Factor Me!

Someone has disputed that the problem is wrong, this a quick proof : Note that

$(a^3+b)(a+b^3)-(a+b)^4= a b (1 - 2 a - 2 b + a b) (1 + 2 a + 2 b + a b).$

$ab\neq 0$ , If $ab+1=2(a+b)$ you'll find $(a-2)(b-2)=3$ which gives $(\pm 1,\mp 1);(5,3);(3,5)$ .

If $1+2a+2b+ab=0$ , then $(a+2)(b+2)=3$ which gives the negative of the above solutions : $(\pm 1,\mp 1);(-5,-3);(-3,-5)$ .

Now, count and avoid repetition to get that the number of solutions is $\boxed{6}$ .