What do we mean by a variety being arithmetically Cohen-Macaulay? Is every such variety also Gorenstein?
[Math] Arithmetically Cohen-Macaulay varieties
ac.commutative-algebraag.algebraic-geometry
Related Solutions
An example was given in Section 3 of this paper by Cutkosky: "A new characterization of rational surface singularities" (The scheme $Z$ in the last page, which is a blow up of some $m$-primary ideal of a regular local ring of dimension $3$, is normal but not Cohen-Macaulay).
The algebraic side of this example has been studied quite a bit, so perhaps more explicit examples are known. I am not an expert here, but you can check out a paper by Huckaba-Huneke here, or papers by Vasconcelos (he has a book called "Arithmetic of Blow-up algebras" which discussed, among other things, Serre's condition $(S_n)$ on Rees algebras), and the references there.
I will argue that the examples you gave are "simplest" in some strong sense, so although they look unnatural, if Martians study commutative algebra they will have to come up with them at some point.
Let's look at the first one $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy,yz,zx)$. Suppose you want
a $0$-dimensional Gorenstein ring which is not a complete intersection (complete intersections are the cheapest way to get Gorenstein but non-regular, for example $k[[x]]/(x^2)$).
Then it would look like $A=R/I$, where $R$ is regular and $I$ is of height equals to $\dim R$. If $\dim R=1$ or $2$, then $I$ would have to be a complete intersection, no good (now, the poor Martian may not know this at the begining, but after trying for so long she will have to give up and move on to higher dimension, or prove that result for herself). Thus $\dim R=3$ at least, and we may assume $R=k[[x,y,z]]$ (let's suppose everything are complete and contains a fied).
Since $I$ is not a complete intersection, it must have more than $3$ minimal generators. If it has $4$ then it would be an almost complete intersection, and by an amazing result by Kunz (see this answer), those are never Gorenstein.
So, in summary, a simplest $0$-dimensional Gorenstein but not complete intersection would have to be $k[[x,y,z]]/I$, where $I$ is generated by at least $5$ generators. At this point our Martian would just play with the simplest non-degenerate generators: quadrics, and got lucky!
(You can look at this from other point of view, Macaulay inverse system or Pfaffians of alternating matrices, see Bruns-Herzog, but because of the above reasons all the simplest examples would be more or less the same, up to some linear change of variables)
On to your second example, $k[[t^3,t^5,t^7]]$. Now very reasonably, our Martian wants
a one dimensional Cohen-Macaulay ring B that is not Gorenstein.
Since $\dim B=1$, being a domain would automatically make it Cohen-Macaulay. The most natural way to make one dimensional domains is to use monomial curves, so $B=k[[t^{a_1},...,t^{a_n}]]$. But if $n=2$ or $n=3$ and $a_1=2$ you will run into complete intersections, so $a_1$ would have to be $3$ and $n=3$ at least, and on our Martian goes...
(Again, you can arrive at this example by looking at things like non-symmetric numerical semi-groups, but again you would end up at the same simplest thing).
Reference: I would recommend this survey for a nice reading on Gorenstein rings.
Best Answer
Arithmetically Cohen-Macaulay means, depending on the source/context, either:
Of course if you are projectively normal in $\mathbb{P}^n$ and the ample line bundle is a very ample line bundle of that embedding, these two definitions coincide.
It doesn't imply anything about Gorenstein-ness. In fact, any Cohen-Macaulay projective variety with $H^i(X, \mathcal{O}_X) = 0$ for $0 < i < \dim X$ is arithmetically Cohen-Macaulay with respect to some embedding into projective space.
To see this, take a sufficiently ample line bundle $L$ such that $H^i(X, \omega_X \otimes L^n) = 0$ and $H^i(X, L^n) = 0$ for all $n \geq 1$ and all $0 < i < \dim X$. In the previous version of this answer, I forgot the Cohen-Macaulay hypothesis on $X$, in which case the first vanishing can't be forced to hold.
If I recall correctly, these notions appear prominently in the study of Linkage (see Eisenbud's book for an introduction).
A related notion is that of arithmetic Macaulayfication of a ring. This means that there exists an ideal $I$ such that the Rees algebra of $I$ (the ring you blow-up to get the blow-up of $I$) is Cohen-Macaulay. These were shown to exist in the last decade by Kawasaki. If I recall correctly, a corollary of this result is that every ring with a dualizing complex is a quotient of a Gorenstein ring (this was previously a conjecture of Sharp). Someone correct me if I'm wrong on this.
EDIT: Added the CM hypothesis on the variety and added an explanation (thanks to Long). EDIT2: Added the two possible definitions (section ring vs coordinate ring). Thanks to J. C. Ottem.