DIVISIBILITY PROPERTIES OF HYPERGEOMETRIC POLYNOMIALS

CLEMENS FUCHS^{†} AND T.N. SHOREY

Abstract. In this paper we give effective upper bounds for the degree k of divisors (over ℚ) of hypergeometric polynomials defined by

∑_{j=0}^{n} a_{j} \frac{(a)_{j}}{(b)_{j}(c)_{j}} x^{j},

where (m)_{j} = m(m + 1) ⋯ (m + j − 1) denotes the Pochhammer symbol and a_{0}, …, a_{n} are integers with |a_{0}| = |a_{n}| = 1, a = −n−r, b = α+1, c ≥ 1 and α = −tn − s − 1, tn + s for integers r ≥ 0, t ≥ 1, s, c bounded in terms of k. These results generalize on earlier results of the authors and others on generalized Laguerre polynomials.

2000 Mathematics Subject Classification: 11C08, 33C45, 11N05. Keywords: Hypergeometric polynomials, irreducibility, p-adic Newton polygons, gaps between primes.

1. Introduction and results

For rational numbers a, b, c the hypergeometric polynomials are defined by

g_{a,b,c}(x) = ∑_{j=0}^{n} \frac{(a)_{j}}{(b)_{j}(c)_{j}} x^{j},

where (m)_{j} = m(m + 1) ⋯ (m + j − 1) denotes the Pochhammer symbol. We mention that such polynomials appear by truncating the infinite series given by generalized hypergeometric functions of type _{2}F_{2}(a, 1; b, c; x) (with the usual notation for such functions). For a = −n, b = α + 1, c = 1 one gets

g_{−n,α+1,1}(x) = \frac{n!}{(α + 1) ⋯ (α + n)} ∑_{j=0}^{n} \frac{(α + n) ⋯ (α + j + 1)}{(n − j)!j!}(−x)^{j}
= \frac{n!}{(α + 1) ⋯ (α + n)}L_{n}^{(α)}(x)

the generalized Laguerre polynomials (up to a constant).

†Corresponding author.

Let n, s and t be integers with n ≥ 2 and |s| ≤ n and

(1)  α = −tn − s − 1 with t ≥ 2

or

(2)  α = tn + s with t ≥ 1.

Such polynomials have been studied extensively, especially the case L_{n}^{(α)}(x), starting with work by Schur [10, 11], Coleman [1] and Filaseta and others (see e.g. [3]). We also mention the papers [2] and [5] since we shall be using their arguments. Now we are additionally assuming that

(3)  a = −n − r, b = α + 1, c ≥ 1

for integers r, c with r ≥ 0. Let α satisfy (1). Then

g_{a,b,c}(x) = \frac{(n + r)!(c − 1)!}{((t − 1)n + s + 1) ⋯ (tn + s)(n + r + c − 1)!} ∑_{j=0}^{n} c_{j}x^{n−j}

with

(4)  c_{j} = \binom{n + r + c − 1}{r + j}((t − 1)n + s + 1) ⋯ ((t − 1)n + s + j)

and therefore we have for m ∈ {0, …, n} that

(5)  \frac{c_{n}}{c_{n−m}} = \frac{(tn + s)!}{(tn + s − m)!} \frac{(n + r − m)!}{(n + r)!} \frac{(m + c − 1)!}{(c − 1)!}.

Now let α satisfy (2). Then we have

g_{a,b,c}(x) = \frac{(−1)^{n}(c − 1)!(n + r)!}{(tn + s + 1) ⋯ ((t + 1)n + s)(n + c + r − 1)!} ∑_{j=0}^{n} c′_{j}x^{n−j}

with

(6) c′_{j} = (−1)^{j}\binom{n + r + c − 1}{r + j}((t + 1)n + s − j + 1) ⋯ ((t + 1)n + s)

and therefore we have

(7)  \frac{c′_{n}}{c′_{n−m}} = (−1)^{m} \frac{(tn + s + m)!}{(tn + s)!} \frac{(n + r − m)!}{(n + r)!} \frac{(m + c − 1)!}{(c − 1)!}

for m ∈ {0, …, n}. For 0 ≤ j ≤ n, we write d_{j} = c_{j} or c′_{j} according as α satisfies (1) or (2). Moreover, we set

f(x) = ∑_{j=0}^{n} d_{j}x^{n−j}

and

F (x) = ∑_{j=0}^{n} a_{j}d_{j}x^{n−j}

for integers a_{0}, …, a_{n}. Here we notice that F (x) is the polynomial stated in the abstract.

Our intention is to generalize the results of [6] to this extended setting. It was proved there that for integers a_{0}, …, a_{n} with |a_{0}| = |a_{n}| = 1 there exist effectively computable absolute constants η_{0} and ε such that for all η_{0} < k ≤ \frac{n}{2} and for all α with t < ε log k, 0 ≤ s < εk log k the polynomial F (x) does not have a factor of degree k. We also mention that for 2 ≤ k ≤ \frac{n}{2}, it was proved in [9, Theorem 1.3] that if for given ε > 0 the hypergeometric polynomial g_{−n−r,α+1,c} with 0 ≤ α ≤ k and r+c < (1/3−ε)k has a divisor of degree k, then k is bounded by an effectively computable constant depending only on ε, and in [9, Theorem 1.4] that g_{−n,α+1,1} with α = −n − s − 1 and 0 ≤ s ≤ 0.95k has no factor of degree k at all.

In the sequel we will denote by η_{1}, η_{2}, … effectively computable absolute positive real constants.

Theorem 1. Let a_{0}, …, a_{n} be any integers with |a_{0}| = |a_{n}| = 1. Then there exist constants ε > 0 and η_{1} such that for all η_{1} < k ≤ \frac{n}{2} and for all α satisfying (1) with t ≥ 4 or (2) with t ≥ 3 and for

t < ε log k, max{r, c} < k, |s| < εkϑ,

where ϑ = log k, the polynomial F (x) does not have factor of degree k.

Moreover, under the abc-conjecture, the statement holds true with ϑ = log n.

For small values of t in both the cases for α we also get results, but under slightly stronger restrictions. We state them separately in the following theorem.

Theorem 2. The statement of Theorem 1 holds true for

max{r, c, |s|} < k, r + c + |s| < n^{6/11+ε} if α satisfies (1) with t = 2,
max{r, c, |s|} < k, r + c < n^{6/11+ε}  if α satisfies (2) with t = 1,

and for

max{r, c} < k, |s| < εkϑ, r + c < n^{6/11+ε}

if α satisfies (1) with t = 3 or (2) with t = 2.

The two theorems imply [6, Theorem 1,2] apart from the values of ε. Further we observe that we cover the negative values of s in contrast to the situation in [6].

In the proof we will again use the p-adic Newton polygon, where the prime p satisfies certain properties. Let us write v_{p} for the p-adic valuation and v_{p}(0) = ∞. Then we use the following lemma, which we take from [2]:

Lemma 1. Let k and l be integers with k > l ≥ 0 and k ≤ \frac{n}{2}. Suppose that

g(x) = ∑_{j=0}^{n} b_{j}x^{n−j} ∈ ℤ[x]

and p is a prime such that p ∤ b_{0}, p|b_{j} for all j ∈ {l + 1, …, n} and the slope of the right-most edge of the Newton polygon for f(x)

max_{1≤m≤n} \{\frac{v(b_{n}) − v(b_{n−m})}{m}\}

is < 1/k. Then for any integers a_{0}, …, a_{n} with |a_{0}| = |a_{n}| = 1, the polynomial

G(x) = ∑_{j=0}^{n} a_{j}b_{j}x^{n−j}

cannot have a factor with degree in the interval [l + 1, k].

The existence of such primes is the main challenge (and also the most significant difference to our results in [6]) and this will be guaranteed by tools from analytic number theory. The result on primes that we are needing is the following lemma:

Lemma 2. There exists a constant η_{2} such that for all x > η_{2} and for all \frac{6}{11} < θ ≤ 1 we have

0.969\frac{y}{log x} ≤ π(x) − π(x − y)

for y = x^{θ}, where π(x) is the prime counting function.

This result is taken from [7]. Moreover, for the conditional result in Theorem 1 we recall the abc-conjecture that we will use.

Lemma 3 (abc-Conjecture). For every ϵ > 0 there exists a constant γ = γ(ϵ) depending only on ϵ such that for all coprime nonzero integers a, b, c with a + b = c the inequality

max{|a|, |b|, |c|} < γN(abc)^{1+ϵ}

holds, where N(m) denotes the product over all different prime divisors of m.

Now we have everything ready to give the proof of Theorem 1 and Theorem 2 that will be done simultaneously in the next section.

2. Proof of Theorem 1 and 2

For the proof we assume that F (x) has a factor of degree k such that k ≤ \frac{n}{2} and k exceeds a sufficiently large constant η_{1}. Let ϑ = log n if the abc-conjecture holds and ϑ = log k otherwise. Moreover, we put δ = 1/4.

By Lemma 2 there exists ℓ with

n^{13/22} ≤ ℓ < ((t + 1)n + s)^{13/22}

such that (t − 1)n + s + ℓ or (t + 1)n + s − ℓ + 1 is a prime p according as (1) or (2) holds, respectively. Then it follows from (4) and (6), respectively, that p∥d_{j} for j ∈ {ℓ, …, n} (here we use, as usual, d∥d_{j} for d|d_{j} and d^{2} ∤ d_{j}). Next we show that p > n + c + r. For this we have to take special care of the small values of t. We have

(t − 1)n + s + ℓ ≥ n − |s| + n^{6/11+ε} > n + c + r  if (1) with t = 2,
(t − 1)n + s + ℓ ≥ n + n^{6/11+ε} > n + c + r  if (1) with t = 3,

(t + 1)n + s − ℓ + 1 > n + (n/2 − n^{7/11}) + 1

> n + n^{6/11+ε} > n + c + r  if (2) with t = 1,
(t + 1)n + s − ℓ + 1 > n + (n − n^{7/11})
> n + n^{6/11+ε} > n + c + r  if (2) with t = 2

and finally p > 2n ≥ n + c + r in all other cases. This implies p ∤ d_{0}. Therefore, the right-most edge of the p-adic Newton polygon for f(x) has slope < 1/k. By Lemma 1 we conclude that k ≤ ℓ ≤ (3εn log n)^{13/22} ≤ n^{7/11}.

Now we will first consider the case (1), i.e. that α = −tn − s − 1. We write z = 6εkϑ. Observe that every prime p > z ≥ k that divides ((t − 1)n + s + 1) ⋯ ((t − 1)n + s + k) divides exactly one of the factors, so p|(t − 1)n + s + 1 + ℓ for some 0 ≤ ℓ ≤ k − 1. We shall show that a prime with this property exists. For this purpose we use the following lemma (cf. [4, Lemma 6] and [6, Lemma 5]).

Lemma 4. Let z be a positive real number. For each prime p ≤ z, let d_{p} ∈ {n, n − 1, …, n − k + 1} with v_{p}(d_{p}) maximal. Define

Q_{z} = Q_{z}(n, k) = ∏_{p>z} p^{v_{p}(A)}

with A = n(n − 1) ⋯ (n − k + 1). Then

Q_{z} ≥ \frac{n(n − 1) ⋯ (n − k + 1)}{(k − 1)! ∏_{p≤z} p^{v_{p}(d_{p})}} ≥ \frac{(n − k + 1)^{k−π(z)}}{(k − 1)!} ,

where π(z) denotes the number of primes ≤ z.

By the above lemma we get for ϑ = log k that

Q_{z}((t − 1)n + s + k, k) ≥ \frac{((t − 1)n + s + 1)^{k}}{(k − 1)!((t − 1)n + s)^{π(z)}} ≥ n^{k−2π(z)−7k/11}
≥ n^{(4/11−12ε(1+δ)^{2})k} > 1,

where we have used the inequality (k − 1)! ≤ k^{k} ≤ n^{7k/11} and the estimate

π(z) ≤ \frac{(1 + δ)6εkϑ}{log(6εkϑ)} ≤ 6ε(1 + δ)^{2}k,

that follows at once from the prime number theorem. It remains to show that we also have a prime p > η_{3}k log n > z for some η_{3} and for ε small enough, dividing ((t − 1)n + s + 1) ⋯ ((t − 1)n + s + k), if we assume the abc-conjecture to be true. For this we just have to follow the arguments of [8, Theorem 1]. We give the proof for the readers convenience (and since the statement that is proved there, at first sight, does not seem to be connected to what we need). For a prime p dividing two different factors of this product of k consecutive terms we have p ≤ k. Thus

∏_{i=1}^{k} N((t − 1)n + s + i) ≤ ( ∏_{p≤P} p) ∏_{p≤k} p^{⌊k/p⌋} ≤ η_{4} exp (η_{5}(P + k log k)) ,

where P denotes the largest prime divisor of ((t − 1)n + s + 1) ⋯ ((t − 1)n + s + k) and N(m) the product over all primes dividing m. Now let j_{1}, j_{2} with N((t − 1)n + s + j_{1}) ≤ N((t − 1)n + s + j_{2}) be the smallest two values in the set {N((t − 1)n + s + j); 1 ≤ j ≤ k}. It follows

N((t − 1)n + s + j_{2}) ≤ (∏_{i=1}^{k} N((t − 1)n + s + i))^{1/(k−1)}
= exp (η_{6}(P/k + log k)) .

We apply Lemma 3 with ϵ = 1 to the equation

\frac{(t − 1)n + s + j_{1}}{d} − \frac{(t − 1)n + s + j_{2}}{d} = \frac{j_{1} − j_{2}}{d}

and get

n ≤ η_{7} (N (\frac{(t − 1)n + s + j_{1}}{d} ) N (\frac{(t − 1)n + s + j_{2}}{d} ) \frac{|j_{1} − j_{2}|}{d})^{2}
≤ exp (η_{8}(P/k + log k)) ,

where d denotes the greatest common divisor of (t − 1)n + s + j_{1} and (t − 1)n + s + j_{2}. Finally, this implies P > η_{9}k log n.

Thus there is a prime p > z dividing ((t−1)n+s+1) ⋯ ((t−1)n+s+k), say p divides (t − 1)n + s + 1 + ℓ with 0 ≤ ℓ ≤ k − 1. We may assume that p ∤ n + c + i for every 0 ≤ i ≤ r − 1, since assuming the contrary we have p|n+c+i, which implies that p divides |(t−1)n+s+1+ℓ−(t−1)(n+c+i)| ≤ |s|+1+ℓ+tc+tr ≤ εkϑ+k+2εk log k ≤ 4εkϑ ≤ z, a contradiction. It follows that p satisfies p|c_{j} for ℓ+1 ≤ j ≤ n and p ∤ c_{0}. Define m = m(p) ∈ {1, …, n} such that

\frac{v_{p}(c_{n}) − v_{p}(c_{n−m(p)})}{m(p)} = max_{1≤m≤n} \{\frac{v_{p}(c_{n}) − v_{p}(c_{n−m})}{m}\}

is the slope of the right most edge of the p-adic Newton polygon for f(x) with respect to p. Then by Lemma 1 and (5) we conclude

\frac{1}{k} ≤ \frac{v_{p}(c_{n}) − v_{p}(c_{n−m})}{m}

≤ \frac{1}{m} [v_{p} (\frac{(tn + s)!}{(tn + s − m)!}) − v_{p} \begin{aligned}( \\ ( & {n + r \atop m} & ) \\ & ) \end{aligned} + v_{p} \begin{matrix}( \\ ( & {m + c − 1 \atop c − 1} & ) \\ ) \end{matrix}] .

For estimating the third summand we may assume that m > 5εkϑ, since otherwise m + c − 1 ≤ 6εkϑ < p and so this summand is zero, and therefore we get

\frac{1}{m}v_{p} \begin{matrix}( \\ ( & {m + c − 1 \atop c − 1} & ) \\ ) \end{matrix} ≤ \frac{1}{m}v_{p}((m + c − 1)!) ≤ \frac{m + c − 1}{m(p − 1)}
= \frac{1}{p − 1} + \frac{c − 1}{m(p − 1)} < \frac{1}{5εkϑ} + \frac{k}{5εkϑ5εkϑ} ≤ \frac{1}{4k} .

If p does not divide (tn + s − m + 1) ⋯ (tn + s) then we immediately get a contradiction. Thus we may assume that p divides tn + s − i with 0 ≤ i ≤ m − 1. But then it also divides t((t − 1)n + s + ℓ + 1) − (t − 1)(tn + s − i) = t(ℓ + 1) + s + (t − 1)i and therefore p ≤ 2εkθ + εm log k ≤ 2εkϑ + 2εm log k.

Since p > z = 6εkϑ, this implies that

(8)  \frac{2kϑ}{log k} < m.

Moreover we get

\frac{3}{4k} ≤ \frac{1}{m}v_{p} (\frac{(tn + s)!}{(tn + s − m)!}) ≤ \frac{1}{m} ∑_{j=1}^{∞} (⌊\frac{tn + s}{p^{j}}⌋ − ⌊\frac{tn + s − m}{p^{j}}⌋)

≤ \frac{1}{m} ∑_{j=1}^{J} (\frac{m}{p^{j}} + 1) ≤ \frac{1}{p − 1} + \frac{J}{m} ≤ \frac{1}{12k} + \frac{J}{m},

where

J := ⌊\frac{log(tn + s)}{log p}⌋ .

This gives m ≤ 3kJ/2 and thus

(9)  m ≤ \frac{3k}{2} \frac{log(tn + s)}{log p} < \frac{3(1 + δ)k log n}{2 log k}.

If ϑ = log n, then it follows from (8) and (9) that 2 log n = 2ϑ < \frac{3}{2}(1+δ) log n. This contradiction proves the result if we assume that the abc-conjecture is true. Now we give the proof for the case ϑ = log k. In fact all the above statements are true for every prime p > z dividing (t − 1)n + s + 1 + ℓ for some 0 ≤ ℓ ≤ k − 1. Especially this is the case for the inequalities (8) and (9). Next we will prove the existence of such a prime with even stronger assumptions.

Let U be the set of numbers (t − 1)n + s + 1 + j with 0 ≤ j ≤ k − 1, where for every prime q ≤ z we have removed those numbers d_{q} ∈ {(t − 1)n + s + 1, …, (t − 1)n + s + k − 1} with v_{q}(d_{q}) maximal. We mention that all elements of U are ≥ n if t > 2 and ≥ n/2 if t = 2. Now let Ω be the set of all primes q > z with v_{q}(u) > 0 for some u ∈ U and q^{v_{q}(u)} ≤ (2k + m)ε log k for all u ∈ U. Observe that all such q divide exactly one u ∈ U, since q > z ≥ k. Thus we have

log ( ∏_{u∈U} ∏_{q∈Ω} q^{v_{q}(u)}) ≤ log ( ∏_{z<q≤(2k+m)ε log k}(2k + m)ε log k)

≤ π((2k + m)ε log k) log((2k + m)ε log k) ≤ (1 + δ)ε(2k + m) log k
≤ ε(1 + δ)(2k log k + 3k log n) ≤ 5ε(1 + δ)k log n,

where for the second summand (9) was used. It follows that

log ( ∏_{u∈U} ∏_{q≤z} q^{v_{q}(u)}) + log ( ∏_{u∈U} ∏_{q∈Ω} q^{v_{q}(u)})

≤ \frac{(1 + δ)6εk log k}{log(6εk log k)} log k + 5ε(1 + δ)k log n ≤ \frac{2}{3}k log n,

since k ≤ n^{7/11}. On the other side we have

log ( ∏_{u∈U} u) ≥ log(((t − 1)n + s + 1) ⋯ ((t − 1)n + s + k − π(z)))

≥ (k − π(z)) log \frac{n}{2} > \frac{2}{3}k log n.

By comparing the lower and upper bound just obtained we conclude that there is a prime q > z that divides some element u ∈ U with the additional property that q^{v_{q}(u)} > (2k + m)ε log k. We write u = (t − 1)n + s + ℓ + 1, 0 ≤ ℓ ≤ k − 1 and define f by q^{f−1} ≤ (2k + m)ε log k < q^{f} and such that q^{f}|u. Observe that 1 ≤ f ≤ v_{q}(u).

Now if q^{f} divides tn + s − i for some 0 ≤ i ≤ m − 1, then it also divides |t((t − 1)n + s + ℓ + 1) − (t − 1)(tn + s − i)| ≤ tℓ + t + |s| + (t − 1)i < 3εk log k which contradicts the fact that q^{f} > (2k + m)ε log k by (8). Thus q^{f} does not divide tn + s − i for any 0 ≤ i ≤ m − 1 and we conclude

\frac{3}{4k} ≤ \frac{1}{m}v_{q} (\frac{(tn + s)!}{(tn + s − m)!}) ≤ \frac{1}{m} ∑_{j=1}^{f−1} (\frac{m}{q^{j}} + 1) ≤ \frac{1}{q − 1} + \frac{f − 1}{m}.

For f = 1 we immediately get a contradiction. For f ≥ 2 we get

2(2k + m)ε log k ≥ q^{f−1} + (2k + m)ε log k log k ≥ (f − 1)6εk log k + 4εk log k,

where we have used (8), and therefore 3(f − 1)k < m, which gives

\frac{3}{4k} ≤ \frac{1}{q − 1} + \frac{f − 1}{m} < \frac{5}{12k} + \frac{1}{3k} = \frac{3}{4k} ,

a contradiction again. This completes the proof in this case.

Now we come to the case (2), i.e. that α = tn + s. Here we can argue in almost the same way. We have to consider primes p > z = 6εkϑ that divide ((t + 1)n + s − k + 1) ⋯ ((t + 1)n + s) and we show by following the arguments from above that such a prime exists. As before we may assume that a prime dividing (t + 1)n + s − ℓ for some 0 ≤ ℓ ≤ k − 1 does not divide any of n + c + i for 0 ≤ i ≤ r − 1, since otherwise we have that p divides |(t + 1)n + s − ℓ − (t + 1)(n + c + i)| ≤ |s| + ℓ + (t + 1)(c + r) ≤ εkϑ + k + 4εk log k ≤ 6εkϑ = z. Therefore it follows that such a p satisfies p|c′_{j} for ℓ + 1 ≤ j ≤ n and p ∤ c′_{0}. Proceeding as in the previous case we conclude from Lemma 1 and (7) that

\frac{1}{k} ≤ \frac{v_{p}(c′_{n}) − v_{p}(c′_{n−m})}{m}

≤ \frac{1}{m} [v_{p} (\frac{(tn + s + m)!}{(tn + s)!}) − v_{p} \begin{aligned}( \\ ( & {n + r \atop m} & ) \\ & ) \end{aligned} + v_{p} \begin{matrix}( \\ ( & {m + c − 1 \atop c − 1} & ) \\ ) \end{matrix}] .

In the same way as before we can estimate the third summand and we may assume that p divides tn + s + m − i with 0 ≤ i ≤ m − 1 and therefore 6εkϑ = z < p ≤ |t((t+1)n+s−ℓ)−(t+1)(tn+s+m−i)| ≤ tℓ+|s|+(t+1)m ≤ 2εkϑ + 2mε log k, which again implies 2kϑ/ log k < m. On the other hand, one shows m ≤ 3kJ/2 with

J := ⌊\frac{log((t + 1)n + s)}{log p}⌋ ,

which gives m < 3(1 + δ)k log n/(2 log k). For ϑ = log n we conclude the proof by comparing the lower and upper bound for m. Thus we may assume that ϑ = log k. By arguing as above we get a prime q > z that divides exactly one element of the form u = (t + 1)n + s − ℓ, 0 ≤ ℓ ≤ k − 1 and with q^{v_{q}(u)} > (2k + m)ε log k (observe that now all such elements u are ≥ n). By defining f as before we conclude that q^{f} does not divide tn + s + 1 + i for any 0 ≤ i ≤ m − 1, since otherwise it would divide t((t + 1)n + s − ℓ) − (t + 1)(tn + s + 1 + i) = −tℓ − ts − (t + 1)(1 + i) that contradicts q^{f} > z being large. Similar as in the case (1) the proof can be finished.

3. Acknowledgment

This paper was partly written during mutual visits of the authors at their home institutions, to which they are indepted for the hospitality and for financial support.

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Clemens Fuchs Department of Mathematics, ETH Zurich Rämistrasse 101, 8092 Zürich, Switzerland Email: clemens.fuchs@math.ethz.ch

T.N. Shorey School of Mathematics, Tata Institute of Fundamental Research Homi Bhabha Road, 400005 Mumbai, India Email: shorey@math.tifr.res.in