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, $\gcd(a,b,c) =1$.
 A: 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?
A: Rewriting, we have:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=2$$
The case of $c=0$ was solved using the fact that the LHS was divisible by $a+b$. Then, we could say that it was divisible by $(a+b+kc)$ since $kc=c=0$. Since the equation is symmetric, we can guess:
$$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=m(a+b+kc)(a+kb+c)(ka+b+c)$$
which turns out to be true for $m=1$ and $k=2$. Now, we have:
$$(a+b+2c)(a+2b+c)(2a+b+c)=2$$
If WLOG $a+b+2c=a+2b+c$ (assuming two of the factors are equal), then we have $b=c$ which gives:
$$(a+3b)^2(2a+2b)=2 \implies (a+3b)^2(a+b)=1$$
We then have $a+b=1$ and $a+3b= \pm 1$ which gives $(a,b)=(1,0),(2,-1)$.
If none of the three factors were equal, it is easy to see that the factors have to be $-2,-1,1$. WLOG, set:
$$a+b+2c=-2,a+2b+c=-1,2a+b+c=1 \implies 4(a+b+c)=-2$$
which is clearly impossible. Thus, the set of all solutions are:
$$(a,b,c)=(1,0,0),(0,1,0),(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)$$
