<s>[INST] Proof that no non-trivial integer solution exists? The equation system
$$2ab+2ac+2bc=4a+4b+4c$$
$$2ab+2ac+2bc=abc$$
In short $$2ab+2ac+2bc=4a+4b+4c=abc$$
has, according to Wolfram Alpha, only the trivial integer solution $a=b=c=0$. I would like to prove this. My approach (not sure whether it is helpful) is using the equation $$(a-2)(b-2)(c-2)=abc-2ab-2ac-2bc+4a+4b+4c-8$$ which simplifies to $$(a-2)(b-2)(c-2)=4(a+b+c-2)$$ if we have $2ab+2ac+2bc=abc$, but I do not see how can I make progress.
Motivation : The circumference and the area of a rectangular with sides $a$ and $b$, if we do not consider units has the same value than the area , when $$2a+2b=ab$$ holds. The only solutions in positive integers are $(3,6)$ $(4,4)$ , $(6,3)$, so besides the square with length $4$ only the rectangular with sides $3$ and $6$ has the property that area and length are equal in the described sense. 
I wanted to extend this to a cuboid with lengths $a,b,c$ , for which the sum of the lengths of the edges ($4a+4b+4c$) is equal to the surface ($2ab+2ac+2bc$) and the volume ($abc$). 
The desired proof would show that no integer cuboid with this proprty exists.
 [/INST] One has $$(ab+bc+ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 + c^2a^2 + 2(2ab+2bc+2ca)(\frac{1}{2}(ab+bc+ca))$$
Then $$a^2b^2 + b^2c^2 + c^2a^2 = -(ab+bc+ca)^2$$
So, one gets $ab=bc=ca=ab+bc+ca=0$,or $a=b=c=0$.
EDIT: i make it clear at the last step as @peter's comment:
One gets from $ab=bc=ca=0$ that $a=b=0$, but note that $c=a+b+c=\frac{1}{2}(ab+bc+ca)=0$.
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