Examples of polynomials $x_1(t), x_2(t)$ such that $(x_1(t))^4 + (x_2(t))^4$ has a double root 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?
 A: 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 $\mathbb{Q}(\sqrt2)[t]$.
A: 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).
A: 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.
A: 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 $x_2$. 
