4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

4 Answers 4

13
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ How did you come up with this example? $\endgroup$ Apr 17, 2015 at 1:29
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $\mathbb{Q}(\sqrt2)[t]$.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $x_2$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.