I want to solve the following equation for $n$ in terms of $P$ and $m$. $$n^5-m^4n+\frac{P}{2m}=0$$

I've bought and read many books, including "Beyond The Quartic Equation" but I've either missed something or do not have enough background or they said, 'such-and-such is used' but did not show how to use such-and-such to solve what I gather is a Bring-Jerrard quintic equation.

I'm just a forklift mechanic 40 years removed from academia with a math hobby. I've been writing a math paper on Pythagorean triples for about $10$ years and, with help, I thought I was almost done with, "On Finding Pythagorean Triples." Then, I thought of a new way to find "Triples On Demand", i.e. how to find a Pythagorean triple, if it exists, given only the product $(P)$ of A,B,C. Using Euclid's formula:

$$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$

the product is $2m^5n-2mn^5=P$. The best I've been able to understand is that the first equation above is in Bring-Jerrard form. The only thing I can add is that $P$ is a multiple of $60$ such as $60, 480, 780$,etc. and $m$ will be one of a range of values to test where $\lfloor\sqrt[6]{P}\rfloor\le m\le \lceil\sqrt[5]{P}\space\rceil$.

  • How do I find the group and know if it is solvable?
  • How does symmetry and/or permutations apply to this equation if at all?
  • How does this equation correspond to an icosahedron?
  • Is there a trig approach like the one here for a cubic equation? $$mn^3-m^3n+D=0$$

Almost any approach would be appreciated. I have so much to learn but none of the answers or comments have been useful so far – the approaches have been self-referential. How do I solve this quintic for $n$ if $P$ and $m$ are known?

Update: I changed an $f$ in the OP to a $P$ so don't be confused by some of the comments.

Also, I'm starting a bounty but not a large one for fear it will be wasted on the less-than-useful answers that have been upvoted already. Hurry, if you have an answer. I'd prefer to award the bounty rather than have it given away by an algorithm.

A comment mentioned I should be more specific about what I want to do. I'm looking for inputs to Euclid's formula (shown above) and which we define here as $F(m,n)$ ––note capitol F. I want one-to-five functions $n_x=f_x(P,m )$ such that, given a number like $4200$ and, knowing

$$\lfloor\sqrt[6]{4200}\rfloor=4\le m\le \lceil\sqrt[5]{4200}\space\rceil=6$$ I can discover $$f(4200,4)=3\Rightarrow F(4,3)=(7,24,25)\qquad f(4200,5)\notin\mathbb{N}\qquad f(4200,6)\notin\mathbb{N}$$

If an integer were not found for any of the $[5]$ solutions in the specified range of $m$-values, then we would know that no Pythagorean triple exists for that value of $P=A\times B\times C$.

Now, I'm told, specific cases are needed before we can find a group. Here are the smallest sample equation values and the "correct" solution of $f(P,m)=n$ for each.

$$n^5-16n+15=0\rightarrow f(60,2)=1\quad n^5-81n+80=0\rightarrow f(480,3)=1\quad n^5-81n+130=0\rightarrow f(780,3)=2\quad n^5-256n+255=0\rightarrow f(2040,4)=1\quad n^5-256n+480=0\rightarrow f(3840,4)=2\quad n^5-256n+525=0\rightarrow f(4200,4)=3\quad n^5-625n+624=0\rightarrow f(6240,5)=1\quad n^5-625n+1218=0\rightarrow f(12180,5)=2\quad n^5-625n+1476=0\rightarrow f(14760,5)=4\quad n^5-1296n+1295=0\rightarrow f(15540,6)=1\quad n^5-625n+1632=0\rightarrow f(16320,5)=3\quad n^5-1296n+2560=0\rightarrow f(30720,6)=2\quad n^5-2401n+2400=0\rightarrow f(33600,7)=1\quad n^5-1296n+3355=0\rightarrow f(40260,6)=5\quad n^5-1296n+3645=0\rightarrow f(43740,6)=3\quad n^5-1296n+4160=0\rightarrow f(49920,6)=4\quad n^5-4096n+4095=0\rightarrow f(65520,8)=1\quad$$

Are these sample equations enough to associate with a Galois group? Once we find the group, how do we proceed?

  • $\begingroup$ so divide by n solve the quartic, and multiply by the solution n values ... $\endgroup$
    – user645636
    Jan 9, 2020 at 0:21
  • $\begingroup$ @Roddy MacPhee The problem with dividing by $n$ is that the answer will include terms of $n$. I want to solve for $n$ only in terms of $f,m$. $\endgroup$
    – poetasis
    Jan 9, 2020 at 0:24
  • $\begingroup$ @Eric Towers I corrected the error. Thanks. $\endgroup$
    – poetasis
    Jan 9, 2020 at 0:53
  • 1
    $\begingroup$ @poetasis I can't tell you what Galois group that equation has (not 'belongs to' — the group is an object that's a property of the polynomial, not a collection that the polynomial is a member of), because there isn't nearly enough information to do so. For instance, absent any further information on $P$ and $m$, your polynomial could be irreducible over the integers, or it could have a linear factor. In fact, this is exactly the question you're trying to answer! But the Galois group is generally computed using that information, not the other way around. $\endgroup$ Jan 31, 2020 at 22:50
  • 1
    $\begingroup$ I don't know if you like this idea, but if you want to find positive integers $n$ such that $n^5-m^4n+\frac{P}{2m}=0$, then by the rational root theorem, every solution (if any) is a divisor of $\frac{P}{2m}$. This reduces the search space considerably. $\endgroup$
    – mathlove
    Feb 4, 2020 at 4:53

4 Answers 4


Solving a quintic is probably not the best idea for your problem.

Your numbers $m,n$ and $f$ are integers and so you should be using techniques for finding integer solutions.

For example, $2mn(m^2-n^2)(m^2+n^2)=f$ means that $m$, $n$, ... are all factors of the integer $f$.

(By the way, you seem to have $m>n$ at first and then $n>m$ later in your post.)

For your work, you may find it helpful to know of the following list of differences of coprime 4th powers https://oeis.org/A147858

  • $\begingroup$ not to mention ${f\over 2mn}+m^4=n^5$ is about as close as you'll get to variable separation. $\endgroup$
    – user645636
    Jan 9, 2020 at 1:06
  • $\begingroup$ @RoddyMacPhee (With $n^4$ on RHS) $\endgroup$
    – user502266
    Jan 9, 2020 at 1:08
  • $\begingroup$ ... and a negative $f$ doh $$(2mn)(m^2-n^2)(m^2+n^2)=f=2mn(m^4-n^4)$$$$\implies -f=2mn(n^4-m^4)$$$$\implies {-f\over 2mn}+m^4=n^4$$ $\endgroup$
    – user645636
    Jan 9, 2020 at 1:24
  • $\begingroup$ @Roddy MacPhee I need to find $n$ solely in terms of $(f,m)$. $\endgroup$
    – poetasis
    Jan 9, 2020 at 1:29
  • $\begingroup$ okay so maybe: $$-2mf{1\over n}+2m^5=2mn^4$$ try and solve it as a quartic ? $\endgroup$
    – user645636
    Jan 9, 2020 at 1:38

I doubt that solving a quintic is the way to go. Consider the congruent number problem, which asks if an integer can be the area of a right triangle with rational sides. In your problem, the area will always be a congruent number. If you read the article, you will notice first that the congruent number problem reduces to solving a cubic, and second that it is unsolved.

Of course, your problem is different. You are requiring integral sides, not rational sides, and you are starting with the product of the sides, not the area. Still, it seems to me that the two problems are somehow related, and the congruent number problem is unsolved, even though it involves a cubic rather than a quintic.

I don't think you are paying sufficient attention to the fact that you require a solution in integers. I think the "number-theory" tag is more appropriate than any of the tags you have applied.

Here are some thoughts of the top of my head. Suppose we are given the easiest case, $ABC=60$. Now $60=2^2\cdot3\cdot5$, and we may assume that $A<B<C.$ We can test the various factors of $60$ as possible values for $C$. For example, can $C=4$? Then $B\leq3$ so $A\leq2$ and $ABC\leq2\cdot3\cdot4<60$, contradiction. In general, we need $$(C-2)(C-1)C^2\geq F$$

Once we have chosen a possible value for $C$, we have the equations $$\begin{align} A^2+B^2 &= C^2\tag1\\ AB &= \frac{F}{C}\tag2 \end{align}$$ so that $$(A+B)^2=C^2+\frac{2F}{C}$$ If the right-hand side is a perfect square, we can solve for $A+B$ and eliminate $B$ from $(2)$.

So, if $F$ is small enough to factor, there's an easy way to test $F$, but I think it will be hard to find a formula that says, "An $F$ of one of these forms is acceptable, and an $F$ of any other form is not."

I know that this is really more of a comment than an answer, but it's too long for a comment box.

  • $\begingroup$ We will have $A>B<C$ or $A<B<C$: $3,4,5\quad 5,12,13\quad 35,12,37\quad 21,20,29...\qquad C\ne 4$ $\endgroup$
    – poetasis
    Jan 9, 2020 at 2:42
  • $\begingroup$ @poetasis I don't understand what you are saying at all. The numbers will always be different. We can call the smallest one $A$, the next smallest $B$ and the largest $C$. As for the comment that $C\neq4$, I don't understand what that relates to. Where did I say that the hypotenuse is alway $4$? $\endgroup$
    – saulspatz
    Jan 9, 2020 at 2:46
  • $\begingroup$ You asked , "For example, can 𝐶=4?". No. Also, an original formula I've developed always has A-odd and B-even. Even with Euclid's formula, B can be smaller than A. $\endgroup$
    – poetasis
    Jan 9, 2020 at 2:51
  • $\begingroup$ @poetasis I was talking about the specific example $F=60$. As I remarked, in general we must have $(C-2)(C-1)C^2\geq F$, so given $F$, you can find a lower bound for $C$. $\endgroup$
    – saulspatz
    Jan 9, 2020 at 3:25
  • $\begingroup$ The solution to the congruent number problem is here and, if I knew the area in advance, this would be a simple problem of solving a quadratic. Any other ideas on solving a quintic? $\endgroup$
    – poetasis
    Jan 26, 2020 at 17:46

$\color{green}{\textbf{Elaborated version (07.02.20).}}$

The issue Diophantine task is presented in the form of quintic over $2D$ set of pairs $(m,n).$

At the same time, from the quihtic should $$P = ABC,\tag{i1}$$

where $$A^2+B^2=C^2,\tag{i2}$$ i.e $(A,B,C)$ is a Pythagorean triple.

If the quintic solution $(m,n)$ exists, then the triple $(A_{mn},B_{mn},C_{mn}),$ where $$A_{mn}=m^2-n^2,\quad B_{mn}=2mn,\quad C_{mn}=m^2+n^2\tag{i3}$$ should belong to the set of the solutions of the Diophantine task $(i1)-(i2).$

This possibility was pointed in OP. Let us apply it.

The approach, proposed below, uses constraints only for unknown $C.$ Then for each possible value of $C$ from $(i1)-(i2)$ calculates the solution triple $(A,B,C).$
If such solution exists, then the solution $(m,n)$ of $(i3)$ is the solution of the given quintic.

Calculations of the pair $(m,n),$ which correspond to the certain solution of $(i1)-(i2),$ are not hard.

For example, if $P=2\,88987\,09840,$ then there are $48$ possible values of $C$ before testing of the required interval and $7$ values after the testing, and only the single value leads to the valid pair $(A,B)$ and to the valid quintic root $(m,n)$.

So the proposed approach looks preferable.



Since $C^2 = A^2 + B^2$ and $A\not=B,$ then $C^2 > 2AB,$ $$C > \sqrt[\large3]{2P\large\mathstrut}.\tag1$$

Example: $A=21, B=20, C=29, P = 12180, \sqrt[3]{24360}\approx 28.988.$


At the same time, $A^2B^2 = (C-k)^2(C^2-(C-k)^2),$

with the least value at $k=1.$

Then $$A^4 B^4 = (C-1)^4(2C-1)^2 = (C-1)^4 (4C^2-4C+1),$$ $$P^4 = A^4B^4C^4 = \dfrac1{256}(4C^2-4C)^4(4C^2-4C+1)\\[4pt] > \dfrac1{256}\Big(4C^2-4C\Big)^5 = \dfrac1{256}\Big((2C-1)^2-1\Big)^5,$$ $$C < \dfrac12\left(\sqrt{(4P)^{^4/_5}+1\ }\ + 1\right).\tag2$$

Example: $A=35, B=12, C=37, P=15540, \dfrac12\left(\sqrt{62160^{0.8}+1}+1\right)\approx41.843.$


Also, is known that $12\,|\,AB,$ then $$C\,\bigg|\,\dfrac P{12}\tag3.$$


Variable $C$ is the sum of two squares. Then should provide conditions, which correspond with the Fermat theorem of sums of two squares (I've used Russian version of the book Harold M. Edwards. Fermat Last Theorem, Schpringer 1977.)

The number is the sum of two squares, if and only if it is

  • a square, or

  • a prime $p$, wherein $p=4t+1,\, t\in\mathbb N,$ or

  • $2,$ or

  • the production of the such numbers.

Therefore, the primary filtration of the possible values of $C$ can be based on the constraints in the form of

\begin{cases} C\in\left[\left\lceil\sqrt[3]{2P\large\mathstrut}\LARGE\mathstrut\right\rceil, \left\lfloor\dfrac12\left(\sqrt{(4P)^{^4/_5}+1\ }\ + 1\right)\right\rfloor\right]\\[4pt] C\,\bigg|\,\dfrac P{12}\\[4pt] C = 2^i s^{2j} \prod\limits_{k=0}^f (4t_k+1)\\[4pt] (i,j,f)\in \mathbb Z_+^3\\[4pt] s-2\in\mathbb N,\ \{t_k\} \in \mathbb Z_+^f.\tag4 \end{cases}

$\color{brown}{\textbf{Secondary filtration and the quintic solution calculations.}}$


From $(i1)-(i2)$ should \begin{cases} C^2+\dfrac{2P}C = S^2\\[4pt] C^2-\dfrac{2P}C = D^2\\[4pt] \dbinom AB \in \left\{\dfrac12\dbinom{S+D}{S-D},\dfrac12\dbinom{S-D}{S+D}\right\} \\[4pt] (A,B,S,D)\in\mathbb N^4\tag5 \end{cases} (secondary filtration).

Easily to see, that $(5)$ has two solutions or nothing.


If the Diophantine system $(5)$ has solution $(A,B,C)$ and $(A_{mn},B_{mn},C_{mn}) = (A,B,C),$ then from $(i3)$ should \begin{cases} 2m^2 = C+A\\ 2mn = B.\tag6 \end{cases}


$(4)-(6)$ define all possible solutions of the given quintic.

$\color{brown}{\mathbf{Example\ P=60.}}$

Equation $(4.3)$ is $C\,|\,5,$ with the single solution $\color{brown}{\mathbf{C=5}}$ in the form of $4\cdot 1+1.$

Required interval is $[5,5].$

Then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 25\pm24\in\{7^2,1^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {3}{4}}}\\ 2m^2 = 5+3\\ 2mn = 4, \end{cases}

with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 2,\ n = 1}}.$

$\color{brown}{\mathbf{Example\ P=480.}}$

Equation $(4.3)$ is $C\,|\,40.$

Required interval is $[10,10].$

The single solution is $\color{brown}{\mathbf{C=10}}$ in the form of $2(4\cdot 1+1).$

Then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 100\pm96\in\{14^2,2^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {8}{6}}}\\ 2m^2 = 10+8\\ 2mn = 6, \end{cases}

with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 3,\ n = 1}}.$

$\color{brown}{\mathbf{Example\ P=780.}}$

Equation $(4.3)$ is $C\,|\,65.$

Required interval is $[12,13].$

The single solution is $\color{brown}{\mathbf{C=13}}$ in the form of $4\cdot3+1.$

Then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 169\pm120\in\{17^2,7^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {5}{12}}}\\ 2m^2 = 13+5\\ 2mn = 12, \end{cases}

with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 3,\ n = 2}}.$

$\color{brown}{\mathbf{Example\ P=2040.}}$

Equation $(4.3)$ is $C\,|\,170.$

Required interval is $[16,18].$

The single solution is $\color{brown}{\mathbf{C=17}}$ in the form of $4\cdot4+1.$

Then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 289\pm240\in\{23^2,7^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {15}{8}}}\\ 2m^2 = 15+17\\ 2mn = 8, \end{cases}

with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 4,\ n = 1}}.$

$\color{brown}{\mathbf{Example\ P=4200.}}$

Equation $(4.3)$ is $C\,|\,350.$

Required interval is $[21,25].$

The single solution is $\color{brown}{\mathbf{C=25}}$ in the form of $5^2 = 4\cdot6+1.$

Then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 625\pm336\in\{31^2,17^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {7}{24}}}\\ 2m^2 = 25+7\\ 2mn = 24, \end{cases}

with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 4,\ n = 3}}.$

$\color{brown}{\mathbf{Example\ P = 2\,88987\,09840.}}$

Required interval is $[3867,13309]$

Equation $(4.3)$ is $C\,|\, 24082\,25820 = 4\cdot3\cdot5\cdot7\cdot11\cdot13\cdot101\cdot397.$

The valid form of $C$ is $$C = 2^i 5^{a-1} 13^{b-1} 101^{c-1} 397^{d-1},$$ where $$i \in \{0,1,2\},\quad (a,b,c,d) \in \{1,2\}^4$$ (totally, $48$ valid productions).

Belong to the required interval $$\color{blue}{\mathbf{C\in\{3970 = 2\cdot5\cdot 397, 5252=4\cdot13\cdot101, 5161=13\cdot397, 6565=5\cdot13\cdot101, {7940=4\cdot5\cdot3}97, 10322=2\cdot13\cdot397, 13130=2\cdot5\cdot13\cdot101\}}}$$ (see also Wolfram Alpha calculations factor1, factor2, factor4).

Therefore, only seven possible values of C belong to the required interval.

If $C=3970,$ then from $(5-6)$ $$\{S^2,D^2\} = 3970^2\pm\dfrac{5\,77974\,19680}{3970}, \{S,D\}\in\varnothing.$$

If $C=5252,$ then from $(5-6)$ $$\{S^2,D^2\} = 5252^2\pm\dfrac{5\,77974\,19680}{5252}, \{S,D\}\in\varnothing.$$

If $\color{brown}{\mathbf{C=5161}},$ then from $(5-6)$ follows \begin{cases} \{S^2,D^2\} = 5161^2\pm\dfrac{5\,77974\,19680}{5161}\in\{6151^2,3929^2\}\\ \color{brown}{\mathbf{\dbinom AB = \dbinom {1111}{5040}}}\\ 2m^2 = 5161+1111\\ 2mn = 5040, \end{cases} with the solution over $\mathbb N$ of the quintic $\color{brown}{\mathbf{m = 56,\ n = 45}}.$

If $C=6565,$ then from $(5-6)$ $$\{S^2,D^2\} = 6565^2\pm\dfrac{5\,77974\,19680}{6565}, \{S,D\}\in\varnothing.$$

If $C=7940,$ then from $(5-6)$ $$\{S^2,D^2\}= 7940^2\pm\dfrac{5\,77974\,19680}{7940}, \{S,D\}\in\varnothing.$$

If $C=10322,$ then from $(5-6)$ $$\{S^2,D^2\} = 10322^2\pm\dfrac{5\,77974\,19680}{10322}, \{S,D\}\in\varnothing.$$

If $C=13130,$ then from $(5-6)$ $$\{S^2,D^2\} = 13130^2\pm\dfrac{5\,77974\,19680}{13130}, \{S,D\}\in\varnothing.$$

Therefore, the single solution over $\mathbb N$ of the quintic is $\color{brown}{\mathbf{m = 56,\ n = 45}}.$

This example demonstrates high effectiveness of the proposed approach.

$\color{blue}{\textbf{Too long for a comment.}}$

1. If $\dfrac Q{60}$ has dividers of the six order, then previosly should be tested the value of $Q$ with the eliminated divider.

This approach can garantee that $\gcd(m,n)=1.$

2. Alternative form

$$\left(A+\dfrac{P}{AC}\right)^2 = C^2+\dfrac{2P}C$$

does not contain $B.$

  • $\begingroup$ @poetasis Totally rewrited. $\endgroup$ Feb 4, 2020 at 9:44
  • $\begingroup$ I'm studying your answer but I still do not know what you mean by "system" and I do not understand the math language you used to frame them. $\endgroup$
    – poetasis
    Feb 5, 2020 at 17:29
  • $\begingroup$ @poetasis Ok,text simplified. Waiting to your comments. $\endgroup$ Feb 5, 2020 at 22:37
  • $\begingroup$ I'm giving you the bounty because you do seem to have a method of solving these equations (though I cannot follow some of it) without being self-referential, i.e. where $n=f(P,m,n)$. I would like to continue this in chat or perhaps we can exchange PDFs in email. Mine is poetasis@gmail.com $\endgroup$
    – poetasis
    Feb 6, 2020 at 17:43
  • $\begingroup$ @poetasis Thank you for awards! At the same time, the calculations are executing by the chain P->C->(A,B)->(m,n). Where have you seen self-referencial? (BTW, I'm registered via the Facebook and am available too.) $\endgroup$ Feb 6, 2020 at 20:11

Upon request from the OP, I'll convert my comment into an answer.

I don't know if you like this idea, but if you want to find positive integers $n$ such that $$n^5−m^4n+\frac{P}{2m}=0$$ then by the rational root theorem, every solution (if any) is a divisor of $\frac{P}{2m}$.

This reduces the search space considerably.

  • $\begingroup$ Thank you for the answer. My experiments didn't work out but they may in the future. Days ago, I also set a bounty on a factoring question here and, by tomorrow, I need to award the bounty. It will probably be to the guy who told me the author of a paper on factoring Bring-Jerrard quintics was wrong. In the mean time, I gave $this$ bounty to Yuri Negometyanov because, though I don't understand his approach yet, he seems to have a working solution, even if not the "one". :) $\endgroup$
    – poetasis
    Feb 6, 2020 at 17:51

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