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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Lemma 2.2, pp. 5--7.

Statement. Let $k\geq2$ and $m\geq2k-1$ be integers, and let $N,d$ be positive integers. Suppose

$$N,N+d,\ldots,N+(m-1)d$$

are $k$-full. Choose nonnegative integers $a_{i,j}$ so that

$$N+jd=\prod_{i=k}^{2k-1}a_{i,j}^{\,i} \qquad(0\leq j<m),$$

and put $t=\gcd(N,d)$. Then

$$\operatorname{Rad}\!\left( \prod_{j=0}^{m-1}\frac{N+jd}{t} \right) \leq C_m \frac{\displaystyle\prod_{j=0}^{m-1}\prod_{i=k}^{2k-1}a_{i,j}} {t^{m/(2k-1)}}, \qquad C_m=\prod_{p\leq m}p.$$

The factorization exists because every integer at least $k$ is a nonnegative integral combination of $k,k+1,\ldots,2k-1$.

Proof. Primes at most $m$ are absorbed by $C_m$. Fix a prime $p>m$ which occurs in the radical on the left. Thus

$$\nu_p\!\left( \prod_{j=0}^{m-1}\prod_{i=k}^{2k-1}a_{i,j}^{\,i} \right)-m\nu_p(t)\geq1. \tag{1}$$

If $2k-1$ divides $m$, the quantity

$$Q=\frac{\prod_{j,i}a_{i,j}}{t^{m/(2k-1)}}$$

is rational and $Q^{2k-1}$ is an integer: for every prime, the exponent in $\prod_{j,i}a_{i,j}^{2k-1}$ is at least its exponent in $\prod_j(N+jd)$, which is at least $m\nu_p(t)$. Also (1) and $i\leq2k-1$ give

$$(2k-1)\nu_p(Q) =\nu_p\!\left(\prod_{j,i}a_{i,j}^{2k-1}\right)-m\nu_p(t) \geq1.$$

Since a rational number whose $(2k-1)$st power is integral is integral, $\nu_p(Q)>0$ implies $\nu_p(Q)\geq1$. Hence $p$ occurs on the right.

Suppose instead that $2k-1\nmid m$. Since $m\geq2k-1$, now $m\geq2k$. The integer

$$M=\frac{\prod_{j,i}a_{i,j}^{\,2k-1}}{t^m}$$

is an integer by the same exponent comparison. Equation (1) shows $\nu_p(M)\geq1$, and $M$ would supply the required factor $p$ if $\nu_p(M)\geq2k-1$. Assume for a contradiction that $1\leq\nu_p(M)\leq2k-2$. The value $\nu_p(t)$ cannot be divisible by $2k-1$, since then so would $\nu_p(M)$. Write

$$\nu_p(t)=(2k-1)q+r, \qquad 1\leq r\leq2k-2.$$

Comparing the weights $i$ and $2k-1$ in the definitions gives

$$\nu_p(M)= \nu_p\!\left(\prod_{j,i}a_{i,j}^{\,i}\right)-m\nu_p(t) +\sum_{i=k}^{2k-2}\sum_{j=0}^{m-1} (2k-1-i)\nu_p(a_{i,j}).$$

The first difference is at least $1$ by (1), so the assumed upper bound on $\nu_p(M)$ gives

$$\sum_{i=k}^{2k-2}\sum_{j=0}^{m-1} (2k-1-i)\nu_p(a_{i,j})\leq2k-3. \tag{2}$$

Consequently at most $2k-3$ indices $j$ have any $p\mid a_{i,j}$ with $i<2k-1$. Since $m\geq2k$, at least three indices $j$ satisfy

$$\nu_p(N+jd)\equiv0\pmod{2k-1}.$$

For each of these indices, $t\mid N+jd$ and the residue of $\nu_p(t)$ modulo $2k-1$ is nonzero. Hence $\nu_p(N+jd)>\nu_p(t)$. This is impossible. Indeed, if $\nu_p(d)>\nu_p(N)$ it never occurs. If $\nu_p(d)<\nu_p(N)$ it forces $p\mid j$, so $p>m$ leaves only $j=0$, not three indices. If the two valuations are equal, two such indices $j_1<j_2$ imply $p\mid j_2-j_1$, although $0<j_2-j_1<m<p$. This contradiction proves $\nu_p(M)\geq2k-1$ and hence the claimed radical bound.

Used by. Theorem 1.1.

Bears on. #937.