A product of divisors lemma

Number Theory Level 3

Let $P_n$ denote the product of all the positive divisors of the positive integer $n$ . For example, all the positive divisors of 6 are $1,2,3,6$ , so $P_6 = 1\times2\times3\times6=36$ .

Below shows a list of values of $P_n$ for even $n\geq4$ .

$n$	$P_n$
$4$	$2^3$
$6$	$6^2$
$8$	$4^3$
$10$	$10^2$
$12$	$12^3$

Based on the pattern above, is the following assertion true?

For all positive integer $m$ , $P_{4m}$ is a perfect cube, and $P_{4m+2}$ a perfect square?

Yes No

1 solution

David Vreken
Nov 10, 2018

The positive divisors of $16$ are $1$ , $2$ , $4$ , $8$ , and $16$ , so $P_{16} = 1 \times 2 \times 4 \times 8 \times 16 = 1024$ , which is not a perfect cube as the assertion predicts, so the assertion is false .

Any proof?

Mr. India - 2 years, 3 months ago

My counterexample is sufficient to show that the assertion is false.

David Vreken - 2 years, 3 months ago

A product of divisors lemma

1 solution

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