On certain biquadratic equations

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1 Arch. Math. 101 (013), c 013 The Author(s). This article is published with open access at Springerlink.com X/13/ published online September 11, 013 DOI /s y Archiv der Mathematik On certain biquadratic equations A. Schinzel and M. Ska lba Abstract. It is proved that if certain biquadratic equations with a prime parameter and four unknowns have a non-trivial solution in the integers, then they have infinitely many such solutions. Keywords. Pell equation, Biquadratic Diophantine equation, Cyclotomic polynomial. The diophantine equation f(x, y, z, t) =c, where f is a quartic form and c 0, has been studied only in the case where f splits over the complex field. The aim of this paper is to prove the following theorems. Theorem 1. If p 3 (mod 4) is a prime, α 1 (mod ), D = p α, and the equation (x + y ) D(z + t ) =1has at least one integer solution with z + t > 0, then it has infinitely many integer solutions. Theorem. If p 3 (mod 8) is a prime, α 1 (mod ), D = p α, and the equation (x + y ) D(z + t ) = has at least one integer solution, then it has infinitely many. Theorem 3. If p 7 (mod 8) is a prime, α 1 (mod ), D = p α, and the equation (x + y ) D(z + t ) =has at least one integer solution, then it has infinitely many. It is natural to ask about the number N i (x) of values D<xcovered by the above Theorem i (i =1,, 3). We cannot prove that lim x N i (x) =, but since Theorem and 3 apply to all odd primes of the form (x + y ) ± and Theorem 1 applies to all odd primes of the form (x + y ) + at least heuristically N i (x) x1/ (i =1,, 3). The method does not seem (log x) 3/ applicable to other equations (x + y ) D(z + t ) = c, where c 0. Lemma 1. Let n be an odd, square-free integer > 3 and Φ n (x) the n-th cyclotomic polynomial. Then Φ n (z) can be written in the form 4Φ n (z) =A n (z) ( 1) (n 1)/ nz B n (z),

2 14 A. Schinzel and M. Ska lba Arch. Math. where A n (z) and B n (z) have integer coefficients and are of degree φ(n)/ and φ(n)/, respectively. A n (z) is symmetric if its degree is even, otherwise it is anti-symmetric. B n (z) is symmetric for n prime. Proof. See [[1], p.330 and 445]. Lemma. For p 3 (mod 4) a prime, p>3, A p (z) is divisible by z 1 and B p (z) is divisible by z +1. Proof. We have [[1], p.330] A p (z)+ pb p (z) = A p (z) pb p (z) = ( r p )=1 (z ζ r p), ( s p )= 1 (z ζ s p). If p 3 (mod 4) is a prime, then ( r r p ) = 1 is equivalent to ( p )= 1. Since for p>3, ( r p )=1 r 0 (mod p), we obtain (1 ζp)= s (1 ζp r )= (1 ζp); r ( s p )= 1 ( r p )=1 ( r p )=1 ( 1 ζp)= s ( 1 ζp r )= (1 ζp), r ( s p )= 1 ( r p )=1 ( r p )=1 thus A p (1) = B p ( 1) = 0 and the lemma follows. Lemma 3. If a form F Q[x, y] satisfies F (x, y) =F (y, x) and d := deg F 0 (mod ), then F = G((x + y),xy), whereg Q[z,t]. Proof. By the theorem on symmetric functions F = H(x + y, xy), where H Q[z,t]. Let F (x, y) = a αβ (x+y) α (xy) β. Since for a αβ 0wehaveα+β = d 0 (mod ), we have a αβ =0forα odd. Lemma 4. Let D = p α,wherep 3 (mod 4) is a prime, α 1 (mod ), and positive integers u, v satisfy u Dv =1.Put and ξ = u + v D, ξ 1 = u v D; L n = ξn ξ n, (1) ξ ξ 1 E n = ξn + ξ n,f n = vl n. () E n and F n are positive integers satisfying En DFn =1. (3) Moreover, for all positive integers n, E p n/e p n 1 is a sum of two squares. (4)

3 Vol. 101 (013) On certain biquadratic equations 15 Proof. We have E n DF n = ξn ++ξ n which proves (3). Moreover, 4 Dv ξn +ξ n 4v D =1, E p n/e p n 1 =Φ p n(ξ,ξ 1 )=Φ p (ξ pn 1,ξ pn 1 )=Φ p ( ξ pn 1,ξ pn 1 ), (5) where Φ n (x, y) is a homogeneous form of Φ n (z). By Lemma 1 for p>3 4Φ p ( ξ pn 1,ξ pn 1 )=A p ( ξ pn 1,ξ pn 1 ) + pb p ( ξ pn 1,ξ pn 1 ). (6) Since A p is anti-symmetric, A p ( ξ pn 1,ξ pn 1 ) is symmetric in ξ,ξ 1, hence is an integer. Now B p (x, y) is, by Lemma, divisible by x+y, and the quotient is of degree p (mod ). By Lemma 3 the quotient is a polynomial in (x + y) and xy. It follows that B p ( ξ pn 1,ξ pn 1 ) = pw with w Q. (4) follows now from (5) and (6). It remains to consider p =3.Letξ 3n 1 = u n + v n 3α.Wehaveby(5) E 3 n/e 3 n 1 =Φ 3 ( ξ 3n 1,ξ 3n 1 ) =(u n +v n 3α ) (u n + v n 3α )(u n v n 3α )+(u n v n 3α ) = u n +(3 α+1 vn ). Lemma 5. In the notation of Lemma 4, F p n/pf p n 1 is a sum of two squares. (7) Proof We have by (1) and () F p n/f p n 1 =Φ p n(ξ,ξ 1 )=Φ p (ξ pn 1,ξ pn 1 ) (8) and by Lemma 1, 4F p n/f p n 1 = A p (ξ pn 1,ξ pn 1 ) + pb p (ξ pn 1,ξ pn 1 ). (9) Since A p (x, y) is anti-symmetric, A p (x, y)/(x y) is symmetric and A p (ξ pn 1,ξ pn 1 ) = pw,w Q. (10) Also B p (x, y) is symmetric, B p (ξ pn 1,ξ pn 1 ) is an integer, and (7) follows from (9) and (10). It remains to consider p = 3. Let again ξ 3n 1 = u n +v n 3α. We have by (8) F 3 n/3f 3 n 1 = 1 ( (u n + v n 3α ) +(u n + v n 3α )(u n v n 3α ) 3 +(u n v n 3α ) ) = u n +(3 α 1 vn ). Proof of Theorem 1. It follows by induction on n from Lemmas 4 and 5 that for n even E p n and F p n are sums of two squares. The formula (3) gives the theorem.

4 16 A. Schinzel and M. Ska lba Arch. Math. Lemma 6. Let ε = ±1, D = p α,wherep 5+ε (mod 8) is a prime, α 1 (mod ), and positive integers u, v satisfy u Dv =ε. Put and εu + Dεv εu Dεv η =,η 1 =,P n = η n η n, (11) η (n,) η (n,) G n = ηn + η n ε,h n = vp n. (1) G n and H n are rational integers satisfying Moreover, for all positive integers n, Proof. We have for odd n G n DH n =ε. (13) G p n/εg p n 1 is a sum of two squares. (14) G n DH n = ηn ++η n ε Dv ηn +η n εdv = 4 ε =ε, which proves (13). P n as a Lehmer number is a rational integer, so is H n and, by (13), G n is an algebraic integer. However, it is also rational being invariant to all automorphisms of Q( ε, D). Moreover, G p n/g p n 1 =Φ p n(η, η 1 )=Φ p (η pn 1,η pn 1 )=Φ p ( η pn 1,η pn 1 ). (15) By Lemma 1 for p>3 4Φ p ( η pn 1,η pn 1 )=A p ( η pn 1,η pn 1 ) + pb p ( η pn 1,η pn 1 ). (16) Since A p (x, y) is anti-symmetric, A p ( x, y) is symmetric and by Lemma, divisible by x + y. The quotient is of degree p (mod ), hence, by Lemma 3, is a polynomial in (x+y) and xy. It follows that A p ( η pn 1,η pn 1 ) =εw1, w 1 Q. Again, by lemmas and 3, B p ( η pn 1,η pn 1 ) =εpw, w Q and (14) follows by (15) and (16). It remains to consider p =3.Let η 3n 1 = εu + ε 3 α v. Then Φ 3 ( η 3n 1,η 3n 1 ) =( εu + ε 3 α v) ( εu + ε 3 α v)( εu ε 3 α v) +( εu ε 3 α v) =ε(u +(3 α+1 v) )=ε(u +3 α+1 v) + ε(u 3 α+1 v). Lemma 7. In the notation of Lemma 6, for all positive integers n, H p n/εph p n 1 is a sum of two squares. (17)

5 Vol. 101 (013) On certain biquadratic equations 17 Proof. We have by (11) and (1) H p n/h p n 1 =Φ p n(η, η 1 )=Φ p (η pn 1,η pn 1 ) (18) and by Lemma 1, 4H p n/h p n 1 = A p (η pn 1,η pn 1 ) + pb p (η pn 1,η pn 1 ). (19) Since A p (x, y) is anti-symmetric, A p (x, y)/(x y) is symmetric of even degree and, by Lemma 3, A p (η pn 1,η pn 1 ) =εpw3,w 3 Q. (0) Also B p (x, y)/(x + y) is symmetric of even degree, hence B p (η pn 1,η pn 1 ) =εw 4,w 4 Q. (1) (17) follows from (18) (1). It remains to consider p =3.Letη 3n 1 = εu + ε 3α v. Then F 3 n/3εf 3 n 1 = 1 ( ( εu + ε 3 3ε α v) +( εu + ε 3 α v)( εu ε 3 α v) +( εu ε 3 α v) ) =(u + v )=(u + v) +(u v). Proof of Theorem 3. It follows by induction on n from Lemmas 6 and 7 that for n even G p n and H p n are sums of two squares. The formula (13) gives the theorems. Open Access. This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. Reference [1] H. Riesel, Prime Numbers and Computer Methods for Factorization, Birkhäuser A. Schinzel Institute of Mathematics, Polish Academy of Sciences, Sniadeckich 8, P.O.Box 1, Warszawa, Poland schinzel@impan.gov.pl

6 18 A. Schinzel and M. Ska lba Arch. Math. M. Ska lba Institute of Mathematics, University of Warsaw, Banacha, Warszawa, Poland Received: April 013

On a Diophantine Equation 1

International Journal of Contemporary Mathematical Sciences Vol. 12, 2017, no. 2, 73-81 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ijcms.2017.728 On a Diophantine Equation 1 Xin Zhang Department