Quartic equation with integer solutions

Question

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.

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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.

Answer 1

For $$f(x) = x^4+22x^3+172x^2+552x+576,$$ notice that if $x\geq0$ then $f(x) \geq 576 > 0,$ so all roots of the equation $f(x) = 0$ are negative. Since the roots are $-k_1, -k_2, -k_3, -k_4,$ this tells us that $k_1, k_2, k_3, k_4$ are all positive.

That leaves only a few possible values of $k_i$ that could occur in the equation $$(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = 5^2 \cdot 7.$$

The factors of $175$ are $1, 5, 5^2 = 35, 7, 5\cdot7 = 35,$ and $5^2 \cdot 7 = 175.$ (I chose this equation to start with because $175$ has fewer factors to consider than either $576$ or $1323$.) Since each factor $k_i - 1$ must be a factor of $175,$ the possible values of any $k_i$ can only be among the numbers $2, 6, 26, 8, 36,$ and $176.$

From $$k_1k_2k_3k_4 = 576 = 2^6 \cdot 3^2, \tag1$$ we know any of the $k_i$ can have only $2$ or $3$ as prime factors. So $k_i \neq 26 = 2 \cdot 13$ and $k_i \neq 176 = 16 \cdot 11.$

From $$(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323 = 3^3 \cdot 7^2,$$ we know $k_i + 1$ must be divisible by $3$ or by $7.$ So $k_i + 1 \neq 37,$ and therefore $k_i \neq 36.$

Therefore the only possible values of any of the $k_i$ are $2,$ $6,$ or $8.$ But Equation $(1)$ implies either that one of the $k_i$ is divisible by $9$ (which we now know cannot be true) or that two of the $k_i$ are each divisible by $3.$ If $k_i$ is divisible by $3,$ its only possible value is $6$ (since neither $2$ and $8$ is divisible by $3$). Without loss of generality, we can set $k_1 = k_2 = 6.$

Dividing both sides of Equation $(1)$ by $k_1k_2 = 36,$ we now have $k_3k_4 = 2^4.$ Each of $k_3$ or $k_4$ can then only be $2$ or $8$ (since $6$ is not a power of $2$), but if one is $2$ then the other is $8$ and vice versa. So without loss of generality we can set $k_3 = 2$ and $k_4 = 8.$

The four roots therefore are $-k_1 = -6,$ $-k_2 = -6,$ $-k_3 = -2,$ and $-k_4 = -8.$ In ascending order they are $-8, -6, -6, -2.$

Answer 2

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.

Answer 3

Lemma: Let $ P(x)= a_nx^n+ a_{n-1}x^{n-1}+ ... + a_1x+ a_0 $ be a polynomail with integral coefficeint;
i.e. $a_i \in \mathbb{Z}$ and $a_n\neq 0$.
Let $\alpha=\dfrac{r}{s}$ be a rational root of this equation, with $\gcd(r,s)=1$;
then we must have: $s \mid a_n$ and $r \mid a_0$ .
Proof: Let $\alpha=\dfrac{r}{s}$ be a rational root of $P(x)$, with $\gcd(r,s)=1$; then we have: $$ 0= P(\alpha)= a_n\alpha^n+ a_{n-1}\alpha^{n-1}+ ... + a_1\alpha+ a_0 \Longrightarrow \\ \ \ \ \ \ \ \ \ \ \ 0= a_n(\dfrac{r}{s})^n+ a_{n-1}(\dfrac{r}{s})^{n-1}+ ... + a_1(\dfrac{r}{s})+ a_0 \Longrightarrow \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0= a_nr^n+ a_{n-1}r^{n-1}s+ ... + a_1rs^{n-1}+ a_0s^n \ \ \ \ \ \ \ \star $$ note that $r$ divides the LHS of $\star$; so it must divides the RHS;
also notice that $r$ divides all terms except $a_0s^n$; so it must divides $a_0s^n$;
by Euclid's lemma we can conclude that $r|a_0$.

note that $s$ divides the LHS of $\star$; so it must divides the RHS;
also notice that $s$ divides all terms except $a_nr^n$; so it must divides $a_nr^n$;
by Euclid's lemma we can conclude that $s|a_n$.

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In the special case of this question we know $576=2^63^2$, so we must have:

$$s|1 \ \ \ \ \text{and} \ \ \ \ r|2^63^2 $$

so it only suffices to check all the integers $\pm d$; where $d$ is a positive divisor of $576$;
one can check by hand that the only posibilities are $-2$ and $-6$.

Answer 4

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 $\color{red}{k_3 =k_4 = 6}$.

Answer 5

Your equations (2) and (3) seem an over-complicated way to attack the problem.

The fact that $k_1k_2k_3k_4 = 576 = 2^6 3^2$ is useful. So is the fact that $k_1 + k_2 + k_3 + k_4 = -22$, which you can prove in a similar fashion. (Note, both of these are general results for any polynomial).

For a third useful result, by inspection the equation has no positive roots, since the sign of every term is positive. This is a particular case of "Descartes' rule of signs."

Those three facts are usually enough to guess the values of $k_i$. If not, it may be simpler to try to find two quadratic factors, rather than four linear ones - the fact that the quadratic factors are not unique can often make it easier to guess one pair of them.

For this specific polynomial, this means that $0< k_i < 22$ for each $k_i$. The only possible integer values in that range which are factors of $576$ are $1, 2, 3, 4, 6, 8, 9, 12, 16, 18$.

By trial and error $1$ does not give a root, but $2$ does. That eliminates $18$, because $18+2+2+2 > 22$.

So to get the $3^2$ factor in $576$, we need to include some of $3, 6, 9$, or $12$. Of those four, the only root is $6$.

So we must have $k_1 = k_2 = 6$ and we already know $k_3 = 2$ gives a root. So $k_4 = 8$.

Answer 6

We set $$f(x)=x^4+22x^3+172x^2+552x+576=0$$

and $$f_0=f(0)=k_1k_2k_3k_4=2^6 3^2=576$$ $$f_{1}=f(1)=(k_1+1)(k_2+1)(k_3+1)(k_4+1)=3^3 7^2=1323$$ $$f_{-1}=f(-1)=(k_1-1)(k_2-1)(k_3-1)(k_4-1)=5^2 7 =175$$

We can check for each of the $42$ negative or positive divisors $d$ of $f_0$ if $-d$ is a zero of the polynomial. Easier is to check for the only $24$ negative or positive divisors $d$ of $f_1$ if $-d+1$ is a zero of the polynomial. More easier is to check for the only $12$ negative or positive divisors $d$ of $f_{-1}$ if $-d-1$ is a zero of the polynomial.

The divisors of $f_{-1}$ can be found in column $d_1$ to $d_{12}$ of line $1$ of the following table. These numbers are possible candidates for the numbers $k_1-1,k_2-1,k_3-1,k_4-1$.

$$\begin{array}{|r|l|c|r|r|r|r|r|r|r|r|r|r|r|r|} \hline \text{line}&\text{pfactors}&\text{type}&d_1&d_2&d_3&d_4&d_5&d_6&d_7&d_8&d_9&d_{10}&d_{11}&d_{12}\\ \hline 1&5^2 7&k-1&-175 & -35& -25& -7& -5& -1& 1& 5& 7& 25& 35& 175\\ \hline 2&2^6 3^2&k&-174& -34& -24&-6& -4& 0& 2& 6& 8& 26& 36& 176\\ \hline 3&&p^e&29&17&&&&0&&&&13&&11\\ \hline 4&3^3 7^2&k+1&-173& -33& -23& -5& -3& 1& 3& 7& 9& 27& 37& 177\\ \hline 5&&p^e&173&11&23&5&&&&&&&37&59\\ \hline 6&&\text{possible}&&&&&\checkmark&&\checkmark&\checkmark&\checkmark&&&\\ \hline 7&5^2 7&k-1&&&&&5&&1&5&7&\\ \hline 8&2^6 3^2&k&&&&&2^2&&2&2\phantom \cdot 3 &2^3&\\ \hline 9&3^3 7^2&k+1&&&&&3&&3&7&3^2&\\ \hline \end{array}$$

If we add $1$ to them we get Line $2$ with the possible candidates for $k_1,k_2,k_3,k_4.$ If we add $2$ to them we get Line $4$ with the possible candidates for $k_1+1,k_2+1,k_3+1,k_4+1.$ The product of four the right $d_j$ values must be equal to the value $\text{pfactors}.$

On the one hand they are $1$ larger than the numbers $k-1$ in line $1$ on the other hand they must be a divisor of $f_0$, which $2^6 3^2.$ For example $-175$ in column $d_1$ and line $1$ is a divisor of $f_{-1}$

So a possible candidate for $k$ satisfies $k_i-1=-175$ and therefore $k_i=-174.$ But $-174$ contains the prime factor $29$ and $29$ is not a prime factor of $f_0$, which $2^6 3^2.$ So $-174$ cannot be a divisor of $f_0$ and so $-174$ is not a possible $k_i.$ We can eliminate column $d_1$ can be eliminated. The prime factor that shows that $k$ is not a divisor of $2^6 3^2$ is shown in line $3.$ Line $5$ shows the prime factor that shows that $k_i+1$ is not a divisor of $f_{+1}$, which is $3^3 7^2.$

All columns can be eliminated except $d_5,d_7,d_8,d_9.$ Line $6$ puts a check mark in each of these columns. Line $7,8,9$ shows the the prime factorization of the values $k-1, k, k+1$ but only the columns that were not eliminated.

Comparing the prime factor $7$ in the columns of line $7$ one sees that $d_9$ is a necessary column. Comparing again the prime factor $7$ in the columns of line $9$ one sees that the divisor $d_8$ is used twice. So the last divisor can only be $d_7$ and the solution is $d_7 d_8^2 d_9.$ So we have $$\begin{eqnarray} k_1&=&2\\ k_2&=&6\\ k_3&=&6\\ k_4&=&8. \end{eqnarray}$$