Can we remove four terms from deg-$n$ equations using deg-$m$ Tschirnhausen transformations with $m>n$?

Question

The Bring quartic transformation to eliminate three consecutive terms from the general equation of deg $n\geq5$ is well-known. It can't be applied to $n=4$ (though a cubic transformation can actually do it) nor $n=3$, since one ends up with the trivial $x^3=0$.

It seems a nonic transformation (with coefficients also in radicals) can remove four consecutive terms from the general equation of deg $n\geq10$. User Thinh Dinh showed a clever process that can be generalized in this post.

The question now is: what happens if we apply this nonic method on equations of deg $n<10$? There is a greater span but, like the Bring, it will not work on the degree equal to itself $n=9$ or if the result becomes trivial like for $n\leq5$. (To remove four terms from $n=9$, apparently there are other ways by Garver and Hilbert, as this post asks.) But there is still $n=6,7,8$. I tested it on $n=6$, doesn't work. However, for $n=(7,8)$, I got the system,

$$A=B=C=0$$

of three equations (2 quadratic and 1 cubic) in five unknowns, a similar system which Thinh managed to solve in radicals for the case $n=10$.

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Question:

Can we remove four consecutive terms from this sample septic,

$$x^7 - 2x^3 - x^2 - x + 1 = 0$$

using a nonic transformation with radical coefficients,

$$y=x^9+ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+mx+n$$

such that the septic is further reduced to the form,

$$x^7+px^2+qx+r = 0$$

a form discussed by Hilbert's 13th Problem?

Answer 1

I. Coefficients $\pmb{C_k}$

Given the sample septic above, $$x^7 - 2x^3 - x^2 - x + 1 = 0$$ and the deg-$9$ Tschirnhaus transformation, $$y=x^9+ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+mx+n$$ Using resultants to eliminate $x$ between the two yields a new septic, $$y^7+C_1y^6+C_2y^5+C_3y^4+C_4y^3+C_5y^2+C_6y+C_7=0$$ Collecting coefficient with respect to $\color{red}{m}$, the $C_k$ above have the form, $$C_1=\alpha_1\\ C_2=\alpha_2 m+\alpha_3\\ C_3=\alpha_4 m^2+\alpha_5 m +\alpha_6$$ The objective is to make all six $\alpha_k$ vanish, and that frees up $m$ to solve $C_4=0$. I obtained the following explicit expressions for the six $\alpha_k$, $$\alpha_1=-16a+7b-6c-5d-8e-7n-18\\ \alpha_2=-18a-16b+7c-6d-5e-8f-25\\ \alpha_3=104a^2-117ab+5ac+41ad+102ae+11af-25ag+96an+220a-21b^2-81bc-61bd-45be-25bf-18bg-42bn-174b+5c^2+41cd+23ce-18cf-16cg+36cn+103c+22de-16df+7dg+30dn-d+24e^2+7ef-6eg+48en+105e-3f^2-5fg-26f-4g^2+11g+21n^2+108n+33\\ \alpha_4=-25a-18b-16c+7d-6e-5f-8g+11\\ \alpha_5=220a^2+34ab-14ac+4ad+146ae+76af+22ag+90an+208a-117b^2-37bc-40bd+41be-34bf-50bg+80bn-23b-81c^2-51cd-4ce-2cf-36cg-35cn-72c+41d^2+23de+4df-32dg+30dn+223d+22e^2+32ef+14eg+25en+108e+7f^2-12fg+40fn+66f-5g^2-52g+125n+336\\ \alpha_6=-502a^3+302a^2b+205a^2c+6a^2d-569a^2e-290a^2f+142a^2g-520a^2n-989a^2-326ab^2 +493abc+422abd+350abe-39abf-23abg+585abn+1178ab+313ac^2-195acd+86acf+310acg-25acn-614ac-63ad^2-46ade+105adf+3adg-205adn-196ad-224ae^2-210aef+114aeg-510aen-864ae-43af^2+2afg-55afn-88af+38ag^2+125agn+46ag-240an^2- 1100an-695a-294b^3-300b^2c-191b^2d-13b^2e-243b^2f-174b^2g+105b^2n-464b^2+155bc^2+122bcd+271bce-72bcf-14bcg+405bcn+457bc+62bd^2+211bde+3bdf-43bdg+305bdn+505bd+75be^2+67bef+24beg+225ben+615be-60bf^2-87bfg+125bfn+40bf- 17bg^2+90bgn-115bg+105bn^2+870bn+890b-44c^3-159c^2d+62c^2e+103c^2f+5c^2g-25c^2n-158c^2-103cd^2-105cde+4cdf-40cdg-205cdn-473cd-35ce^2+24cef+112ceg-115cen-368ce-8cf^2+52cfg+90cfn-36cf-cg^2+80cgn+86cg-90cn^2-515cn-718c+40d^3-d^2e+41d^2f-61d^2g+341d^2-15de^2+41def-4deg-110den+11de+26df^2-25dfg+80dfn+282df+2dg^2-35dgn-277dg-75dn^2+5dn+967d-30e^3-45e^2f+23e^2g-120e^2n-188e^2-ef^2+4efg-35efn-14ef+16eg^2+30egn+10eg-120en^2-525en-189e-6f^3-16f^2g+15f^2n+49f^2+7fg^2+25fgn-92fg+130fn+370f-2g^3+20g^2n+33g^2-55gn-351g-35n^3-270n^2-165n+822$$

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II. Equations

Let $\alpha_1=\alpha_2=\alpha_4=0$ and solve for $(f,g,n)$, $$f=(-18a - 16b + 7c - 6d - 5e - 25)/8\\ g=(-110a - 64b - 163c + 86d - 23e + 213)/64\\ n=(-16a + 7b - 6c - 5d - 8e - 18)/7$$ Substitute into $\alpha_3=\alpha_5=\alpha_6=0$, $$A=-242532a^2+141004ac-203736ad-242532ae-650004a-18129c^2-62044cd+70502ce-53586c+48892d^2-101868de+93700d-60633e^2-325002e-68801=0\\ \color{red}B=-59604a^2-242532ac+369032ad-59604ae+678460a+35251c^2-101868cd-121266ce-325002c+8204d^2+184516de+217332d-14901e^2+339230e+386947=0$$

and collecting $C=0$ by the parameter $b$,

$$C=752771623 + 2613786290 a - 946683852 a^2 + 71929048 a^3 + 269736861 c + 481160836 a c + 160630740 a^2 c - 598396411 c^2 - 310195118 a c^2 + 103073423 c^3 + 324027638 d + 2821171896 a d - 503923944 a^2 d + 276847436 c d + 255337592 a c d - 235467818 c^2 d - 233229740 d^2 + 758943688 a d^2 + 57260788 c d^2 - 103188920 d^3 + 1610259593 e - 414771212 a e + 61164036 a^2 e - 14221150 c e - 29514348 a c e - 127460775 c^2 e + 1580974236 d e - 214602856 a d e + 47804284 c d e + 385903780 d^2 e + 29285357 e^2 + 7217250 a e^2 - 54914859 c e^2 + 18679558 d e^2 - 2691253 e^3 + 1568 \color{red}B\, b = 0$$

where the linear term has the factor $B$.

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III. Radicals?

Note that $(A,B)$ have no parameter $b$. But if $B=0$, then $C=0$ has no $b$ as well. Thus, there are only four parameters $(a,c,d,e)$ to solve three non-linear equations (2 quadratics and 1 cubic) which does not seem to be enough to solve in radicals, just like in this related post.

Answer 2

I. Coefficients $\pmb{C_k}$

Given the sample nonic, $$x^9 + x^5 - x^4 + 2x^3 + x^2 - x + 1$$ and the deg-$10$ Tschirnhaus transformation, $$y=x^{10}+ax^9+bx^8+cx^7+dx^6+ex^5+fx^4+gx^3+hx^2+mx+n$$ Using resultants to eliminate $x$ between the two yields a new septic, $$y^10+C_1y^9+C_2y^8+C_3y^7+C_4y^6+\dots+C_{10}=0$$ Collecting coefficient with respect to $\color{red}{m}$, the $C_k$ above have the form, $$C_1=\alpha_1\\ C_2=\alpha_2 m+\alpha_3\\ C_3=\alpha_4 m^2+\alpha_5 m +\alpha_6$$ The objective is to make all six $\alpha_k$ vanish, and that frees up $m$ to solve $C_4=0$. I obtained the following explicit expressions for the six $\alpha_k$, $$\alpha_1=18a-12b+7c+12d-5e+4f-9n-25\\ \alpha_2=-25a+18b-12c+7d+12e-5f+4g+11\\ \alpha_3=60a^2-114ab+146ac+151ad-13ae+7af+4ag+11ah-144an-146a+82b^2-149bc-67bd-5be-44bf+11bg-25bh+96bn+96b+63c^2+19cd-31ce+39cf-25cg+18ch-56cn-73c+74d^2-49de+23df+18dg-12dh-96dn-280d-2ef- 12eg+7eh+40en+60e+2f^2+7fg+12fh-32fn-23f+6g^2-5gh-65g+2h^2+4h+36n^2+200n+222\\ \alpha_4=11a-25b+18c-12d+7e+12f-5g+4h+4\\ \alpha_5=-146a^2+216ab-187ac-134ad+211ae-36af-58ag+8ah+175an+461a-114b^2+310bc+2bd-80be+2bf-40bg+22bh-126bn+26b-149c^2+59cd+14ce-75cf+50cg-50ch+84cn-31c+19d^2+117de-10df-2dg+36dh-49dn+161d-49e^2+23ef+16eg-24eh-84en-315e-2f^2-8fg+14fh+35fn+39f+7g^2+24gh-28gn+54g-5h^2-130h-77n-373\\ \alpha_6=-120a^3+297a^2b+375a^2c+426a^2d+212a^2e-457a^2f-26a^2g+17a^2h-420a^2n+889a^2-153ab^2-456abc-950abd-578abe+200abf-212abg+26abh+798abn-6ab+400ac^2+1073acd-315ace-35acf+186acg-7ach-1022acn-1324ac+454ad^2+5ade-18adf-36adg+120adh-1057adn-577ad-81ae^2+189aef-32aeg+111aeh+91aen-613ae-122af^2+182afg+130afh-49afn+679af+43ag^2+64agh-28agn-324ag-29ah^2-77ahn-399ah+504an^2+1022an-965a-116b^3+480b^2c+533b^2d+272b^2e+99b^2f+172b^2g+96b^2h-574b^2n-421b^2-738bc^2-665bcd-120bce-92bcf-31bcg-187bch+1043bcn+263bc-145bd^2+31bde-328bdf+120bdg-116bdh+469bdn+739bd+121be^2-52bef+129beg-89beh+35ben+177be+12bf^2-170bfg-90bfh+308bfn+306bf+5bg^2-70bgh-77bgn+309bg-20bh^2+175bhn+215bh-336bn^2-672bn-647b+294c^3+120c^2d+17c^2e+125c^2f-73c^2g+146c^2h-441c^2n+15c^2+42cd^2-378cde+237cdf-134cdg +2cdh-133cdn-859cd+82ce^2-13cef-89ceg+113ceh+217cen+756ce-7cf^2+103cfg+26cfh-273cfn-224cf-23cg^2-27cgh+175cgn-192cg+25ch^2-126chn+186ch+196cn^2+511cn+1236c+244d^3-136d^2e+16d^2f+151d^2g-67d^2h-518d^2n-1137d^2-35de^2+15def-80deg+14deh+343den+149de+df^2+26dfg+104dfh-161dfn+67df+76dg^2-38dgh-126dgn-622dg-dh^2+84dhn-60dh+ 336dn^2+1960dn+1345d+20e^3-13e^2f-5e^2g-31e^2h+144e^2-3ef^2-75efg-10efh+14efn+13ef-19eg^2-25egh+84egn+217eg+8eh^2-49ehn+9eh-140en^2-420en+110e-12f^3+39f^2g+23f^2h-14f^2n+54f^2-fg^2+16fgh-49fgn-231fg-4fh^2-84fhn- 244fh+112fn^2+161fn-161f+6g^3-12g^2h-42g^2n-130g^2+7gh^2+35ghn-5gh+455gn+751g+4h^3-14h^2n+27h^2-28hn+373h-84n^3-700n^2-1554n-306$$

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II. Equations

Let $\alpha_1=\alpha_2=\alpha_4=0$ and solve for $(g,h,n)$, $$g=(25a - 18b + 12c - 7d - 12e + 5f - 11)/4\\ h=(81a + 10b - 12c + 13d - 88e - 23f - 71)/16\\ n=(18a - 12b + 7c + 12d - 5e + 4f - 25)/9$$ Substitute into $\alpha_3=\alpha_5=\alpha_6=0$, $$A=142983a^2-255060ab+149688ac-95850ad-153936ae+132030af-322146a+102780b^2-102480bc+61308bd+159456be-95604bf+289356b+22000c^2-21288cd-98368ce+51320cf-179720c-5121d^2+60144de-35706df+141798d+21184e^2-111568ef+205360e+10231f^2-116786f+148567\\ B=111787a^2-136964ab+62680ac-36162ad-218640ae+4934af-192346a+38924b^2-36112bc+28812bd+154720be+17756bf+90396b+11440c^2-20360cd-77888ce-15208cf-27112c+9155d^2+43056de+6894df+10958d+87744e^2-43152ef+218736e-19109f^2+26390f+77499\\ C=67646097a^3-117392247a^2b-19702629a^2c-8508240a^2d-183204855a^2e-7430805a^2f-186999246a^2+66617640ab^2+29432916abc+15318558abd+209731140abe+8228898abf+211237038ab-27381240ac^2+18408870acd+47996280ace-6640218acf+27636606ac-227691ad^2-881766ade-9110880adf+26582742ad+166517424ae^2+19754658aef+347536818ae+1847043af^2+20278404af+163568673a-13550220b^3-4303044b^2c-11559348b^2d-58311900b^2e-1152792b^2f-53905500b^2+10525824bc^2 +1620bcd-51193872bce-12985740bcf-16448796bc-5365035bd^2+9581652bde+17898894bdf-22487382bd-87760224be^2-3386268bef-215926524be+263961bf^2-26655714bf-81046971b-3549520c^3+4227192c^2d+41689680c^2e+11528520c^2f+11625528c^2-438885cd^2-43245864cde-19889514cdf+2358414cd-19917696ce^2+27087672cef-31838088ce+12071499cf^2+2386734cf-16190877c-879930d^3+11077857d^2e+8625501d^2f-5758020d^2+3056688de^2-5230998def-16042950de-8008560df^2+7876998df-11471850d-57158848e^3-26280912e^2f-143293200e^2-11160615ef^2+7797570ef-169645719e-1556695f^3 +10831602f^2-24909303f-42395840$$

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III. Radicals

To solve this in radicals, we first do the following substitution: $$f=1,d=d_1+d_2,e=e_1+e_2$$ And split $A = B = 0$ into the following parts: $$A=A_{20}+A_{11}+A_{02}+A_{10}+A_{01}+A_{00},$$ $$B=B_{20}+B_{11}+B_{02}+B_{10}+B_{01}+B_{00}.$$ Where $A_{ij},B_{ij}$ has $i,j$ being the corresponding degree of $(a,d_1,e_1),(b,c,d_2,e_2)$.

The first pair:

$$A_{20}=142983a^2-95850ad_1-153936ae_1-5121d_1^2+60144d_1e_1+21184e_1^2=0$$ $$B_{20}=111787a^2-36162ad_1-218640ae_1+9155d_1^2+43056d_1e_1+87744e_1^2=0$$

then solve for $\color{blue}{(d_1,e_1)}$. Since each quartic has four roots, the correct pairing should be chosen such that $A_{20} = B_{20} = 0$. One pair is,

$$d_1\approx0.740645701664\,a \\ e_1\approx0.737877453787\,a $$

with "$a$" to be determined later.

The second pair:

$$M_1 = A_{11}+A_{10}=-255060ab+149688ac-95850ad_2-153936ae_2-190116a+61308bd_1+159456be_1-21288cd_1-98368ce_1-10242d_1d_2+60144d_1e_2+106092d_1+60144d_2e_1+42368e_1e_2+93792e_1=0$$

$$N_1 = B_{11}+B_{10}=-136964ab+62680ac-36162ad_2-218640ae_2-187412a+28812bd_1+154720be_1-20360cd_1-77888ce_1+18310d_1d_2+43056d_1e_2+17852d_1+43056d_2e_1+175488e_1e_2+175584e_1=0$$

We then solve for $\color{blue}{(d_2,e_2)}$ in $M_1 = N_1 = 0$ which is easily done since they are just linear.

The third pair:

$$M_2=A_{02}+A_{01}+A_{00}=102780b^2-102480bc+61308bd_2+159456be_2+193752b+22000c^2-21288cd_2-98368ce_2-128400c-5121d_2^2+60144d_2e_2+106092d_2+21184e_2^2+93792e_2+42012=0$$

$$N_2=B_{02}+B_{01}+B_{00}=38924b^2-36112bc+28812bd_2+154720be_2+108152b+11440c^2-20360cd_2-77888ce_2-42320c+9155d_2^2+43056d_2e_2+17852d_2+87744e_2^2+175584e_2+84780$$ We first substitute our expressions for $\color{blue}{(d_2,e_2)}$ into $M_2 = N_2 = 0$. They remain quadratic in $(b,c)$. Then eliminate $c$ using resultants to get a quartic in $b$, then eliminate $b$ to get a quartic in $c$. Substituting $\color{blue}{(d_1,e_1)}$ into the two quartics, we can solve for $(b,c)$. As before, choose the correct pairing of roots so that $M_2 = N_2 = 0$. One pair is,

$$b=c=0$$

which also yields,

$$d_2\approx0.259354298335\\ e_2\approx-0.737877453787$$

where the variable "$a$" disappears from both pairs. Recall that,

$$d_1\approx0.740645701664\,a \\ e_1\approx0.737877453787\,a $$

and we now have all unknowns except $a$. Substitute all values into $C=0$ (which is just a cubic in "$a$") defined in the Equations Section and any of its three roots can set $A=B=C=0$, one of which is,

$$a=1$$

Since the last variable $m$ is used to solve,

$$C_4 = m^4 = 0$$ $$\implies m=0$$

then we eliminated four terms $C_1 = C_2 = C_3 = C_4 = 0$ of the nonic using a nonic Tschirnhausen transformation with coefficients in the radicals and the reduced nonic being, $$y^9=0$$