# Is there alternative factoring of a quintic equation?

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 $$p,q$$?

• I arrived here after Googling "Samuel Bonaya Buya" because as soon as I saw his claim I knew it was unlikely to be correct. Other readers arriving here may find the following compact summary of the current state of play regarding quintics useful, in addition to the discussion here and else where of Math Stack exchange : archive.lib.msu.edu/crcmath/math/math/q/q111.htm – Martin Hansen Jul 13 at 16:55

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

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 $$(p,q)=(-31,30)$$.

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 $$10$$.

• I get the impression from other papers and reviews that Samuel Bonaya Buya is not that careful in checking his own work before publishing. – poetasis Feb 1 at 10:47
• @poetasis: I agree with you. It seems that the step from the equation 13 to 14,15 is incorrect. – mathlove Feb 1 at 10:57