<s>[INST] 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.
 [/INST] 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 $\color{red}{k_3 =k_4 = 6}$.
</s>