Consider the following form of the Riemann Zeta function:
$s\in\mathbb{C}$, such that: $\left |s \right | > 1$
$\zeta \left ( s \right )= 1+2^{-s}+3^{-s}+4^{-s}+5^{-s}...$
Now, due to the unique prime factorisation of the integers, this can be expressed as the following:
$\zeta \left ( s \right )= 1+2^{-s}\zeta \left ( s \right )+3^{-s}\zeta \left ( s \right )+5^{-s}\zeta \left ( s \right )+7^{-s}\zeta \left ( s \right )+...+p^{-s}\zeta \left ( s \right )+...-D(s)$
With $p$ prime.
$-D(s)$ removes all of the double counted terms (e.g $6^{-s}=3^{-s}2^{-s}$ , hence $6^{-s}$ will appear both in $2^{-s}\zeta \left ( s \right )$ and $3^{-s}\zeta \left ( s \right )$).
Hence $D(s)$ has the form:
$D(s)=\sum_{p}^{ }\sum_{q> p}^{ }LCM(p,q)^{-s}$
Where $p$ and $q$ are prime, and $LCM(p,q)$ is the lowest common multiple of $p$ and $q$.
So the Zeta function becomes:
$\zeta (s)=1+\zeta (s)\sum_{p}^{ }p^{-s}-\sum_{p}^{ }\sum_{q> p}^{ }LCM(p,q)^{-s}$
Now to my question: Is the above expression correct or have I made a mistake? As far as I can tell I have just algebraically manipulated the equation, but I've possibly produced something totally incorrect.
My understanding of pure math is very limited, so please let me know if what I have done is wrong.
Thankyou very much for reading.
Unfortunately your expression is not complete, there are some terms missing. To analyse the difference, we give your series a name: $$Z(s) = 1+2^{-s}\zeta \left ( s \right )+3^{-s}\zeta \left ( s \right )+5^{-s}\zeta \left ( s \right )+7^{-s}\zeta \left ( s \right )+\ldots = 1 + \zeta(s)\sum_p p^{-s},\quad\quad(*)$$ We will denote the difference with $\zeta(s)$ by $E(s) = \zeta(s) - Z(s)$. The question now is whether $E(s)$ has indeed the form $E(s) \overset{?}= D(s)$, where $D(s)$ is as in your post $$D(s) = \sum_{p}\sum_{q > p} (pq)^{-s} = 6^{-s} + 10^{-s} + 14^{-s} + 15^{-s} + 21^{-s} + \ldots .$$ (notice that for distinct prime numbers $p, q$ we have $LCM(p,q) = pq$.)
When we look at the number of times the term $30^{-s}$ shows up in the expanded version of $Z(s)$, we see that it appears three times: once in the term $2^{-s}\zeta(s)$ in the form $2^{-s}\cdot 15^{-s}$, once in the term $3^{-s}\zeta(s)$ in the form $3^{-s}\cdot 10^{-s}$, and once in the term $5^{-s}\zeta(s)$ in the form $5^{-s}\cdot 6^{-s}$. That is two more terms $30^{-s}$ than the $\zeta$-function has, so the correction $E(s)$ contains a term $2\cdot 30^{-s}$. Likewise, counting terms $12^{-s}$ shows that it appears two times in equation $(*)$, one time coming from $2^{-s}\zeta(s)$ and one time coming from $3^{-s}\zeta(s)$. Therefore, $E(s)$ has a term $12^{-s}$. Both of these terms are missing in $D(s)$, so we must conclude that $E(s) \neq D(s)$, or in other words, $$\zeta (s)\neq1+\zeta (s)\sum_{p}^{ }p^{-s}-\sum_{p}^{ }\sum_{q> p}^{ }(pq)^{-s}.$$
Is there a way to fix this? Yes, there is, we just have to repair the expression for $D(s)$. To do that, we must first analyse your series $Z(s)$ a bit more. After all, if we want $D(s)$ to correct for all `extra' terms in $Z(s)$, we need to know which terms occur in $Z(s)$, and with which multiplicity. To do that, we will analyse the terms occuring in $Z(s)$ in exactly the way we did before with $30^{-s}$ and $12^{-s}$, but this time we are going to do it for general $n$.
Let $n$ be any natural number. We are going to find out how often $n^{-s}$ appears in $Z(s)$. For this, we first factorize $n$ as $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$, where the $p_i$ are distinct primes. Then we get a term $n^{-s}$ in $Z(s)$ from the term $p_1^{-s}\zeta(s)$, in the form $p_1^{-s}\cdot (p_1^{e_1-1}p_2^{e_2}\cdots p_r^{e_r})^{-s}$, another term $n^{-s}$ from the term $p_2^{-s}\zeta(s)$, in the from $p_2^{-s} \cdot (p_1^{e_1}p_2^{e_2-1}\cdots p_r^{e_r})^{-s}$, and so on until the term $n^{-s}$ coming from $p_r^{-s}\zeta(s)$ in the form $p_r^{-s} \cdot (p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r-1})^{-s}$. Thus, the total number of times $n^{-s}$ appears in $Z(s)$ is exactly $r$, the number of distict prime divisors of $n$. (Make sure you understand this part. This is the difficult part of the story, but I believe it will become clear once you try the argument for a few numbers like $n = 600 = 2^3\cdot 3\cdot 5^2$). Note that this matches what we found in the specific cases $n = 12$ and $n = 30$ we looked at before: the term $12^{-s}$ occurs 2 times and $30^{-s}$ occurs 3 times, and indeed, $12$ has 2 different prime divisors and $30$ has 3 different prime divisors. There is only one exception, and that is the case $n = 1$. This number has zero prime divisors, but the term $1^s = 1$ occurs one time. This is just because of how we defined $Z(s)$, so that is nothing to worry about, but it is good to keep in mind anyways.
So now we now which terms we need to subtract from $Z(s)$ to get to $\zeta(s)$. Namely, we need to subtract every term of the form $(p_1^{e_1}p_2^{e_2})^{-s}$ once (these are the doubly counted terms that you wrote about, but note that we now include product of distinct prime powers, not just distinct primes), every term of the form $(p_1^{e_1}p_2^{e_2}p_3^{e_3})^{-s}$ two times, every term of the form $(p_1^{e_1}p_2^{e_2}p_3^{e_3}p_4^{e_4})^{-s}$ three times, and so on. We can put that into formulas in the same way you did in your question, and then we will end up with the following formula: $$\zeta(s) = Z(s) - \sum_{p_1}\sum_{p_2>p_1}\sum_{e_1=1}^\infty\sum_{e_2=1}^\infty (p_1^{e_1}p_2^{e_2})^{-s} - \sum_{p_1}\sum_{p_2>p_1}\sum_{p_3>p_2}\sum_{e_1=1}^\infty\sum_{e_2=1}^\infty\sum_{e_3=1}^\infty 2(p_1^{e_1}p_2^{e_2}p_3^{e_3})^{-s} - \ldots$$ Well, that doesn't look very nice. Perhaps we should think this over a bit. How are we going to clean up this formula? Well, the first iterated summation is over all products of two distinct prime powers. Which numbers are products of two distinct prime powers? Well, those are exactly all numbers with two different prime divisors. Likewise, the second iterated summation is over all numbers with exactly three different prime divisors. Thus, we might define the sets $$ A_k = \{n : \textrm{ $n$ has exactly $k$ distinct prime divisors } \}.$$ for each $k$. So $A_2$ consists of those numbers with exactly 2 distinct prime numbers, etc. Then we can rewrite the horrible formula above to this form: $$\zeta(s) = Z(s) - \sum_{n \in A_2} n^{-s} - \sum_{n \in A_3} 2n^{-s} - \sum_{n \in A_4} 3n^{-s} - \ldots = Z(s) - \sum_{k = 2}^\infty \sum_{n \in A_k}(k-1)n^{-s},$$ which at least looks better. But we still have a double summation, perhaps we can get rid of that as well? Indeed we can. For this, we first note that the outer sum, which now start at 2, might just as well start at 1, since al those terms will be 0 anyway. Also, lets define $\omega(n)$ to be the number of distinct prime divisors $n$ has, so that $\omega(n) = k$ if and only if $n \in A_k$. Then we can rewrite the previous formula to $$\zeta(s) = \sum_{k = 1}^\infty \sum_{n \in A_k} (\omega(n)-1) n^{-s}.$$ We see that the individual terms in the double summation no longer directly depend on $k$. Thus, the sum only depends on which values of $n$ occur. Well, we first sum over all numbers with exactly one prime divisor, then over all numbers with exactly two prime divisors, then all numbers with exactly three prime divisors, and so one. Thus, in the end, the summation is just over all numbers except $1$. So we might as well write $$\zeta(s) = Z(s) - \sum_{n = 2}^\infty (\omega(n) - 1)n^{-s}.$$ Ah, but that looks good. We now have a single summation left, and moreover, when we split off the $-1$-part in each term, we almost get a $\zeta$-function (but be careful, the summation runs from $n = 2$, not $n=1$). Rewriting using this idea in our mind, we end up with \begin{align}\zeta(s) &= Z(s) - \sum_{n = 2}^\infty \omega(n)n^{-s} + \sum_{n=2}^\infty n^{-s} \\&= Z(s) - \sum_{n = 2}^\infty \omega(n)n^{-s} + \sum_{n=1}^\infty n^{-s} \,- 1 \\&= Z(s) + \zeta(s) - 1 - \sum_{n = 2}^\infty \omega(n)n^{-s}.\end{align} But now we have $\zeta(s)$ on both sides of the equation, so it cancels, and after that we are left with the equation (bringing the summation to the other size) $$Z(s) - 1 = \sum_{n = 2}^\infty \omega(n)n^{-s}.$$
Note that this last equation expresses precisely that for any $n > 1$ the term $n^{-s}$ will show up in $Z(s)$ with multiplicity exactly $\omega(n)$, the number of prime divisors of $n$. This is exactly what we already had discovered a few paragraphs back. Apparently we have come back to were we started. Well, at least it suggests that our computations and reasoning were correct.
On a related note, the idea to use the unique factorisation property of the integers to rewrite $\zeta(s)$ is a good one, and it also occured to Euler. However, he did it in a slightly different way, leading to a way to express $\zeta(s)$ as an infinite product instead of an infinite sum. For details of the derivation, I urge you to read this wiki page.