Don't forget to count the negatives!

Number Theory Level 3

If $N$ is an integer such that it is divisible by precisely 4 distinct integers.

Then, $N^2$ is divisible by precisely $\text{\_\_\_\_\_\_\_}$ distinct integers.

10 8 12 6

1 solution

Brian Charlesworth
Feb 25, 2017

Any integer $N$ such that $|N| \ne 1$ will be divisible, at the very least, by 4 distinct integers, namely $-N, -1, 1$ and $N$ . If these are the only divisors, then as the only distinct positive divisors of $|N|$ are $1$ and $|N|$ we can conclude that $|N|$ is prime, in which case $N^{2}$ will only be divisible by $-N^{2}, -N, -1, 1, N$ and $N^{2}$ , a total of $\boxed{6}$ distinct integer divisors.

Don't forget to count the negatives!

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