How can I solve, for example, $n^5-625n+1632=0$ for $n$ if it is solvable?

Question

I understand that some quintics in Bring-Jerrard form are solvable but first one must identify a solvable group for it or any quintic. I don't know how to identify a group for this equation or how to proceed from there. For the examples below, I know what the solution(s) will be only because I constructed them from known values.

The equations come from the product of sides of Pythagorean triples generated by Euclid's formula where:

$P=2m^5n-2mn^5\quad $and it becomes $\quad n^5-m^4n+\frac{P}{2m}=0\quad$ that I hope to solve for $n$.

I will "know" a narrow range of $m$ values to test for any given $P$ and

I seek one-to-five functions $f(b,m )$ to test to see which $m$, if any, yields an integer $n$.

Here are the samples showing also the correct values of $n$ that I seek. In the form $\qquad\qquad\qquad n^5-an+b=0\rightarrow f(b,m)=n$.

$$ n^5-16n+15=0\rightarrow f(15,2)=1\quad n^5-81n+80=0\rightarrow f(80,3)=1\quad n^5-81n+130=0\rightarrow f(130,3)=2\quad n^5-256n+255=0\rightarrow f(255,4)=1\quad n^5-256n+480=0\rightarrow f(480,4)=2\quad n^5-256n+525=0\rightarrow f(525,4)=3\quad n^5-625n+624=0\rightarrow f(624,5)=1\quad n^5-625n+1218=0\rightarrow f(1218,5)=2\quad n^5-625n+1632=0\rightarrow f(1632,5)=3\quad n^5-625n+1476=0\rightarrow f(1476,5)=4\quad n^5-1296n+1295=0\rightarrow f(1295,6)=1\quad $$

The monomial $(n-r)$, where $r$ is a root, will always factor the polynomial. For example, $n^5-625n+1632=(n - 3) (n^4 + 3 n^3 + 9 n^2 + 27 n - 544)$ but I have no use for the quartic once I find $n=3$.

I can find $n$ by testing the factors of $b$ but I want to learn what is usually done in solving a Bring-Jerrard quintic, starting with groups. I don't know group theory well enough to figure out how it relates to polynomials.

Can someone help me find the "group" for one or all of these and then show me how such is used to solve this kind of equation?

Answer 1

You seem to think that finding the Galois group of one of these polynomials would lead to a shortcut to finding eventual integer roots. In my humble opinion this is quite the wrong approach to your real problem of finding integer roots to a quintic with integer coefficients in Bring-Jerrard form.

• Figuring out the Galois group takes more work than checking for integer roots via the rational root test (or by other means if you cannot factor the constant term). If only for the reason that checking for the presence of rationa roots would be the first step in my approach to the task of finding in the Galois group.
• Knowing the Galois group $G$ (as a group of permutations of the roots, a subgroup of $S_5$), when it happens to be solvable, only helps in finding ALL the five roots. If we know $G$ then we already know whether the polynomial has integer roots WITHOUT SOLVING THEM. For example, unless $G$ is a subgroup of a point stabilizer, we can say right away that there are no integer roots.

In many comments you have indicated that you are not interested in the quartic factor. That is fine - for the purposes of finding integer roots you are right. But it also further underlines the fact that you are really not interested in the Galois group. This is because when there is an integer root, the Galois group only contains information about the zeros of that quartic factor.

---

To give you something useful let me demonstrate the use of modular reduction as a technique for finding eventual integer roots of a quintic in Bring-Jerrard form. If you can factor the constant $b$ of $x^5+ax+b$, then the rational root test already leads to a problem of testing a finite number of candidates. As do elementary estimates from calculus and friends. Modular reduction will simply further cut down the number of remaining alternatives.

Consider the (not quite random) quintic equation $$ f(x)=x^5-23232x-69277=0. $$ The task is to find integer solutions. Here it would not be difficult to factor the constant term, but let's assume that you cannot do that (or that it would leave to uncomfortably many alternatives to be tested). A point is that if $f(n)=0$, then the residue class of $n$ modulo any prime $p$ will also be a solution to the congruence $f(x)\equiv \pmod p$. Another point is that finding modular roots modulo a prime much smaller than $b$ is a lot easier (there are efficient algorithms for that when $p$ is a bit too large for paper and pencil work).

This $f(x)$ is a case in point. We see that $\root4\of{23232}$ is a bit over ten. We can then check that $f(15)$ is safely positive, and that the function is increasing from that point on. Similarly $f(-15)<0$ and $f$ is increasing in the interval $(-\infty,-15)$. Therefore any integer zeros $n$ will satisfy $|n|<15$. It would be easy to test all the integers in this range (either as zeros of $f(x)$ or as factors of $b$. But, again, I'm assuming that in a real version of this question we would instead be left with too many alternatives for a brute force check.

On we go by reducing modulo small primes $p$.

• Reducing the coefficients of $f$ modulo $p=2$ we arrive at $f(x)\equiv x^5+1$. The only modular zero here is $x\equiv1$. This is basically restating the obvious fact an integer root must be odd.
• Reduction modulo $p=3$ gives $f(x)\equiv x^5-1$. From this we can deduce that if $f(n)=0$ we must have $n\equiv1\pmod3$.
• With the question settled modulo two or primes we invoke the Chinese Remainder Theorem (=CRT). We know that $n\equiv1\pmod2$ and $n\equiv1\pmod3$. CRT then tells us that $n\equiv1\pmod6$. Basically knowing the remainder of $n$ modulo $2$ and $3$ leaves only a single alternative for its remainder modulo $2\cdot3=6$.
• Next in line is $p=5$. Reducing the coefficients modulo $5$ shows that $$f(x)\equiv x^5+3x+3\pmod5.$$ Testing $x=0,1,2,3,4$ (with or without using that by Little Fermat $x^5\equiv x$) quickly reveals that we must have $x\equiv3\pmod5$.
• Another round of CRT is due. We will get information modulo $6\cdot5=30$. Combining the pieces $n\equiv1\pmod6$ and $n\equiv3\pmod5$ leaves $n\equiv13\pmod{30}$ as the only possibility.

Because we initially constrained the solutions to the range $(-15,15)$ we can conclude that the only possible integer zero of $f(x)$ is $x=13$. Of course, I set up the example carefully so that, in fact, $f(13)=0$ – something you can verify easily enough.

As a further observation we see that $69277=13\cdot73^2$. This time the rational root test would have immediately given us that the only possible integer roots are $\pm1,\pm13,\pm73,\pm13\cdot73,\pm 73^2,\pm 13\cdot73^2$.

The product of small primes grows quickly enough so that you can, in a reasonable time, cover a relatively wide range of possible roots. If, instead, you wanted to find the roots of the quintic $g(x)=x^5-23232x-69276$ then, as above, $p=2,3,5$ will give us the information that an integer root $n$ must satisfy the congruence $n\equiv24\pmod{30}$. One alternative would then be that as $x=-6$ does not work, there are no integer solutions. Another possibility would be to include reduction modulo $p=7$. For we see that $$g(x)\equiv x^5+x+3\pmod{7}$$ and, more importantly, $n^5+n+3$ is not divisible by $7$ for any $n=0,1,2,3,4,5,6$. This also implies that $g(x)$ has no integer roots.

---

Conclusions:

• Combining the rational root test (when feasible), range estimation (trivial) and reductions modulo several small primes leaves a relatively short list of possible candidate zeros that can be tested.
• Quite often considerations modulo a single prime may leave several alternatives modulo that prime. Then you either accept a longer list of candidates or reduce modulo some other prime (this is inevitable when the original quintic has several integer roots).
• For information on the technique of finding the roots of a low degree polynomial modulo a larger prime (when testing $n=0,1,2,\ldots,p-1$ feels too painful) you can take a look at this old thread.
• I don't see a way of using Galois theory to your end. I do want to point out that if your quintic is irreducible over $\Bbb{Q}$ (when we can immediately conclude that there are no integer roots), then it will also be irreducible modulo some prime $p$. This fact is special to quintics ($5$ is a prime), and requires a result known as Chebotaryev's density theorem (or a lighter version thereof). This is in contrast to quartics – we can easily find quartic polynomials with integer coefficients, irreducible over $\Bbb{Q}$ but reducible modulo every prime.

---

Doing the quintic in the question title as another example. This time we easily see that any real zero of $$ h(x)=x^5-625x+1632 $$ must satisfy $|x|<8$.

• Reduction modulo $p=2$ does not give us anything for $h(n)$ is even for all integers $n$.
• Modulo $p=3$ we have $h(x)\equiv x^5-x$. This isn't very useful either for we can only exclude $n\equiv2\pmod3$.
• Modulo $p=5$ we get $h(x)\equiv x^5+2$. By Little Fermat this implies that an integer root $n$ must satisfy $n\equiv3\pmod 5$.
• Modulo $p=7$ we get $h(x)\equiv x^5+5x+1$. Testing with $x=0,1,\ldots,6$ shows that $x^5+5x+1$ is divisible by seven if and only if $x\equiv1$ or $x\equiv 3\pmod 7$.
• Combining the modulo $5$ and $7$ pieces of information implies (CRT) that an eventual integer zero $n$ must be congruent to either $8$ or $3$ modulo $35$.
• This means that in our range $n=3$ is the only possibility (and actually a solution as per OP's design).

This time the constant term $1632=2^5\cdot3\cdot17$ has quite a few factors, so using the rational root test would not have helped much.

Answer 2

$$n^5-16n+15=$$ $$=n^5-n^4+n^4-n^3+n^3-n^2+n^2-n-15n+15=$$ $$=(n-1)(n^4+n^3+n^2+n-15),$$ which gives $n=1$ or $$n^4+n^3+n^2+n-15=0.$$ The last equation for any real $k$ gives $$\left(n^2+\frac{1}{2}n+k\right)^2-\frac{1}{4}n^2-k^2-kn-2kn^2+n^2+n-15=0$$ or $$\left(n^2+\frac{1}{2}n+k\right)^2-\left(\left(2k-\frac{3}{4}\right)n^2+(k-1)n+k^2+15\right)=0.$$ Now, we'll choose a value of $k$, for which the last side of the last equation is a difference of squares,

for which we need $$ 2k-\frac{3}{4}>0$$ and $$(k-1)^2-4\left(2k-\frac{3}{4}\right)(k^2+15)=0$$ or $$4k^3-2k^2+61k-23=0,$$ which by the Cardano's formula ( https://en.wikipedia.org/wiki/Cubic_equation ) gives: $$k=\frac{1}{6}\left(1+\sqrt[3]{\frac{1}{2}\left(695+3\sqrt{2733033}\right)}+\sqrt[3]{\frac{1}{2}\left(695-3\sqrt{2733033}\right)}\right)$$ and since for the last value we have $$2k-\frac{3}{4}>0,$$ we obtain: $$\left(n^2+\tfrac{1}{2}n+k-\sqrt{2k-\tfrac{3}{4}}\left(n+\tfrac{k-1}{2\left(2k-\tfrac{3}{4}\right)}\right)\right)\left(n^2+\tfrac{1}{2}n+k+\sqrt{2k-\tfrac{3}{4}}\left(n+\tfrac{k-1}{2\left(2k-\tfrac{3}{4}\right)}\right)\right)=0.$$ Id est, it remains to solve two quadratic equations.