SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS COURSE NOTES, 2015

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1 SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS COURSE NOTES, Squarefree values of olynomials: History In this section we study the roblem of reresenting square-free integers by integer olynomials. It is conjectured that a searable olynomial (that is, without reeated roots) f Z[x] takes infinitely many square-free values, barring some simle excetional cases, in fact that the integers a for which f(a) is square-free have a ositive density. A clear necessary condition is that the sequence f(n) has no fixed square divisor; the conjecture is that this is the only obstruction: Conjecture 1. Let f(x) Z[x] be a searable olynomial (i.e. with no reeated roots) of ositive degree. Assume that gcd{f(n) : n Z} is squarefree 1. Then there are infinitely many square-free values taken by f(n), in fact that a ositive roortion of the values are square-free: with #{1 n X : f(n) is square-free } c f X, as X, (1.1) c f = where (1 ρ f ( 2 ) 2 ), (1.2) ρ f (D) = #{c mod D : f(c) = 0 mod D}. The roblem is most difficult when f is irreducible. Nagell ([6] 1922) showed the infinitude of squarefree values in the quadratic case. Estermann ([2] 1931) gave ositive density for the case f(x) = x 2 + k. The general quadratic case was solved by Ricci in 1933 [7]. For cubics, Erdös ([1], 1953) showed that there are infinitely many square-free values, and Hooley ([4], 1967) gave the result about ositive density. Beyond that nothing seems known unconditionally for irreducible f, for instance it is still not known that a is infinitely often square-free. Date: March 29, In fact one can even allow fixed, square divisors of f(n), rovided we divide them out in advance, by relacing f(n) by f(n)/b, where B is the smallest divisor of B := gcd{f(n) : n Z} so that B/B is square-free, and if we relace c f by (1 ω f () 2+q ), where for each rime, we denote by q the largest ower of dividing B, and by ω f () the number of a mod 2+q for which f(a)/b = 0 mod 2. 1

2 2 A roblem which has recently been solved is to ask how often an irreducible olynomial f Z[x] of degree d attains values which are free of (d 1)-th owers, either when evaluated at integers or at rimes, see [8] The ABC conjecture. Granville [3] showed that the ABC conjecture comletely solves the conjecture 1. The ABC conjecture states that for every ε > 0, there exist only finitely many triles (a, b, c) of ositive corime integers, with a + b = c, such that c > rad(abc) 1+ε. Here the radical of an integer is the roduct of all distinct rimes dividing it: rad(n) := N. Equivalently, for every ε > 0, there exists a constant K ε such that for all triles (a, b, c) of corime ositive integers, with a + b = c, we have c < K ε rad(abc) 1+ε. 2. The density c f We ause to analyze the conjectural density c f of squarefree values of f, given by (1.1). Exercise 1. Assume that f(n) admits no common square factor. Show that c f > 0, i.e. that ρ f ( 2 ) < 2 for all rimes. By the Chinese remainder theorem, D ρ f (D) is a multilicative function The slit quadratic case f(x) = x(x + 1). Lemma 2.1. Suose f(x) = x(x + 1). Then for all rime, and k 1, ρ( k ) = 2. Proof. We want to count solutions modulo k of c(c + 1) = 0 mod k. But since is rime, and c, c+1 have no common factors, this means that either c = 0 mod k or c + 1 = 0 mod k and each case has exactly one solution. Thus ρ f ( k ) = The irreducible quadratic case f(x) = x Lemma 2.2. Suose f(x) = x i) If 2 then ρ( k ) = ρ() for all k 1. iii) For 2, { 2, = 1 mod 4 ρ() = 0, = 3 mod 4 iii) ρ(4) = 0. Proof. Part (i) follows from Hensel s Lemma, and is valid for any olynomial f Z[x], for disc(f). Part (ii) is secific to f(x) = x and is due to Fermat. Part (iii) is a direct comutation.

3 SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS 3 Note: The above shows that for f(x) = x 2 + 1, our density c f is ( ) (2.1) c f = (1 ρ(2 ) 2 ) = (1 2 ) = The quadratic case Our goal here is to treat the quadratic case, in fact below we will secialize to the simle cases of f(x) = x(x + 1) (the slit case) and f(x) = x (the irreducible case). For X 1, we set and N(X) := #N (X). N (X) := {n X : f(n) squarefree} Theorem 3.1. Let f(x) = x(x + 1) or f(x) = x Then N(X) = c f X + O(X 2/3 log X), as X with c f = C slit = (1 2 2 ) in the slit case f(x) = x(x + 1), and c f = 2 (1 1+( 1 ) 2 ) = in the irreducible case f(x) = x Note that in the slit case, since n(n + 1) is squarefree if and only if both n and n + 1 are squarefree (because n, n + 1 are corime), the result says that the robability that both n and n + 1 are squarefree is C slit = (1 2 2 ) = , which is smaller than 1/ζ(2) 2 = (1 2/2 + 1/ 4 ) = , which would be the case if these were indeendent events The strategy. We use the sieve of Eratosthenes and Legendre: Recall that the indicator function of the squarefrees is Hence N(X) = n X 1 SF (f(n)) = n X 1 SF (m) = d 2 m µ(d). d 2 f(n) µ(d) = µ(d)#{n X : d 2 f(n)} d X Note that we can constrain d X because d 2 divides the quadratic olynomial f(n), which is X 2 if n X. We ick a arameter Y (eventually taken to be Y = X 1/3 ) and decomose the sum into two arts, a sum N (X) over small divisors d < Y, and a sum N (X) over large divisors Y < d < X: N(X) = N (X) + N (X), N (X) = µ(d)#{n X : d 2 f(n)} and N (X) = µ(d)#{n X : d 2 f(n)} Y <d X

4 4 We will show that (3.1) N (X) = c f X + O( X Y log Y + Y log Y ) and (3.2) N (X) X2 Y 2 Taking Y = X 1/3 we obtain giving Theorem 3.1. N(X) = c f X + O(X 2/3 log X) 4. The main term: small divisors We will estimate N (X) (the main term) by using inclusion-exclusion. Recall N (X) = µ(d)#{n X : d 2 f(n)}. Lemma 4.1. Proof. We decomose Using we get #{n X : D f(n)} = Xρ(D) ( ) D + O ρ(d). #{n X : D f(n)} = C mod D f(c)=0 mod D #{n X : n = C mod D}. #{n X : n = C mod D} = X D + O(1) #{n X : D f(n)} = C mod D f(c)=0 mod D X D + O(1) = Xρ(D) D + ρ(d). (4.1) Hence we obtain N (X) = = X ( Xρ(d 2 ) ) µ(d) d 2 + O(ρ(d 2 )) µ(d)ρ(d 2 ) d 2 ( + O ) µ(d) ρ(d 2 ).

5 We have SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS 5 µ(d)ρ(d 2 ) d 2 = d=1 µ(d)ρ(d 2 ) d 2 ( + O Using multilicativity of ρ (and of µ) gives µ(d)ρ(d 2 ) d 2 = (1 ρ(2 ) 2 ) = c f. d=1 µ(d) ρ(d 2 ) ) d 2 By Lemmas 2.1 and 2.2, ρ( 2 ) 2 for rime, and thus for d squarefree ρ(d 2 ) = d ρ( 2 ) d 2 = τ(d) where τ is the divisor function. Hence the tail of the sum is bounded by µ(d) ρ(d 2 ) d 2 τ(d) d 2 log Y Y and the remainder in (4.1) is bounded by µ(d) ρ(d 2 ) τ(d) Y log Y. Therefore as claimed. N (X) = c f X + O( X Y log Y ) + O(Y log Y ) Exercise 2. Using n x τ(n) = x(log x + C) + O(x1/2 ), show that n>y τ(n) n 2 = log Y + C + 2 Y 1 + O( Y 3/2 ) 5. Bounding the contribution of large divisors We write the condition d 2 f(n) as f(n) = d 2 D for some integer D 1. Then N (X) = µ(d) #{n X : f(n) = d 2 D}. n X d 2 f(n) We now interchange the roles of d and D: If d > Y then D = f(n)/d 2 X 2 /Y 2. Hence ignoring the size and squarefreeness restriction on d, (5.1) N (X) #{u, v X : f(u) = v 2 D}. 1 D X 2 /Y 2 Now take f(x) = x Then the equation f(u) = Dv 2 becomes which is a Pellian equation. u 2 Dv 2 = 1

6 6 The main new arithmetic ingredient we need now is a bound on the number of solutions of the Pellian equation x 2 Dy 2 = 1 lying in a box of side X: Let (5.2) S D (X) := #{(x, y) [1, X] 2 : x 2 Dy 2 = 1} Proosition 5.1. Suose 1 < D < X is not a erfect square. Then S D (X) log X log D. If D = is a erfect square, then there are are no solutions of x 2 Dy 2 = 1 if D > 1, while for D = 1 there are 2 solutions. Proof. Suose D > 1 is not a erfect square. By the theory of Pell s equation, if the equation x 2 Dy 2 = 1 is solvable in integers, then all integer solutions (x, y) are of the form x + Dy = ±ɛ 2n+1 D, n Z, where ɛ D = x 1 + y 1 D is the fundamental solution, with x1, y 1 1. Hence if 1 x, y X then x + y D = ɛ 2n+1 D for some n 0 and then 0 n log(x + Dy) = log(x + x 2 + 1) log X. 2 log ɛ D 2 log ɛ D log ɛ D Since ɛ D = x 1 + y 1 D > D, we obtain S D (X) log X log D. For D = C 2 a erfect square, the equation x 2 Cy 2 = 1 becomes x 2 (Cy) 2 = 1 or (Cy x)(cy+x) = 1, which forces Cy x = Cy+x = ±1, so that x = 0, and then C 2 y 2 = 1 is solvable only for C = 1 in which case there are two solutions. Inserting Proosition 5.1 into the bound (5.1) for N 2 gives N 2 log X S D (X) 1 + log D X2 Y 2 1 D<X 2 /Y 2 1<D<X 2 /Y 2 as claimed, on using 1<D<Z 1 Z log D 2 1 log t dt Z log Z Other quadratic olynomials. The considerations above extend to the case when f(x) = Ax 2 +Bx+C Z[x] is any quadratic olynomial, say f(x) = x(x + 1) (the slit case). All we have to do is rewrite the equation f(u) = Dv 2 : Multilying by 4A and comleting the square gives 4ADv 2 = (2Au + B) 2 f where f = B 2 4AC is the discriminant of f, which is nonzero if and only if f has no reeated roots. Thus the equation f(u) = Dv 2 becomes (2Au + B) 2 AD(2v) 2 = f

7 SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS 7 and we need to bound the number of solutions of U 2 (AD)V 2 = f with U, V X. For instance, in the slit case f(x) = x 2 +x we get = +1 and the equation becomes U 2 DV 2 = 1, to which we aly a version of Proosition 5.1. Remark. When > 1 there may be more than one orbit of the unit grou {±ɛ n D } and one has to account for that. References [1] P. Erdös. Arithmetical roerties of olynomials. J. London Math. Soc. 28, (1953) [2] T. Estermann, Einige Sätze über quadratfreie Zahlen. Math. Ann. 105 (1931), [3] A. Granville, ABC allows us to count square-frees. Internat. Math. Res. Notices 1998, no. 19, [4] C. Hooley, On the ower free values of olynomials. Mathematika [5] C. Hooley, On the square-free values of cubic olynomials. Journal für die reine und angewandte Mathematik (Crelles Journal). (1968), 229, [6] T. Nagell, Zur Arithmetik der Polynome, Abhandl. Math. Sem. Hamburg 1 (1922), [7] G. Ricci, Ricerche aritmetiche sui olinomi. Rend. Circ. Mat. Palermo 57 (1933), [8] T. Reuss, Power-Free Values of Polynomials, Bull. London Math. Soc. (2015) doi: /blms/bdu116. arxiv: [math.nt]