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If we have the general quintic equation $$ax^5+bx^4+cx^3+dx^2+ex+f=0$$ we can vanish the quartic term by doing the substitution $x=y-b/5a$. The question I wanna ask is if there is a possible way to do a substitution so the quarticc and the quadratic term is vanished so that we are left with the equation $$ay^5+by^3+cy+d=0$$ ($a$,$b$,$c$,$d$ not the same of course)
Thank you for your time.
If someone can fix the latex it would be great, thank you.

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  • $\begingroup$ I don't know if it's any help, but Mathematica is also doing research in this kind of subject: library.wolfram.com/examples/quintic/main.html $\endgroup$ – Nick May 29 '14 at 21:57
  • $\begingroup$ Thank you for the information. I am pretty sure that there is no such substitution cause if there was, then the quintic would be solvable, which has been proved to be impossible. $\endgroup$ – user154129 May 29 '14 at 21:58
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    $\begingroup$ It's possible to reduce quintic equations to Bring Quintic Form which has a quintic term, but then just a linear term. $\endgroup$ – Peter Woolfitt May 29 '14 at 22:02
  • $\begingroup$ @user154129 Sure. This is generally known to us as the Brioschi quintic form $\endgroup$ – Balarka Sen May 31 '14 at 12:07
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Let us see directly why this is impossible. I divide through by $a$ (or equivalently, consider $a = 1$) for ease of notation.

We start with $x^5 + bx^4 + cx^3 + dx^2 + ex + f$, and we will be doing a substitution $x \mapsto x - \lambda$. Let's see what constraints we get from wanting to remove the quartic term. So we collect the coefficients of $x^4$.

The $x^5$ term gives us $-5\lambda x^4$, and the $bx^4$ term gives us $bx^4$ (i.e. it doesn't change). We want these to cancel, so we want $bx^4 - 5\lambda x^4 = 0$, or for $\lambda = \frac{b}{5}$. This is completely forced, and no other substitution will remove the quartic term in general.

So doing any other substitution will not remove the quartic term, and this substitution does not in general remove the quadratic term (which I do not show, because it's an annoying computation - alternatively, choose just about any quintic and it will give a counterexample).

So the answer is no, it is not possible to do a generic substitution such that both the quartic and quadratic terms vanish.

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  • $\begingroup$ This answer and Peter Woolfit's comment seem to be contradictory. It would be helpful to explain why they are not. $\endgroup$ – bubba May 29 '14 at 23:38
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    $\begingroup$ It depends on what transformations you allow. I was assuming you wanted a linear shift, as was stated by $x \mapsto x - \lambda$. Tshirnhausen transformations are much more complicated. More on those can be found in this post. $\endgroup$ – davidlowryduda May 29 '14 at 23:51
  • $\begingroup$ @mixedmath I think you meant something else. It's possible to do a transformation that will reduce a general quintic to the required form, which will essentially be a rational transformation rather than linear. But "it is not possible to do a generic substitution" is a confusing statement. $\endgroup$ – Balarka Sen May 31 '14 at 12:10

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