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.
Best Answer
This is an attempt to show how the method suggested in the question can be progressed.
Look at $1323$ with factors $3^3\times 7^2$. The size of $7$ will restrict the possibilities to investigate more quickly than the smaller primes. We have $576=24^2$
The available multiples of $7$ are $7, 21, 49, 63, 147$ (others are too large) giving possible factors of $576$ which differ by $1$ (can't tell the sign at this stage).
So the possible factors of $576$ are $6 , 8; 20, 22; 48, 50; 62, 64; 146, 148$ and the only ones which are actually factors are $6, 8, 48$
And the possible factors of $175$ differ by $2$ (in the same direction as the factors of $576$) so can be $5, 9; 47$ and this time only $5$ is possible, with two factors to match the two $7$s.
This means the factors we have for $576$ have numerical value $6$, and in fact we can tell that the roots are $-6$ by attention to signs [NB]. Then there are lots of ways to finish - having two solutions, we can, for example, identify and solve the quadratic for the remaining roots. Or alternatively the remaining factor $7$ from $175$ gives $-8$ and the final factor has to be $1$ which gives $-2$ (signs to be allocated with care, but since coefficients are all positive, roots must all be negative see NB below).
NB The $k_i$ are the negatives of the roots so the $k_i$ here would be $6$ with the factors $(x+6)^2$ and the double root $x=-6$. And then $8$ and $2$ similarly.