Odd and Even Factors of a Perfect Square

Question

If N is a perfect square, then we know that N has an odd number of distinct factors (because the square root gets counted twice).

However, can we prove that if N is a perfect square, then it will always have an odd number of odd factors and even number of even factors? If N is an odd perfect square, then each factor must be odd and therefore N has an odd number of odd factors and no even factors. However, I can't prove the same for even squares, since factors could be either odd or even, as long as one of the numbers in each pair is even..

The number of odd factors for a perfect square of the form $a^2\cdot 2^{2k}$ is the same as the number of odd factors of $a^2$. — Michael Burr, Nov 29 '16 at 09:11

score 0 · Accepted Answer · answered Nov 29 '16 at 09:13

You can prove your claim as follows: if $n$ is a square, for some prime numbers $p_1,\dots,p_m$ and some naturals $a_1,\dots,a_m$, you can write $$n=\Pi_{i=1}^m p_i^{2a_i}.$$ Since every divisor of $n$ has, as prime factors, a subset of $\{p_1,\dots,p_m\}$ with exponent no larger than $a_i$, $n$ has exactly $\Pi_{i=1}^m(2a_i+1)$ divisors. Now, let us suppose that $n$ is even, so (say) $p_1=2$. Then, to count odd divisors of $n$, you can simply ignore the factor $2$, which means that $n$ has $\Pi_{i=2}^m(2a_i+1)$ odd divisors, and this is an odd number (it is product of odd numbers).

Great, thanks very much. – dontchaworrychile Nov 29 '16 at 09:44 — dontchaworrychile, Nov 29 '16 at 09:44

Odd and Even Factors of a Perfect Square

1 Answers1