Calculating the bi-tangents of x 4 + y 3 + 1 = 0 x^4+y^3+1=0

Algebra Level 5

For how many ordered pairs of complex numbers ( a , b ) (a,b) is the polynomial P ( x ) = x 4 + ( a x + b ) 3 + 1 P(x)=x^4+(ax+b)^3+1 equal to a square of a complex polynomial?


The answer is 27.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Dennis Gulko
Mar 6, 2014

Assume that P ( x ) = x 4 + ( a x + b ) 3 + 1 = x 4 + a 3 x 3 + 3 a 2 b x 2 + 3 a b 2 x + b 3 + 1 P(x)=x^4+(ax+b)^3+1=x^4+a^3x^3+3a^2bx^2+3ab^2x+b^3+1 is a square, i.e. P ( x ) = ( x 2 + α x + β ) 2 = x 4 + 2 α x 3 + ( α 2 + 2 β ) x 2 + 2 α β x + β 2 P(x)=(x^2+\alpha x+\beta)^2=x^4+2\alpha x^3+(\alpha^2+2\beta)x^2+2\alpha\beta x+\beta^2 .

We get a system of equations: a 3 = 2 α 3 a 2 b = α 2 + 2 β 3 a b 2 = 2 α β = ( 1 ) a 3 β b 3 + 1 = β 2 \begin{aligned} & a^3=2\alpha\\ &3a^2b=\alpha^2+2\beta\\ &3ab^2=2\alpha\beta\overset{(1)}{=}a^3\beta\\ &b^3+1=\beta^2 \end{aligned} Consider equation ( 3 ) (3) :

  1. If a = 0 a=0 then, by ( 1 ) (1) we have α = 0 \alpha=0 and then, by ( 2 ) (2) also β = 0 \beta=0 , so by ( 4 ) (4) , b 3 + 1 = 0 b^3+1=0 - and we have three solutions.

  2. Otherwise, a 0 a\neq0 , so we can divide by a a to get β = 3 b 2 a 2 \beta=3\frac{b^2}{a^2} (from ( 3 ) (3) ) and also α = 1 2 a 3 \alpha=\frac12a^3 . Substituting those into equtions ( 2 ) , ( 4 ) (2),(4) we get: 3 a 2 b = a 6 4 + 6 b 2 a 2 a 8 12 a 4 b + 24 b 2 = 0 b 3 + 1 = 9 b 4 a 4 \begin{aligned} &3a^2b=\frac{a^6}4+6\frac{b^2}{a^2} \hspace{5pt}\Rightarrow\hspace{5pt} a^8-12a^4b+24b^2=0\\ &b^3+1=9\frac{b^4}{a^4} \end{aligned} By solving the first equation for a 4 a^4 we get: a 4 = 12 b ± 144 b 2 96 b 2 2 = 6 b ± 12 b 2 = ( 6 ± 2 2 ) b a^4=\frac{12b\pm\sqrt{144b^2-96b^2}}{2}=6b\pm\sqrt{12b^2}=(6\pm2\sqrt{2})b . Substituting this to the second equation we have: 9 b 4 = ( b 3 + 1 ) a 4 = ( b 3 + 1 ) ( 6 ± 2 2 ) b ( 3 2 3 ) b 4 = ( 6 ± 2 3 ) b 9b^4=(b^3+1)a^4=(b^3+1)(6\pm2\sqrt{2})b \hspace{5pt}\Rightarrow\hspace{5pt} (3\mp2\sqrt{3})b^4=(6\pm2\sqrt{3})b So either b = 0 b=0 or b 3 = 6 ± 2 3 3 2 3 b^3=\frac{6\pm2\sqrt{3}}{3\mp2\sqrt{3}} (the signs match each other).

The option b = 0 b=0 is impossible since it implies a = 0 a=0 but that implies b 3 = 1 b^3=-1 as we have seen before. So we have the later option - which gives 6 6 solutions for b b . Each of those gives 4 4 solutions for a a (since the signs have to match). Hence we hve 6 4 = 24 6\cdot4=24 solutions here.

So there are 3 + 24 = 27 3+24=\boxed{27} pairs.

Patrick Corn
Mar 6, 2014

Any smooth plane quartic curve has 28 28 bitangents (see e.g. http://en.wikipedia.org/wiki/Bitangents of a_quartic). So the projective curve x 4 + z y 3 + z 4 = 0 x^4 + zy^3 + z^4 = 0 has 28 28 bitangents; the given curve is just the affine piece of it. The "line at infinity" z = 0 z = 0 is indeed a bitangent to this curve (it actually hits the curve at ( 1 , 0 , 0 ) (1,0,0) with multiplicity 4 4 ), so there are 27 \fbox{27} affine bitangents.

Computing them explicitly is a bit of a mess!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...