Extensions of Schur's irreducibility results - Tata Institute of ...

Shanta Laishram and T. N. ShoreyBy using estimates of π(ν) from Lemma 4.1, we obtainTherefore we have(π(4k) − π(k)) log kk 4 log k 1.2762(1 +log 4k log 4k ) − log klog k − 1( 4 1 − log 4 ) (1 + 1.2762log 4k log 4k() 4 1 − − 1 3.log 4 − 1.2762log 4k)− 10 > log(0.4) − 18k− .5 log k+ log( 335.7k2.83 ) − k −14 log(2.83) − 4.5.The right hand side of the above inequality is an increasing function of k and the inequality is notvalid at k = 10 6 . This is a contradiction.8. Proof of Theorems 5By Theorem 4, we restrict to those triplets (n, a, k) given in the statement of Theorem 4 witha 12. We now factorize f a (x) with a 0 a n = ±1, a 1 = a 2 = . . . = a n−1 = 1 to find that these f a (x)are irreducible. Hence the assertion follows.9. Proof of Theorem 1For the proof of Theorem 1, we put α = a through out this section. As remarked in Section1 after the statement of Theorem 1, we may assume that 10 < a 40. For n 18 and n ∈{24, 25, 27, 30, 32, 36, 45, 48, 54, 60, 64, 72, 75, 80, 90, 112, 120}, we find that L (a)n (x) is irreducible exceptfor (n, a) listed in Theorem 1. Thus we assume n > 18, n /∈ {24, 25, 27, 30, 32, 36, 45, 48, 54, 60,64, 72, 75, 80, 90, 112, 120}. Assume that L (α)n (x) is reducible. Then L (α)n (x) has a factor of degree kwith 1 k n 2. First we prove the following lemma.Lemma 9.1. Let k 2. Then L (a)n (x) has no factor of degree k.Proof. Let k 2 and a 40 if k = 2. We may restrict to those (n, k, a) given in the exception ofTheorem 4. For each of these triplets (n, k, a), we first check if there is a prime p k + 2 with (10)such that either (8) or (9) is satisfied and they are excluded by Lemma 1. We are now left withtriples (n, k, a) given by k = 2, (n, a) ∈ {(100, 21), (40, 24), (256, 33), (42, 40)}. For these (n, a), wecheck that L (a)n (x) are irreducible.Let k = 2 and 40 < a 50. Suppose n /∈ N 1 (23) and n+a /∈ N 1 (23). Then P 1 = P (n(n−1)) > 23and P 2 = P ((n + a)(n + a − 1)) > 23. Further either P 1 ∤ (a + 1)(a + 2) or P 2 ∤ (a + 1)(a + 2))and the assertion follows by Lemma 1. Therefore we may assume that either n = N ∈ N 1 (23)or n + a = N ∈ N 1 (23). Further we may also suppose that P (n(n − 1)(n + a)(n + a − 1)) P ((a + 1)(a + 2)) otherwise assertion follows by Lemma 1. For N ∈ N 1 (23) and N > 10000, wecheck that P ((N −a)(N −a−1)) > P ((a+1)(a+2)) and P ((N +a)(N +a−1)) > P ((a+1)(a+2))except when (a, N) = (45, 10648), (46, 12168) where P (N(N − 1)) = 13, 23, respectively. Observethat N(N − 1)|n(n − 1)(n + a)(n + a − 1). By taking p = P (N(N − 1)), the assertion follows fromLemma 1. We now consider n 10000. Let a be given. By Lemma 1, we first restrict to those n forwhich P (n(n − 1)(n + a)(n + a − 1)) P ((a + 1)(a + 2)). Further we check that there is a primep|n(n − 1)(n + a)(n + a − 1), p > 7 and p ∤ (a + 1)(a + 2). Lemma 1 implies the assertion now.By Lemma 9.1, we need to consider k = 1. If there is a prime p|n(n + a), p ∤ (a + 1) with either14