# Evaluating a series with the Möbius function and greatest common divisor.

Problem: Let $\gcd(a,b,c,d)$ refer to the largest integer $r$ such that $r$ divides each of $a,b,c,d$. Evaluate the series $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty}\sum_{d=1}^{\infty}\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^{2}b^{2}c^{2}d^{2}}\gcd(a,b,c,d)^{4},$$ where $\mu(n)$ is the Möbius function.

I tried several tricks, but I eventually got stuck. I think it should be possible to rewrite the entire thing as an Euler Product. It looks very similar to the double series $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{\mu(a)\mu(b)}{a^{2}b^{2}}\gcd(a,b)^{2}=\frac{6}{\pi^2}.$$

Any help is appreciated.

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Can you rewrite the double sum as an Euler product? – Gerry Myerson Jun 26 '12 at 5:54
@GerryMyerson Well, the double sum equals $\frac{6}{\pi^2}=\frac{1}{\zeta(2)}$ which is just $\prod_p \left(1-\frac{1}{p^2}\right)$. – Eric Naslund Jul 3 '12 at 10:58
That's cheating. You can write pretty much any number as an Euler product. If you didn't know the double sum came to $6/\pi^2$, could you write it as an Euler product? – Gerry Myerson Jul 3 '12 at 12:51
@GerryMyerson: That is how you get $\frac{6}{\pi^2}$ in the first place. In general, if you have infinite series with multiplicative functions, and it evaluates to something with the zeta function, it is almost always by using Euler products. – Eric Naslund Jul 3 '12 at 13:52
OK, so what you're saying is that you can twiddle the double sum into an Euler product, but the same techniques don't work for the 4-fold sum. Are you able to get a good numerical estimate of the 4-fold sum and then check to see if it seems to be a simple combination of some Euler products? – Gerry Myerson Jul 4 '12 at 0:20

Your sum can be re-written in terms of an Euler product:

$$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})\approx .16544\cdots$$

A proof can be given as follows:

First note that,

$$d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4\iff d\mid \gcd(x_1,x_2,x_3,x_4)$$

So we get:$$\sum_{d=1}^\infty f(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid \gcd(x_1,x_2,x_3,x_4)}$$ $$=\sum_{d\mid \gcd(x_1,x_2,x_3,x_4)}f(d)$$

Where $1_{A}=[A]$ in Iverson bracket notation

Then define:

$$\phi_s(n)=n^s\prod_{p\mid n}(1-\frac{1}{p^s})$$

So that we have: $(\phi_s*1)(n)=n^s$

Now set $f=\phi_4$ in the aforementioned equality and we get that:

$$\sum_{d=1}^\infty \phi_4(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\gcd(x_1,x_2,x_3,x_4)^4$$

Now we note that:

$$\sum_{x_i=1}^\infty\frac{\mu(x_i)}{x_i^2}1_{d\mid x_i}=\sum_{n=1}^\infty\frac{\mu(dn)}{(dn)^2}=\frac{6}{\pi^2}\frac{\mu(d)}{\phi_2(d)}$$

Then multiplying both sides of the previous series by $\frac{\mu(x_1)}{x_1^2}\frac{\mu(x_2)}{x_2^2}\frac{\mu(x_3)}{x_3^2}\frac{\mu(x_4)}{x_4^2}$ and rearranging gives:

$$\sum_{d=1}^\infty \phi_4(d)(\frac{\mu(x_1)}{x_1^2}1_{d\mid x_1})(\frac{\mu(x_2)}{x_2^2}1_{d\mid x_2})(\frac{\mu(x_3)}{x_3^2}1_{d\mid x_3})(\frac{\mu(x_4)}{x_4^2}1_{d\mid x_4})$$ $$=\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$$

Thus:

$$\sum_{x_1=1}^\infty\sum_{x_2=1}^\infty\sum_{x_3=1}^\infty\sum_{x_4=1}^\infty\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$$

$$=(\frac{6}{\pi^2})^4\sum_{d=1}^\infty \phi_4(d)\frac{\mu(d)^4}{\phi_2(d)^4}=\frac{1296}{\pi^8}\sum_{n=1}^\infty\frac{\phi_4(n)}{\phi_2(n)^4}|\mu(n)|=\frac{1296}{\pi^8}\prod_{p}(1+\frac{\phi_4(p)}{\phi_2(p)^4})$$

So we have:

$$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})$$

A similar argument will give your second sum as $\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{a^2b^2}\gcd(a,b)^2=\frac{6}{\pi^2}$.

In addition to formula for other such similar generalizations.

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As an update to readers on the status of this answer: (a) Ethan has informed me that the strange bs are in fact ks, and (b) I have pointed out an issue in going from the last line of the first image to the first line of the second image: $(a,b,c,d,k)=1$ does not entail $$\mu(ak)\mu(bk)\mu(ck)\mu(dk)=\mu(a)\mu(b)\mu(c)\mu(d)\mu(k)^4.$$ (Counterexample: $a=2,b=3,c=5,d=7,k=2$.) It is possible a symmetry argument however can show that these counterexamples' contributions to the sum cancel each other out. – anon May 16 '13 at 6:26
@anon Fixed, this is an entirely new solution. – Ethan Apr 5 '14 at 19:03