Determine whether there exist two reals $x,y$ and a sequence $\{a_n\}_{n=0}^{\infty}$ of nonzero reals such that $a_{n+2}=xa_{n+1}+ya_n$ for all $n\ge0$ and for every positive real number $r$, there exist positive integers $i,j$ such that $|a_i|<r<|a_j|$.

[i]Alex Zhu.[/i]
1. **Define the sequence and parameters:**
   Let \(\lambda = 1.5 e^{\pi i \alpha}\) where
   \[
   \alpha := \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^{2^2}} + \frac{1}{2^{2^{2^2}}} + \cdots.
   \]
   Since \(\alpha\) is an infinite sum of fractions with denominators that are powers of 2, it is irrational. If \(\alpha\) were rational, its binary expansion would be periodic, which is not the case here.

2. **Define \(x\) and \(y\):**
   Let \(x = \lambda + \bar{\lambda}\) and \(y = -\lambda \bar{\lambda}\). Since \(\lambda\) is a complex number, \(\bar{\lambda}\) is its complex conjugate. Therefore, \(x\) and \(y\) are real numbers.

3. **Define the sequence \(\{a_n\}\):**
   Let \(a_n = \Im(\lambda^n)\), where \(\Im(z)\) denotes the imaginary part of \(z\).

4. **Verify the recurrence relation:**
   We need to show that \(a_{n+2} = x a_{n+1} + y a_n\). Note that
   \[
   \lambda^n = (\lambda + \bar{\lambda}) \lambda^{n-1} + (-\lambda \bar{\lambda}) \lambda^{n-2} = x \lambda^{n-1} + y \lambda^{n-2}.
   \]
   Taking the imaginary part of both sides, we get
   \[
   \Im(\lambda^n) = x \Im(\lambda^{n-1}) + y \Im(\lambda^{n-2}),
   \]
   which implies
   \[
   a_n = x a_{n-1} + y a_{n-2}.
   \]
   Thus, the sequence \(\{a_n\}\) satisfies the given recurrence relation.

5. **Show that \(a_n\) is never zero:**
   Since \(\alpha\) is irrational, \(\lambda^n\) will never be a purely real number for any integer \(n\). Therefore, \(\Im(\lambda^n) \neq 0\) for all \(n\), ensuring that \(a_n \neq 0\).

6. **Show the existence of \(i\) such that \(|a_i| < r\):**
   Define the function \(f(n) = 2^{f(n-1)}\) with \(f(1) = 2\). Then \(\alpha = f(1)^{-1} + f(2)^{-1} + \cdots\). Note that
   \[
   \chi(n) := f(n)^{-1} + f(n+1)^{-1} + \cdots \le \frac{1}{f(n)} + \frac{1}{2f(n)} + \frac{1}{4f(n)} + \cdots = \frac{2}{f(n)}.
   \]
   Combining the first \(n\) terms of \(\alpha\), we get
   \[
   \alpha = \frac{g(n)}{f(n)} + \chi(n+1).
   \]
   Thus,
   \[
   a_{f(n)} = \Im(1.5^{f(n)} e^{\pi i (g(n) + \chi(n+1) f(n))}) = \pm 1.5^{f(n)} \sin(\pi \chi(n+1) f(n)).
   \]
   Note that
   \[
   \theta(n) := \pi \chi(n+1) f(n) \le 2\pi \frac{f(n)}{f(n+1)} = 2\pi \frac{f(n)}{2^{f(n)}}.
   \]
   For large \(n\), \(\theta(n)\) becomes very small, so \(\sin(\theta(n)) \le \theta(n)\) by the Taylor expansion for \(\sin\). Therefore,
   \[
   0 < |a_{f(n)}| \le 1.5^{f(n)} \cdot \left(2\pi \frac{f(n)}{2^{f(n)}}\right).
   \]
   The right-hand side becomes arbitrarily small as \(f(n)\) grows larger, so \(|a_i|\) can be made arbitrarily small.

7. **Show the existence of \(j\) such that \(|a_j| > r\):**
   Since \(\alpha\) is irrational, \(\{\{j \alpha : j \in \mathbb{N}\}\}\) is dense in \([0, 1)\). Therefore, there are infinitely many \(j\) with \(\frac{1}{6} \le \{j \alpha\} \le \frac{5}{6}\). For such \(j\),
   \[
   |a_j| = |\Im(1.5^j e^{\pi i j \alpha})| = |1.5^j \sin(\pi \alpha)| \ge \frac{1}{2} 1.5^j \sin(\pi \alpha).
   \]
   The right-hand side grows arbitrarily large, so \(|a_j|\) can be made arbitrarily large.

Conclusion:
The construction shows that for every positive real number \(r\), there exist positive integers \(i\) and \(j\) such that \(|a_i| < r < |a_j|\).

The final answer is \(\boxed{\text{yes}}\).