only if (F ) > 0 An indefinite definite form F (a, b, c) of discriminant is said to be reduced if a < b < Gauss defined the group [ action of GL(,Z) which is the multiplicative r s group of matrices g such that r, s, t, u Z with det(g) ±1 t u on the set of forms as (11) gf (x, y) F (rx + ty, sx + uy) MATH REPORTS 19(69), (017), 63 79

2 64 Ahmet Tekcan [ r s for g GL(, Z) If there exists a g GL(, Z) such that gf G, t u then F and G are called equivalent If det(g) 1, then F and G are called properly equivalent and if det(g) 1, then F and G are called improperly equivalent An element g GL(, Z) is called an automorphism of F if gf F If det g 1, then g is called a proper automorphism and if det g 1, then g is called an improper automorphism Let Aut(F ) + denote the set of proper automorphisms and let Aut(F ) denote the set of improper automorphisms of F The right neighbor R(F ) of an integral indefinite form F (a, b, c) of discriminant is the form (A, B, C) determined by four conditions: A c, b + B 0 (mod A), A < B < and B 4AC It is clear that (1) R(F ) where [ δ (a, b, c), (13) δ b + B c The left neighbor L(F ) of F is defined as [ 0 1 (14) L(F ) R(c, b, a) 1 0 So F is properly equivalent to its right and left neighbor Let ρ(f ) denote the normalization of (c, b, a) Let F F 0 (a 0, b 0, c 0 ) and let r i sign(c i ) bi c i for c i or r i sign(c i ) bi + c i for c i < with i 0 Then the reduction of F is ρ i+1 (F ) (c i, b i + c i r i, c i ri b i r i +a i ) Then the proper cycle of F is the sequence (ρ i (G)) for i Z, where G is a reduced form which is properly equivalent to F and the cycle of F is the sequence ((τρ) i (G)) for i Z, where G (A, B, C) is a reduced form with A > 0 which is equivalent to F for τ(f ) ( a, b, c) The cycle of a reduced integral form F is computed as follows: Let F 0 F, s i and let F i+1 ( c i, b i + s i c i, a i b i s i c i s i ) bi + c i Then the cycle of F is F 0 F 1 F F l 1 of length l If l is odd, then the proper cycle of F is F 0 τ(f 1 ) F τ(f 3 ) τ(f l ) F l 1 τ(f 0 ) F 1 τ(f ) F l τ(f l 1 ) of length l and if l is even, then the proper cycle of F is F 0 τ(f 1 ) F τ(f 3 ) F l τ(f l 1 ) of length l (for further details see [ 4)

3 3 Indefinite quadratic forms and Pell equations involving quadratic ideals 65 Mollin considered the arithmetic of ideals in his book [6 Let D 1 be a square free integer and let 4D, where r if D 1(mod 4) or r 1 r otherwise Then is called a fundamental discriminant with fundamental radicand D If we set K Q( D), then K is called a real quadratic number field of discriminant Thus, there is one to one correspondence between quadratic fields and square free rational integers D 1 A complex number is an algebraic integer if it is the root of a monic polynomial with coefficients in Z The set of all algebraic integers in the complex field C is a ring which we denote by A Therefore A K O is the ring of integers of the quadratic field K of discriminant A real number γ is called a quadratic irrational associated with the radicand D, if γ can be written as γ P + D Q, where P, Q, D Z, D > 0, Q 0 and P D(mod Q) We denote the continued fraction expansion of γ by γ [m 0 ; m 1, m,, γ i, where (for i 0 and γ γ 0, P 0 P, Q 0 Q) we recursively define γ i P i+ D Q i, P i + D (15) m i, P i+1 m i Q i P i and Q i+1 D P i+1 Q i Q i An infinite simple continued fraction γ is called periodic if γ [m 0 ; m 1, m,, where m n m n+l for all n k with k, l N In this case we use the notation [m 0 ; m 1, m,, m k 1 ; m k, m k+1,, m l+k 1 An infinite simple continued fraction γ is called purely periodic if γ [m 0 ; m 1,, m l 1 with period length l If γ is a quadratic irrational, then I γ [Q, P + D is a quadratic ideal and its cycle is I γ0 I γ1 I γl 1 of length l In [8, Mollin considered the Jocabi symbols, ambiguous ideals and continued fractions In [9, Mollin and Cheng derived some results on palindromy and ambiguous ideals In [10, we considered the proper cycles of a reduced form F and its right neighbors We proved that the proper cycle of a reduced form F can be given by its consecutive right neighbors, namely, Lemma 11 ([10, Theorem 1) Let F 0 F 1 F l 1 be the cycle of a reduced form F of length l, and let R i (F 0 ) be the consecutive right neighbors of F F 0 for i 0 Then (1) If l is odd, then the proper cycle of F is F 0 R 1 (F 0 ) R l (F 0 ) R l 1 (F 0 ) of length l () If l is even, then the proper cycle of F is F 0 R 1 (F 0 ) R l (F 0 ) R l 1 (F 0 ) of length l Later in [11, the present author and collaborators considered the same problem for the left neighbors and proved that

4 66 Ahmet Tekcan 4 Lemma 1 (11, Theorem 4) Let F 0 F 1 F l 1 be the cycle of a reduced form F of length l, and let L i (F 0 ) be the consecutive left neighbors of F F 0 for i 0 Then (1) If l is odd, then the proper cycle of F is F 0 L l 1 (F 0 ) L (F 0 ) L 1 (F 0 ) of length l () If l is even, then the proper cycle of F is F 0 L l 1 (F 0 ) L (F 0 ) L 1 (F 0 ) of length l Now let p be a prime number such that p 1(mod 4) Let γ P + p Q be a quadratic irrational, let I γ [Q, P + p be a quadratic ideal and let F γ (Q, P, Q) be an indefinite quadratic form of discriminant 4p for some positive integers P and Q depending on p In the present paper, we demonstrate (1) how to use the cycle of I γ to determine the left and right neighbors of F γ () how to use the continued fraction expansion of γ to determine the continued fraction expansion of p, the set of proper automorphisms of F γ, the fundamental solution of the Pell equation x py ±1, the set of all positive integer solutions of the equation x py ±p Recall that the equation (16) x dy ±n is called a norm form equation since N(x + y d) x dy is called the norm of x + y d, where d is any positive non square integer and n is any fixed integer When n 1, (16) is known as the Pell equation after John Pell ( ), who actually had little to do with its solution The Pell equation x dy ±1 has infinitely many integer solutions (In particular, x dy 1 has infinitely many solutions when the length of the continued fraction expansion of d is odd) The first non trivial positive integer solutions (x 1, y 1 ) is called the fundamental solution from which all integer solutions can be derived Namely, if (x 1, y 1 ) is the fundamental solution of x dy 1, then the other solutions are (x n, y n ), where x n + y n d (x1 + y 1 d) n for n 1 and if (x 1, y 1 ) is the fundamental solution of x dy 1, then the other solutions are (x n+1, y n+1 ), where x n+1 + y n+1 d (x1 + y 1 d) n+1 for n 0 (see [1, 5, 7) Let α [q 0 ; q 1,, q l for l N be a finite continued fraction expansion Define two sequences A 0, A 1 1, A k q k A k 1 + A k and B 1, B 1 0, B k q k B k 1 + B k for a nonnegative integer k Then C k

5 5 Indefinite quadratic forms and Pell equations involving quadratic ideals 67 A k B k is the k th convergent of α for any nonnegative integer k l Then the fundamental solution is given below Lemma 13 ([7, Corollary 57) If D > 0 is not a perfect square and D has continued fraction expansion of period length l, then the fundamental solution of x Dy 1 is given by (x 1, y 1 ) (A l 1, B l 1 ) if l is even or (A l 1, B l 1 ) if l is odd If l is odd, then the fundamental solution of x Dy 1 is given by (x 1, y 1 ) (A l 1, B l 1 ) MAIN RESULTS Let p be a prime number such that p 1(mod 4) Then it is known that p can be written of the form p a + b, where a is odd, b is even and a + b 1(mod 4) Now we set (1) P b, Q a and D p Then () γ P + p Q is a quadratic irrational since P p (mod Q), and so (3) I γ [Q, P + p is a quadratic ideal and (4) F γ (x, y) Q(x + γy)(x + γy) Qx + P xy Qy is an indefinite quadratic form of discriminant 4p Here we note that for some values of p, we have Q 1; but for some values of p, we have Q 1 (For instance, for p 13, we have P, Q 3, but for p 17, we have P 4, Q 1) Therefore, we will consider all results in two cases: Q 1 or Q 1 Theorem 1 Let I γ be the ideal in (3) (1) If Q 1, then the cycle of I γ is I γ0 [Q 0, P 0 + p I γ1 [Q 1, P 1 + p I γ [Q, P + p I γ l 3 I γ l+1 [Q l 3, P l 3 + p I γ l 1 [Q l 3, P l 1 + p I γ l+3 [Q l 1, P l 1 + p [Q l 5, P l 3 I γl [Q 1, P + p I γl 1 [Q 0, P 1 + p + p of length l () If Q 1, then the cycle of I γ is I γ0 [1, P 0 + p of length 1

6 68 Ahmet Tekcan 6 Proof (1) Let Q 1 Then from (15), we deduce the values in Table 1 TABLE 1 Cycle of I γ i 0 1 l 3 l 1 l+1 l+3 l l 1 P P 1 P i P 0 P 1 P l 3 Q i Q 0 Q 1 Q l 3 m i m 0 m 1 m l 3 P l 1 Q l 1 m l 1 P l 1 Q l 3 m l 3 P l 3 Q l 5 m l 5 Q 1 Q 0 m 1 m 0 So the the cycle of I γ is I γ0 [Q 0, P 0 + p I γ1 [Q 1, P 1 + p I γ [Q, P + p I γ l 3 I γ l+1 [Q l 3, P l 3 + p I γ l 1 [Q l 3, P l 1 + p I γ l+3 [Q l 1, P l 1 + p [Q l 5, P l 3 I γl [Q 1, P + p I γl 1 [Q 0, P 1 + p + p of length l () Let Q 1 Then from (15), we get m 0 P and hence P 1 P P 0 and Q 1 1 Q 0 So the cycle of I γ is I γ0 [1, P 0 + p of length 1 Note that in Lemma 11, we proved that the proper cycle of a reduced form F can be given by its consecutive right neighbors and in Lemma 1, we showed that the proper cycle of a reduced form F can be given by its consecutive left neighbors Now by virtue of Theorem 1, we show that the right and left neighbors of F γ can be obtained from the cycle of I γ as follows Theorem Let I γ I γ0 I γ1 I γl 1 be the cycle of I γ of length l (1) If Q 1, then the right neighbors of F γ in (4) are where R 1 (F γ ), R (F γ ),, R l 1 (F γ ), R l (F γ ), R l+1 (F γ ),, R l 1 (F γ ), R i (F γ ) (( 1) i+ Q i 1, P i, ( 1) i+1 Q i ) for 1 i l 1 R l (F γ ) ( Q 0, P 0, Q 0 ) R l+i (F γ ) (( 1) i+1 Q i 1, P i, ( 1) i+ Q i ) for 1 i l 1 and the left neighbors of F γ are L 1 (F γ ), L (F γ ),, L l 1 (F γ ), L l (F γ ), L l+1 (F γ ),, L l 1 (F γ ),

7 7 Indefinite quadratic forms and Pell equations involving quadratic ideals 69 where L i (F γ ) (( 1) i+ Q i, P i, ( 1) i+1 Q i 1 ) for 1 i l 1 L l (F γ ) ( Q 0, P 0, Q 0 ) L l+i (F γ ) (( 1) i+1 Q i, P i, ( 1) i+ Q i 1 ) for 1 i l 1 () If Q 1, then F γ has one right and one left neighbor and they are same, that is R 1 (F γ ) L 1 (F γ ) ( 1, P, 1) Proof (1) Let Q 1 and let F γ F γ0 (Q 0, P 0, Q 0 ) Then for the first right neighbor R 1 (F γ ) (A, B, C), we have A Q 0 Also B + P 0 0(mod Q 0 ) is satisfied for B P 1 in the range 4(P0 + Q 0 ) Q 0 < B < 4(P 0 + Q 0 ) and hence C Q 1, that is, R 1 (F γ ) ( Q 0, P 1, Q 1 ) Similarly, we get R (F γ ) (Q 1, P, Q ), R 3 (F γ ) ( Q, P 3, Q 3 ),, R l 1 (F γ ) (Q l, P l 1, Q l 1 ), R l (F γ ) ( Q 0, P 0, Q 0 ), R l+1 (F γ ) (Q 0, P 1, Q 1 ),, R l 1 (F γ ) ( Q l, P l 1, Q l 1 ), R l (F γ ) (Q 0, P 0, Q 0 ) F γ Applying (14), the first left neighbor of F γ is [ L (F γ ) R( Q 1 0 0, P 0, Q 0 ) ( Q 1, P 1, Q 0 ) Similarly, we find that L (F γ ) (Q, P, Q 1 ),, L l (F γ ) ( Q l, P l, Q l 3 ), L l 1 (F γ ) (Q l 1, P l 1, Q l ), L l (F γ ) ( Q 0, P 0, Q 0 ), L l+1 (F γ ) (Q 1, P 1, Q 0 ),, L l 1 (F γ ) ( Q l 1, P l 1, Q l ), L l (F γ ) (Q 0, P 0, Q 0 ) F γ as we wanted () Let Q 1 Then R 1 (F γ ) ( 1, P, 1), R (F γ ) (1, P, 1) F γ, L 1 (F γ ) ( 1, P, 1), L (F γ ) (1, P, 1) F γ So F γ has one right and one left neighbor and they are same, namely, R 1 (F γ ) L 1 (F γ ) ( 1, P, 1) Example 3 Let p 13 Then the cycle of I γ [3, + 13 is I γ0 [3, + 13 I γ1 [4, I γ [1, I γ3 [4, I γ4 [3, of length 5 So the right neighbors of F γ (3, 4, 3) are R 1 (F γ ) ( 3,, 4), R (F γ ) (4, 6, 1), R 3 (F γ ) ( 1, 6, 4), R 4 (F γ ) (4,, 3), R 5 (F γ ) ( 3, 4, 3), R 6 (F γ ) (3,, 4), R 7 (F γ ) ( 4, 6, 1), R 8 (F γ ) (1, 6, 4), R 9 (F γ ) ( 4,, 3)

8 70 Ahmet Tekcan 8 and the left neighbors of F γ are L 1 (F γ ) ( 4,, 3), L (F γ ) (1, 6, 4), L 3 (F γ ) ( 4, 6, 1), L 4 (F γ ) (3,, 4), L 5 (F γ ) ( 3, 4, 3), L 6 (F γ ) (4,, 3), L 7 (F γ ) ( 1, 6, 4), L 8 (F γ ) (4, 6, 1), L 9 (F γ ) ( 3,, 4) Here, we note that R 5 (F γ ) L 5 (F γ ) ( 3, 4, 3) From Theorem, we can give the following result Corollary 4 If Q 1, then we have (1) R i (F γ ) τ(r i l (F γ )) and L i (F γ ) τ(l i l (F γ )) for l i l 1 () R i (F γ ) L l i (F γ ) for 1 i l 1 and so the l th right and left neighbors of F γ are the same, that is, R l (F γ ) L l (F γ ) ( Q 0, P 0, Q 0 ) For the second part of this work, we can give the following results Theorem 5 Let γ be the quadratic irrational in () (1) If Q 1, then the continued fraction expansion of γ is [m 0, m 1, m,, m l 3, m l 1, m l 3,, m, m 1, m 0 of length l and the continued fraction expansion of p is [ m l 1 ; m l 3, m l 5,, m 1, m 0, m 0, m 1,, m l 3, m l 1 of length l () If Q 1, then the continued fraction expansion of γ is [P of length 1 and the continued fraction expansion of p is [P ; P of length 1 Proof (1) Let Q 1 Then from Table 1, we easily seen that the continued fraction expansion of γ is [m 0, m 1,, m l 3, m l 1, m l 3,, m 1, m 0 Notice that m l 1 p p m l 1 So we get ( p m l 1 + ) m l m l 1 p+ ( m ) l 1 p m l 1 + m l ( m l 1 m ) l 1 p+ m l 3 p ( m ) l 1 p

9 9 Indefinite quadratic forms and Pell equations involving quadratic ideals 71 m l 1 + m l ( m l 1 m ) l 1 p+ m l 3 p ( m ) l 1 p Hence [ m l 1 p m l 1 ; m l 3 +, m l 5 m l m l m l 1 + ( p m l 1,, m 1, m 0, m 0, m 1,, m l 3, m l 1 ) of length l () Let Q 1 Then m 0 P, P 1 P P 0 and Q 1 1 Q 0, we get γ [P Also since p P + 1, we get 1 1 p P + ( p P ) P + p+p P + P + ( p P ), p P that is, p [ P ; P of length 1 as we claimed Remark 6 We note in (1) that, we take P b and Q a If we take P a and Q b, then we cannot deduce the continued fraction expansion of p from the continued fraction expansion of γ Indeed, for p 73, we have { [5, 1, 1, 16, 1, 1, 5 if P 8, Q 3 γ [1,, 3, 1, 7, 1, 3,, 1 if P 3, Q 8 Note that 73 [8; 1, 1, 5, 5, 1, 1, 16 Similarly for p 137, we have { [1,,, 1,, 1,,, 1 if P 4, Q 11 γ [5, 1,, 11,, 1, 5 if P 11, Q 4 But 137 [11; 1,,, 1, 1,,, 1, Q a So that is why we take P b and We can deduce the set of proper automorphisms of F γ in (4) by using the continued fraction expansion of γ For the matrix defined in (1), we set [ 1 [ 0 1 δ 1 T (δ) 1 δ 1 0 and define g F,n T (δ 0 )T (δ 1 ) T (δ n 1 ), where δ is defined in (13) Then we can give the following theorem

10 7 Ahmet Tekcan 10 Theorem 7 Let the continued fraction expansion of γ be as in Theorem 5 (1) If Q 1, then the set of proper automorphisms of F γ is where g Fγ,l l 1 Aut + (F γ ) {±(g Fγ,l) t : t Z}, T (δ i ) and δ i { ( 1) i+1 m i for 0 i l 1 ( 1) i l m i l for l i l 1 () If Q 1, then the set of proper automorphisms of F γ is Aut + (F γ ) {±(g Fγ,) t : t Z}, where [ 4P g Fγ, 1 P P 1 Proof (1) Let Q 1 We proved in Theorem that R i (F γ ) (( 1) i+ Q i 1, P i, ( 1) i+1 Q i ), R l+i (F γ ) (( 1) i+1 Q i 1, P i, ( 1) i+ Q i ) for 1 i l 1 and R l (F γ ) ( Q 0, P 0, Q 0 ), δ i ( 1) i+1 s i for 0 i l 1 and δ i ( 1) i l s i l for l i l 1 For the form F i (Q i, P i, Q i ) of discriminant 4p, we have s i b i + c i Pi + 4p Q i Pi + p m i So δ i ( 1) i+1 m i for 0 i l 1 and δ i ( 1) i l m i l for l i l 1 Thus g F,l T (δ 0 )T (δ 1 ) T (δ l 1 ) by [4, Theorem 94 since R l (F γ ) F γ Therefore, the set of proper automorphisms of F γ is Aut + (F γ ) {±(g Fγ,l) t : t Z} () Let Q 1 Since R (F γ ) F γ for F γ (1, P, 1), we get [ 4P g F, T (δ 0 )T (δ 1 ) 1 P P 1 Q i and hence Aut + (F γ ) {±(g Fγ,) t : t Z} Example 8 1) Let p 13 Then γ [1, 1, 6, 1, 1 So 9 [ g Fγ,10 T (δ i ) Hence Aut + (F γ ) {±(g Fγ,10) t : t Z} for F γ (3, 4, 3)

11 11 Indefinite quadratic forms and Pell equations involving quadratic ideals 73 ) Let p 73 Then γ [5, 1, 1, 16, 1, 1, 5 So g Fγ,14 13 T (δ i ) [ Hence Aut + (F γ ) {±(g Fγ,14) t : t Z} for F γ (3, 16, 3) 3) Let p 17 Then γ [8 So [ 65 8 g F, T (δ 0 )T (δ 1 ) 8 1 Hence Aut + (F γ ) {±(g Fγ,) t : t Z} for F γ (1, 8, 1) From Theorems 1 and 5, we can give the following result Corollary 9 If Q 1, then in the cycle of I γ, we have Q i Q l 1 i for 0 i l 1 and P i P l i for 1 i l 1, and in the continued faction expansion of γ, we have m i m l 1 i for 0 i l 1 Now we can consider the Pell equations solution of the Pell equation Recall that the fundamental x py ±1 is very important to find all other integer solutions In the following theorem, we prove that the fundamental solution of the Pell equation can be obtained from the continued fraction expansion of γ Theorem 10 Let the continued fraction expansion of γ be as in Theorem 5 (1) If Q 1, then the fundamental solution of x py 1 is (A l 1, B l 1 ), where l 1 A l 1 + B l 1 p and the fundamental solution of x py 1 is (A l 1, B l 1 ), where γ i l 1 A l 1 + B l 1 p γ i () If Q 1, then the fundamental solution of x py 1 is (x 1, y 1 ) (P +1, P ) and the fundamental solution of x py 1 is (x 1, y 1 ) (P, 1)

12 74 Ahmet Tekcan 1 Proof First we note that N(γ i ) P i p Q i Therefore l 1 N( Q i Q i γ i ) N(γ 0 )N(γ 1 ) N(γ l 1 ) ( 1) l 1 1 for γ i P i+ p Q i (1) Let Q 1 Then from Theorem 5, the continued fraction expansion of p is [ m l 1,, m 1, m 0, m 0, m 1,, m l 3, m l 1 ; m l 3 Since l is odd, the fundamental solution of x py 1 is (x 1, y 1 ) (A l 1, B l 1 ) by Lemma 13 On the other hand, it can be easily seen that l 1 γ i l 1 A l 1 + B l 1 p Similarly, it can be shown that γ i A l 1 + B l 1 p () Let Q 1 Since p [ P ; P, we get A 0 P, A 1 P + 1, B 0 1 and B 1 P So the result is obvious Example 11 1) Let p 53 Since 53 [7; 3, 1, 1, 3, 14, we get A 4 18, A , B 4 5 and B So the fundamental solution of x 53y 1 is (x 1, y 1 ) (6649, 9100) and the fundamental solution of x 53y 1 is (x 1, y 1 ) (18, 5) Note that 9 γ i and 4 γ i for γ [1, 3, 14, 3, 1 ) Let p 113 Since 113 [10; 1, 1, 1,,, 1, 1, 1, 0, we get A 8 776, A , B 8 73, B So the fundamental solution of x 113y 1 is (x 1, y 1 ) (104353, 11396) and the fundamental solution of x 113y 1 is (x 1, y 1 ) (776, 73) Note that 17 γ i and 8 γ i for γ [, 1, 1, 1, 0, 1, 1, 1, 3) Let p 101 Then 101 [10; 0 and hence A 0 10, A 1 01, B 0 1, B 1 0 Therefore the fundamental solution of x 101y 1 is (x 1, y 1 ) (01, 0) and the fundamental solution of x 101y 1 is (x 1, y 1 ) (10, 1) Also 1 γ i and 0 γ i for γ [0

13 13 Indefinite quadratic forms and Pell equations involving quadratic ideals 75 For an integral quadratic form F, we set Aut (F ) {g GL(, Z) : gf F with det(g) 1} From above theorem, we can give the following result Theorem 1 Let F γ be the form defined in (4) and let A l 1, B l 1 be as in Theorem 10 (1) If Q 1, then Aut (F γ ) {±(gγ) t+1 : t Z}, where [ gγ Al 1 P B l 1 QB l 1 QB l 1 A l 1 + P B l 1 () If Q 1, then Aut (F γ ) {±(gγ 1 ) t+1 : t Z}, where [ gγ P Proof (1) Let Q 1 First we note that det(g γ) (A l 1 P B l 1 )(A l 1 + P B l 1 ) (QB l 1 ) A l 1 (P + Q )B l 1 A l 1 pb l 1 1 since (A l 1, B l 1 ) is the fundamental solution of the Pell equation x py 1 From (11), we get g γf γ F γ ((A l 1 P B l 1 )x + QB l 1 y, QB l 1 x + (A l 1 + P B l 1 )y) Q((A l 1 P B l 1 )x + QB l 1 y) + P ((A l 1 P B l 1 )x + QB l 1 y) (QB l 1 x + (A l 1 + P B l 1 )y) Q(QB l 1 x + (A l 1 + P B l 1 )y) x { Q(A l 1 P B l 1 ) + P QB l 1 (A l 1 P B l 1 ) Q 3 Bl 1 3 } { Q + xy } B l 1 (A l 1 P B l 1 ) + P (A l 1 P B l 1 ) (A l 1 + P B l 1 ) + P Q Bl 1 Q B l 1 (A l 1 + P B l 1 ) + y { Q 3 Bl 1 + P QB l 1(A l 1 + P B l 1 ) Q(A l 1 + P B l 1 ) } (A l 1 pb l 1 )(Qx + P xy Qy ) Qx P xy + Qy F γ (x, y) So gγ Aut (F γ ) It can be proved by induction on t that ±(gγ) t+1 Aut (F γ ) for t Z So Aut (F γ ) {±(gγ) t+1 : t Z} Statement () can be proved similarly

14 76 Ahmet Tekcan 14 Remark 13 (1) In the above theorem, we note that odd powers of gγ are the elements of Aut (F γ ), that is, Aut (F γ ) {±(gγ) t+1 : t Z} In fact, even powers of gγ are the proper automorphisms of F γ () There is a connection between g Fγ,l and gγ and also g Fγ, and gγ 1 obtained in Theorems 7 and 1 which is given below Theorem 14 For the matrices g Fγ,l, g γ and g Fγ,, g 1 γ, we have (g γ) g Fγ,l and (g 1 γ ) g Fγ, Proof For gγ, we easily have [ (gγ) A l 1 pbl 1 P A l 1B l 1 QA l 1 B l 1 QA l 1 B l 1 A l 1 pb l 1 + P A l 1B l 1 On the other hand, it can be proved by induction on l that [ A g Fγ,l l 1 pbl 1 P A l 1B l 1 QA l 1 B l 1 QA l 1 B l 1 A l 1 pb l 1 + P A l 1B l 1 So (gγ) g Fγ,l Similarly for gγ 1, we have [ (gγ 1 ) 4P 1 P g P 1 Fγ, as we wanted Example 15 Let p 13 Then [ g Fγ,10 and gγ Here (gγ) g Fγ,10 Let p 73 Then [ g Fγ,14 and gγ Again (g γ) g Fγ,14 Finally, we can consider the equation x py ±p [ [ Before considering all integer solutions we need some notations: Let be a non square discriminant Then the order O is defined for non square discriminants to be the ring O {x + yρ : x, y Z}, where ρ 4 if 0(mod 4) or ρ 1+ if 1(mod 4) So O is a subring of Q( ) {x + y : x, y Q} The unit group O is defined for non square discriminants to be the group of units of the ring O

15 15 Indefinite quadratic forms and Pell equations involving quadratic ideals 77 The module M F of an integral form F is M F {xa + y b+ : x, y Z} Q( ) So we get (u + vρ )(xa + y b+ ) x a + y b+, where (5) [x y Therefore, there is a bijection [ u b [x y v av [ cv u + b v u + 1 b [x y v av cv u + 1+b v if 0(mod 4) if 1(mod 4) Ψ : Ω {(x, y) : F (x, y) m} {γ M F : N(γ) am} for solving the equation F (x, y) m The action of O,1 {α O : N(α) 1} on the set Ω is the most interesting when is a positive non square since O,1 is infinite So the orbit of each solution will then be infinite and hence the set Ω is either empty or infinite Since O,1 can be explicitly determined, Ω is satisfactorily described by the representation of such a list, called a set of representatives of the orbits Let ε be the smallest unit of O that is greater than 1 and let τ ε if N(ε ) 1; or ε if N(ε ) 1 Then every O,1 orbit of integral solutions of F (x, y) m contains a solution (x, y) Z such that 0 y U, where U amτ 1 (1 1 τ ) if am > 0 or U amτ 1 (1 + 1 τ ) if am < 0 So for finding a set of representatives of the O,1 orbits of F (x, y) m, we must determine for which values of y, y +4am is a perfect square in the range 0 y U since y + 4am (ax + by) We note that we can determine A l 1 and B l 1 in Theorem 10 from the continued fraction expansion of γ Thus we can give the following theorem Theorem 16 For the Pell equation x py ±p, we have (1) If Q 1, then the set of all positive integer solutions of x py p is Ω {(x n, y n )}, where [x n y n [ p(u + 1) UM n for n 1, and the set of all positive integer solutions of x py p is Ω {(x n, y n )}, where [x n y n [0 1M n [ Al 1 B for n 1 with M l 1 pb l 1 A l 1 () If Q 1, then the set of all positive integer solutions of x py p is Ω {(x n, y n )}, where [x n y n [p P M n for n 1, and the set of all positive integer solutions of x py p

16 78 Ahmet Tekcan 16 is Ω {(x n, y n )}, where for n 1 with M [x n y n [0 1M n [ P + 1 P P 3 + P P + 1 Proof (1) Let Q 1 Then we proved in Theorem 10 that the fundamental solution x py ±1 can be obtained from the continued fraction expansion of γ, that is, we can determine the integers A l 1 and B l 1 depending on γ For the equation x py p, we have τ A l 1 + B l 1 p and y + 4am 4p(y + 1) is a square only for y U in the range 0 y U, where U 1 A l 1 1+B l 1 p So x ± p(u + 1) and hence Al 1 +B l 1 p {[± p(u + 1) U} is a set of representatives and thus [ p(u + 1) UM n generates the solutions (x n, y n ) for n 1, where M is defined as above So the set of all positive integer solutions of x py p is Ω {(x n, y n )}, where [x n y n [ p(u + 1) UM n for n 1 For the equation x py p, we see that {[0 1} is a set of representatives and [0 1M n generates the solutions (x n, y n ) for n 1 Thus the set of all positive integer solutions of x py p is Ω {(x n, y n )}, where [x n y n [0 1M n for n 1 () Let Q 1 Then for the equation x py p, we have τ P P p and in the range 0 y P, y + 4am is a square only for y P Hence we get x ±p So {[±p P } is a set of representatives and [p P M n generates the solutions (x n, y n ) for n 1, where M is defined as above For the equation x py p, we see that {[0 1} is a set of representatives and [0 1M n generates the solutions (x n, y n ) for n 1 This completes the proof Example 17 1) Let p 73 Then A , B and hence U 1068 In the range 0 y 1068, 9(y + 1) is square only for y 1068 and hence x ±915 So {[± } is a set of representatives Therefore, the set of all positive integer solutions of x 73y 73 is Ω {(x n, y n )}, where [ n [x n y n [ for n 1 For the equation x 73y 73, 9(y 1) is square only for y 1 in the range 0 y 1068 and hence x 0 So {[0 1} is a set of representatives So the set of all positive integer solutions of x 73y 73 is Ω {(x n, y n )}, where [ n [x n y n [ for n 1

17 17 Indefinite quadratic forms and Pell equations involving quadratic ideals 79 ) Let p 37 Then in the range 0 y 6, 148(y +1) is square only for y 6 and hence x ±37 So {[±37 6} is a set of representatives Therefore, the set of all positive integer solutions of x 37y 37 is Ω {(x n, y n )}, where [ n 73 1 [x n y n [ for n 1 For the equation x 37y 37, 148(y 1) is square only for y 1 in the range 0 y 6 and hence x 0 So {[0 1} is a set of representatives Thus Ω {(x n, y n )}, where [ n 73 1 [x n y n [ for n 1 REFERENCES [1 E Barbeau, Pell s Equation Springer-Verlag, 003 [ J Buchmann and U Vollmer, Binary Quadratic Forms: An Algorithmic Approach Springer-Verlag, Berlin, Heidelberg, 007 [3 DA Buell, Binary Quadratic Forms, Classical Theory and Modern Computations Springer-Verlag, New York, 1989 [4 DE Flath, Introduction to Number Theory Wiley, 1989 [5 M Jacobson and H Williams, Solving the Pell Equation, CMS Books in Mathematics Springer, 010 [6 RA Mollin, Quadratics CRS Press, Boca Raton, New York, London, Tokyo, 1996 [7 RA Mollin, Fundamental Number Theory with Applications Second Edition Discrete Math Appl, Chapman & Hall/ CRC, Boca Raton, London, New York, 008 [8 RA Mollin, Jocabi symbols, ambiguous ideals, and continued fractions Acta Arith 85 (1998), [9 RA Mollin and K Cheng, Palindromy and ambiguous ideals revisited J Number Theory 74 (1999), [10 A Tekcan, Proper Cycles of Indefinite Quadratic Forms and their Right Neighbors Appl Math 5(5) (007), [11 A Tekcan, A Özkoç and İN Cangül, Indefinite Quadratic Forms and their Neighbours Numerical Anal Appl Math ICNAAM 011 AIP Conf Proc 181 (010), Received 5 November 015 Uludag University, Faculty of Science, Department of Mathematics, Bursa Turkiye tekcan@uludagedutr