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Squares in products in arithmetic progression with at most one term omitted and common difference a prime power

机译:算术级数乘积中的平方,最多省略一个术语,并且共同的差为素数幂

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For an integer x > 1, we denote by P(x) and ω(x) the greatest prime factor of x and the number of distinct prime divisors of x, respectively. Further, we put P(1) = 1 and ω(1) = 0. Let p_i be the ith prime number. Let k≥ 4, t ≥k — 2 and γ1 < ... < γt be integers with 0 ≤γi < k for 1 ≤ i ≤ t. Thus t ∈{k, k — 1, k — 2}, γt ≥ k — 3 and γi = i — 1 for ≤ i ≤ t if t= k. We put φ = k — t. Let b be a positive squarefree integer and we shall assume, unless otherwise specified, that P(b) ≤ k. We consider he equation 1.1)Δ =Δ(n,d,k) = (n + γ1d) ... (n + γtd) = by~2 in positive integers n, d, k, b, y, t. It has been proved (see [SaSh03] and MuSh04]) that (1.1) with φ = 1, k ≥ 9, d n, P(b) < k and ω(d) = 1 does not hold. Further, it has been shown in [TSH06] that the assertion ontinues to be valid for 6 ≤ k ≤ 8 provided b = 1. We show THEOREM 1. Letφ = 1, k ≥ 7 and d n. Then (1.1) with ω(d) = 1 does of hold.
机译:对于x> 1的整数,我们分别用P(x)和ω(x)表示x的最大素数和x的不同素数除数的数量。此外,我们将P(1)= 1和ω(1)=0。令p_i为第i个质数。令k≥4,t≥k_2,且γ1<... <γt是1≤i≤t的0≤γi

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