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Are there examples of polynomials $x_1(t), x_2(t) \in \mathbb{Q}[t]$ of equal degree at least one, with $\gcd(x_1(t), x_2(t)) = 1$, such that the sum $(x_1(t))^4 + (x_2(t))^4$ is divisible by the square of a polynomial $z(t)$ which is defined over $\mathbb{Q}$? If this question is too much to ask for, can we find examples such that $(x_1(t))^4 + (x_2(t))^4$ has a double root in $\mathbb{C}$?

Edit: in view of the answer given by Jarek Kuben, I am refining the question to the following: suppose we want the degree of $x_1, x_2$ to be at most $3$. Can one produce examples still?

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  • $\begingroup$ I presume you don't want $x_1$ and $x_2$ have a common divisor? $\endgroup$ Apr 16, 2015 at 19:46
  • $\begingroup$ Yes; I forgot to include that in the body of the question. Thank you for pointing that out. $\endgroup$ Apr 16, 2015 at 19:58
  • $\begingroup$ If some polynomial $f(t)\in\mathbb{Q}[t]$ has a multiple root in $\mathbb{C}$, then it's divisible by the square of (non-constant) polynomial $\operatorname{GCD}(f(t)/\operatorname{GCD}(f(t),f'(t)),f'(t))\in\mathbb{Q}[t]$. $\endgroup$ Apr 16, 2015 at 21:31

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Here's an example: $x_1(t)=3t^4 + 4t - 1$, $x_2(t)=t^4 - 4t^3 - 3$. Then

$\begin{align*}(x_1&(t))^4 + (x_2(t))^4=\\ &82t^{16} - 16t^{15} + 96t^{14} + 176t^{13} + 136t^{12} + 144t^{11} + 288t^{10} + 336t^9 +\\ &108t^8 + 336t^7 + 288t^6 + 144t^5 + 136t^4 + 176t^3 +96t^2 - 16t + 82=\\ &2\cdot(t^4 + 1)^2 \cdot (41t^8 - 8t^7 + 48t^6 + 88t^5 - 14t^4 + 88t^3 + 48t^2 - 8t + 41). \end{align*}$

Also $x_1(t)$ and $x_2(t)$ are coprime since $\operatorname{Res}(x_1(t),x_2(t))=-3072\ne0$.

EDIT: Such examples can be found by factorization described by Joe Silverman. Simply choose some $f(t)\in\mathbb{Q}(\zeta)[t]$, where $\zeta=e^{2\pi i/8}$ (in the example above it's $f(t)=t-\zeta$) and look for non-zero $g(t)\in\mathbb{Q}(\zeta)[t]$ such that $f(t)^2g(t)$ is of the form $x_1(t)-\zeta x_2(t)$ for some $x_1(t),x_2(t)\in\mathbb{Q}[t]$ (in this example it turns out that $g(t)$ must be at least quadratic).

As for the refined question, for quadratic $x_1(t)$ and $x_2(t)$ it's easy to calculate that $x_1(t)-\zeta x_2(t)$ has multiple root only if both $x_1(t)$ and $x_2(t)$ have the same multiple root, thus the answer is no. For cubic polynomials it seems much more difficult, it leads to the question whether certain hypersurface in $\mathbb{P}^2$ of degree $8$ has any rational points.

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  • $\begingroup$ How did you come up with this example? $\endgroup$ Apr 17, 2015 at 1:29
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    $\begingroup$ I've edited the answer to address this. $\endgroup$ Apr 17, 2015 at 15:45
  • $\begingroup$ Are there any examples possible with $\deg g = 1$? $\endgroup$ Apr 18, 2015 at 23:11
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Let $\zeta$ be a primitive 8'th root of unity. Assuming $x_1$ and $x_2$ have no common roots (which you certainly want to assume or else there's a trivial solution), the factors in $$ x_1^4+x_2^4 = (x_1-\zeta x_2)(x_1-\zeta^3 x_2)(x_1-\zeta^5 x_2)(x_1-\zeta^7 x_2) $$ are relatively prime in $\mathbb{C}[t]$, so the only way to get a square factor in $x_1^4+x_2^4$ is for at least one of the factors $x_1-\zeta^k x_2$ to have a square factor.

So now suppose (WLOG) $x_1-\zeta x_2 = y^2z$. If you require $x_1,x_2\in\mathbb{Q}[t]$, then we can take complex conjugates to get $x_1-\overline \zeta x_2 = \overline y^2\overline z$, which lets us solve $$ x_1=\frac{\overline \zeta y^2 z-\zeta\overline y^2\overline z}{\overline\zeta-\zeta} \quad\text{and}\quad x_2=\frac{y^2z-\overline y^2\overline z}{\overline\zeta-\zeta}. $$ Of course, if you choose $y$ and $z$ to have real coefficients, you get $x_1=1$ and $x_2=0$. But if you choose $y$ to satisfy $\overline y^2\ne y^2$, then you'll get an answer to your question, at least over $\mathbb C$.

ADDENDUM: Alex seems to have posted something similar while I was typing, but I'll post my answer, too.

2nd ADDENDUM: Take $y\in\mathbb{Q}(i)[t]$ and $z=1$, then I think you get a solution in $\mathbb{Q}(\sqrt2)[t]$.

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For the second question, the condition that the discriminant is zero describes a hypersurface in $\mathbb{C}^{2n},$ and you are asking for a rational point on this hypersurface away from the hypersurface which specifies that the polynomials be relatively prime. This is probably a priori undecidable, but Magma might work for low degrees.

The first question is similar (saying that $p^2$ of a certain degree divides your form is an algebraic equation; you will have one for each possible degree).

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Over $\Bbb C$ this is easy; over $\Bbb Q$ it seems impossible. Let $\epsilon$ be a $4$-th root of $-1$. Then you have $x_1^4+x_2^4=(x_1-\epsilon x_2)(x_1+\epsilon x_2)(x_1-\epsilon^3x_2)(x_1+\epsilon^3x_2)$, and either one of the factors has a double root $t_0$, or two factors have a common root $t_0$. In the latter case, we have $x_1(t_0)=x_2(t_0)=0$, which you probably don't want. The former case is quite thinkable over $\Bbb C$ but, since $\epsilon$ is irrational, a rational root would still be common for $x_1$ and $x_2$.

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  • $\begingroup$ In fact, the OP asks for a double irreducible factor, which always exists even if $\epsilon$ is irrational (just take the minimal polynomial of the root). $\endgroup$ Apr 16, 2015 at 20:06
  • $\begingroup$ Oops. Yes, I guess you are right. Not always, of course, but one can probably find a suitable pair. $\endgroup$ Apr 16, 2015 at 20:08

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