# 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.

• Just verify the divisors of $576$ (including the negative ones!) whether they are a root. The leading coefficient is $1$, so every rational root must be an integer dividing the constant term – Peter Aug 28 '17 at 19:22
• @Benjamin With the quadratic term added back, there are indeed integer roots. To do it by hand, I'd search the small factors of $576$ first (noting that all roots are negative). There is a small one, and once you have found it you can factor to get a cubic. Now repeat. – lulu Aug 28 '17 at 19:27
• The fourth root of $576$ is less than $5$, so there must be a root less than $5$ in absolute value. That gives eight to try, so finding $-2$ shouldn't be hard. Then you can divide out $x+2$ and you are looking for three numbers with product $288$. The cube root of that is less than $7$ so there must be another factor less than $7$ in absolute value. Divide that out and you have a quadratic. – Ross Millikan Aug 28 '17 at 19:31
• And here, you are successfull in the second(!) try, $-1$ fails to be an integer root, the next possible root , namely $-2$ is one. You can apply polynomial division or continue the search and soon you will find $-6$. If the numbers were larger, you could save much time by using the three equations, but $576$ is still so small that result $1$ is sufficient even for a solution by hand. – Peter Aug 28 '17 at 19:43
• There are lots of comments here on efficient ways to solve the quartic knowing that the results are integers. I have tried in my solution to show how the preliminary results in the question can be used to make progress. This does not involve substituting values back into the original equation. – Mark Bennet Aug 28 '17 at 20:00

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 $-8, -6, -6, -2.$

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.

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

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 $s|a_n$.

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 $-6$.

• suffices to check all the integers ±d where d is a positive divisor of 576 The roots must be negative (since all coefficients are positive), must be even (since all coefficients other than the leading one are even), and must be no smaller than $-16$ (since their sum is $-22 = -16 + 3 \cdot (-2)$). This leaves just $-2, -4, -6, -8, -12, -16$ to check. Once you find the first root, the possibilities narrow further. – dxiv Aug 28 '17 at 20:00
• @dxiv ; excelent, that was perfect. – Davood KHAJEHPOUR Aug 28 '17 at 20:09

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 $k_4 = 8$.

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 $$\begin{eqnarray} k_1&=&2\\ k_2&=&6\\ k_3&=&6\\ k_4&=&8. \end{eqnarray}$$