<s>[INST] Integer solutions of $\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $ (AusPol 1994) Find all integer solutions of
$$\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $$
Attempt:
I noticed that $a,b,c>0$ or $a,b,c<0$ can't happen. Besides, if one of them is zero, we can find some solutions. Suppose $c=0$, we get
$$(a+b)\frac{ab}{2} + (a+b)^3 = 1 $$
$$\implies (a+b)(ab+ 2(a+b)^2) = 2 $$
So, we have $(a+b) \in \{\pm 1,\pm 2\}$. Just $(a+b) = 1$ has solution, with $a= 1$ and $b= 0$ or $a= 0$ and $b=1$.
So, due to the symmetry of the equation,  just the case $a,b>0$ and $c<0$ is missing. Besides, $\gcd(a,b,c) =1$.
 [/INST] It's $$(a+b+c)(ab+ac+bc)+abc+2(a+b+c)^3=2$$ or
$$(a+b+c)^3+\sum_{cyc}a(a+b+c)^2+\sum_{cyc}ab(a+b+c)+abc=2$$ or
$$\prod_{cyc}(a+b+c+a)=2$$ or $$(2a+b+c)(2b+a+c)(2c+a+b)=2.$$
Can you end it now?
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