Is there alternative factoring of a quintic equation?

Question

In a paper here the author appears to be able to factor a Bring-Jerrard quintic making

$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$ become $$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$ but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:

$$n^5-m^4n+\frac{P}{2m}=0$$

given that I will know the values of $P$ and $m$.

Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?

Answer 1

It seems that the claim of the paper is not true.

The author finally got the following equation $$b^4-2b^3\bigg(\frac{q-p+\sqrt{q^2-q}}{q-\sqrt{q^2-q}}\bigg)+q-p+\sqrt{q^2-q}=0$$

Here, let us consider one example. We have $$x^5-31x+30=(x^3+3x^2+7x+15)(x^2-3x+2)$$ This means that one of the possible values of $b$ is $3$ for $(p,q)=(-31,30)$.

However, the above equation does not have a solution $b=3$ for $(p,q)=(-31,30)$.

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Since we have $$(x^3+bx^2+cx+d)(x^2+ex+f)$$ $$=x^5+(b+e)x^4+(eb+c+f)x^3+(bf+ec+d)x^2+(cf+ed)x+df=0$$ if we compare this with $x^5+px+q$, then we get the following system $$\begin{cases}b+e=0 \\eb+c+f=0 \\bf+ec+d=0 \\cf+ed=p \\df=q\end{cases}$$ from which we want to represent $b,c,d,e,f$ by $p,q$.

Now, we have $$\begin{align}&\begin{cases}b+e=0 \\eb+c+f=0 \\bf+ec+d=0 \\cf+ed=p \\df=q\end{cases}\\\\&\stackrel{\text{eliminating $e$}}{\implies} \begin{cases}e=-b \\(-b)b+c+f=0 \\bf+(-b)c+d=0 \\cf+(-b)d=p \\df=q\end{cases} \\\\&\stackrel{\text{eliminating $f$}}{\implies}\begin{cases}e=-b \\df=q \\-b^2d+cd+q=0 \\bq-bcd+d^2=0 \\bd^2=cq-pd \end{cases} \\\\&\stackrel{\text{eliminating $b$}}{\implies} \begin{cases}e=-b \\df=q \\bd^2=cq-pd \\c^2dq^2-cqd^2p+d^3p^2-d^2pcq-cd^5-qd^4=0 \\c^2dq^2-cq^3+dpq^2-cd^2pq-d^4q=0 \end{cases} \\\\&\stackrel{\text{eliminating $c$}}{\implies} \begin{cases}e=-b \\df=q \\bd^2=cq-pd \\c(-d^2pq-d^5+q^3)=dpq^2-d^3p^2 \\d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0 \end{cases}\end{align}$$

So, we have to solve the following equation for $d$ : $$d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0$$ whose degree is $10$.

In conclusion, if we want to find $b,c,d,e,f$ such that $$x^5+px+q=(x^3+bx^2+cx+d)(x^2+ex+f)$$ then, in general, we have to solve an equation whose degree is $10$.

Answer 2

The usual way to solve that would be to expand

$(x^3 +bx^2 +cx+ d)(x^2 + ex + f)=x^5 + (e + b)x^4 + (eb+c+f)x^3+(bf + d + ce)x^2 + (ed+cf)x + fd $

Want this to equal to $x^5 + px +q$ for all $x$ which means the two ploynomails are equal and thats only true if the coefficients for the same powers are equal so

$$\begin{array}{cccCC} e+b &=&0 \Rightarrow &e&=&-b \\ eb+c+f &=& 0\Rightarrow &c+f&=&b^2\\ bf+d+ce&=&0 \Rightarrow &d+b(f-c)&=&0 \\ ed+cf &=&p\Rightarrow &-bd+cf&=&p\\ fd &=& q\Rightarrow &f &= &\frac{q}{d} \end{array}$$

$$\begin{array}{ccc} cd + q &=& db^2 \\ d^2 + bq-bdc&=&0 \\ -bd^2+cq&=& dp\end{array}$$

Lets eliminate $c$

$$\begin{array}{ccc} cbdq + q^2b & = & qdb^3 \\ -bdcq + qd^2+bq^2 & = & 0 \\ cqbd - b^2d^3 & = & d^2bp \\ \end{array}$$

Then $$\begin{array}{ccc} q^2b+qd^2+bq^2 &=& qdb^3\\ qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$

We need to solve the two equations $$\begin{array}{ccc} 2qb + d^2 &=& db^3 \\ qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$

Answer 3

Yes the paper has an error and was not properly checked. There is an alternative approach here:

http:/?utm_medium=email&utm_source=researchgate&utm_campaign=re322&utm_term=re322_x&utm_content=re322_x_p2&cp=re322_x_p2&uid=cx4ZWZ6hcdiX0hKERkrM99zJiDWztOEoeI4v&ch=reg

Answer 4

Yes there were some technical errors in that paper. I wrote some other papers possible factorization methods but did not actually succeed in getting a general solution. There was need for change in approach. When the general quintic equation is changed to identity for one actually succeeds and obtains an algebraic solution. Try this link:

http:/?utm_medium=email&utm_source=researchgate&utm_campaign=re322&utm_term=re322_x&utm_content=re322_x_p2&cp=re322_x_p2&uid=cx4ZWZ6hcdiX0hKERkrM99zJiDWztOEoeI4v&ch=reg