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Is there an equation that relates the parity of $\sigma_0$ (the number of divisors) to the parity of $\sigma_1$ (the sum of divisors)?

Added February 21 2017

From S.C.B.'s comment: "$\sigma_{0}(n)$ is odd iff $n$ is a square. $\sigma_{1}(n)$ is odd iff $n$ is a square or two times a square." Therefore, $\sigma_{0}(n)$ and $\sigma_{1}(n)$ are both odd when $n$ is a square.

I would still be interested in receiving answers to the following question:

Earlier Version of Question

Is there an equation that relates $\sigma_0$ (the number of divisors) to $\sigma_1$ (the sum of divisors)?

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    See [The number of divisors of a number shose sum of divisors is a perfect square](http://math.stackexchange.com/questions/1120521/the-number-of-divisors-of-a-number-whose-sum-of-divisors-is-a-perfect-square?rq=1) – 2017-02-20
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    Thanks @amWhy! I am checking that out now. – 2017-02-20
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    $\sigma_{0} (n)$ is odd iff $n$ is a square. $\sigma_{1}( n)$ is odd iff $n$ is square or two times a square. – 2017-02-20
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    @S.C.B., that was exactly what I needed! So I guess $\sigma_{0}(n)$ and $\sigma_{1}(n)$ are both odd when $n$ is a square? – 2017-02-20
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    @JoseArnaldoBebitaDris Indeed, I believe so. – 2017-02-20
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    @S.C.B., I would still be interested in an answer to my *Earlier Version of (the) Question*. – 2017-02-20
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    Obviously no. There are pairs of numbers having the same $\sigma_0$ and different $\sigma_1$, and then there are pairs of numbers having the same $\sigma_1$ and different $\sigma_0$. – 2017-02-22

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