All the statements below are considered over local rings, so by regular, I mean a regular local ring and so on;


It is well-known that every regular ring is Gorenstein and every Gorenstein ring is Cohen-Macaulay. There are some examples to demonstrate that the converse of the above statements do not hold. For example, $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy, yz, xz)$ where $k$ is a field, is Gorenstein but not regular, or $k[[x^3, x^5, x^7]]$ is C.M. but not Gorenstein. 
Now, here is my question:


I want to know where these examples have come from, I mean, have they been created by the existence of some logical translations to the Algebraic combinatorics (like Stanley did), or even algebraic geometry, or they are as they are and they are some kind of lights that have been descended from heaven to their creators by any reason!



Regular, Gorenstein and Cohen-Macaulay 
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 



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> 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 $k[[x]]/(x^2)$). 
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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](https://mathoverflow.net/questions/36396/length-of-i-i2-versus-anni-anni2-in-artinian-rings/41527#41527)), 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 



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> a one dimensional Cohen-Macaulay ring B that is not Gorenstein. 
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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](http://arxiv.org/abs/math/0209199) for a nice reading on Gorenstein rings. 


