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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Lemma 2.1, pp. 4--5.

Statement. Let $k\geq2$, let $n$ be $k$-full, and let $t$ be a $k$-full divisor of $n$. Then

$$\operatorname{Rad}(n/t)\leq \frac{n^{1/k}}{t^{1/k^2}}.$$

Here $\operatorname{Rad}(u)$ is the product of the distinct primes dividing $u$, with $\operatorname{Rad}(1)=1$.

Proof. It is enough to compare the exponent of each prime $p$. Put $e=\nu_p(n)$ and $f=\nu_p(t)$. If $e=k$, the $k$-fullness of $t$ forces $f=0$ or $f=k$. In the first case $p\mid n/t$ and

$$1\leq e/k-f/k^2=1;$$

in the second case $p\nmid n/t$ and $e/k-f/k^2=1-1/k\geq0$.

Now suppose $e\geq k+1$. If $f=0$, then $1\leq e/k-f/k^2$. If $k\leq f\leq e-1$, then $p\mid n/t$ and

$$\frac ek-\frac f{k^2} \geq \frac ek-\frac{e-1}{k^2} =\frac{(k-1)e+1}{k^2}\geq1.$$

Finally, if $f=e$, then $p\nmid n/t$ and $e/k-f/k^2=e(k-1)/k^2\geq0$. Thus the exponent on the right is at least $1$ whenever $p$ occurs in the radical on the left and is nonnegative for every other prime. Multiplication over $p$ proves the claim.

Used by. Theorem 1.1.

Bears on. #937.