Quartic equation with integer solutions I am trying to find the solutions of the following quartic equation given that all the solutions are integers.
$$x^4+22x^3+172x^2+552x+576=0$$
Below is the original phrasing of the problem with hints building up to this equation. I can prove all of the results it asks for before the final part of the question, but I am struggling to actually find the solutions to the equation and require help with explaining the process as well.



I then know that:
(1) $k_1k_2k_3k_4 = 576 = (1)(2^6)(3^2)$
(2) $(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323=(1)(3^3)(7^2)$
(3) $(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = (1)(5^2)(7)$
However, I do not know how to proceed from here and the solution to this problem didn't explain in enough detail for me to either understand the solution, or the approach.

Observe that the given equations imply that $k_1,k_2,k_3$ and $k_4$ should be non-zero even integers. Now consider $$(k_1-1)(k_2-1)(k_3-1)(k_4-1)=1\times 5^2 \times 7.$$
Since, $k_i$'s are integers, one of the $k_i-1$ must be $1$ ($k_i-1 \neq -1$ as $k_i \neq 0$). Let $k_1-1=1 \implies \color{red}{k_1 =2}$. 
We are now left with $$(k_2-1)(k_3-1)(k_4-1)= 5^2 \times 7.$$
Since we have $5^2$ on the RHS, at least one of the $k_i-1$ (say $k_2-1$) cannot be a multiple of $5$. Consequently, $k_2-1 = 1$ or $k_2-1=\pm 7$ (i.e., $k_2 =2,8$ or $-6$). But $k_2 + 1$ must be a multiple of $3$ or $7$ (due to the second constraint). So $k_2 \neq -6$. Therefore, $k_2=2$ or $k_2=8$.
Now suppose $k_2=2$. Then the constraints simplify to
$$k_3 k_4 = 2^4 \times 3^2$$
$$(k_3+1)(k_4+1) = 3 \times 7^2$$
$$(k_3-1)(k_4-1) = 5^2 \times 7$$
It can be verified that these equations do not have any solution. 
Therefore, we are left with $\color{red}{k_2 = 8}$ only. In that case, the equations simplify to
$$k_3 k_4 = 2^2 \times 3^2$$
$$(k_3+1)(k_4+1) = 7^2$$
$$(k_3-1)(k_4-1) = 5^2$$
These equations can easily be solved to result in $\color{red}{k_3 =k_4 = 6}$.
