Find all pairs of positive integers $(a,b)$ satisfying the following conditions:
[list]
[*] $a$ divides $b^4+1$,
[*] $b$ divides $a^4+1$,
[*] $\lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor$.
[/list]

[i]Yang Liu[/i]
1. **Assume $\gcd(a, b) = 1$:**
   - Suppose $p$ is a prime that divides both $a$ and $b$. Then $p$ would divide both $b^4 + 1$ and $a^4 + 1$. This implies $p$ divides 1, which is a contradiction. Therefore, $\gcd(a, b) = 1$.

2. **Analyze the divisibility conditions:**
   - Since $a \mid b^4 + 1$, we have $a \mid (b - a)^4 + 1$.
   - Similarly, since $b \mid a^4 + 1$, we have $b \mid (a - b)^4 + 1$.
   - Given $\gcd(a, b) = 1$, it follows that $ab \mid (b - a)^4 + 1$.

3. **Express $a$ and $b$ in terms of $x$:**
   - Let $a = x^2 + a_1$ and $b = x^2 + b_1$ with $0 \leq a_1, b_1 \leq 2x$.
   - Then $(b - a)^4 + 1 \leq 16x^4 + 1$.
   - Therefore, $\frac{(b - a)^4 + 1}{ab} \leq \frac{16x^4 + 1}{x^4} \leq 17$.

4. **Consider the possible values of $\frac{(b - a)^4 + 1}{ab}$:**
   - If $\frac{(b - a)^4 + 1}{ab} = 17$, then $x = 1$, $a - b = 2$, and $ab = 1$, which is impossible.
   - Therefore, $\frac{(b - a)^4 + 1}{ab} \leq 16$ and must be an integer. Let $k = \frac{(b - a)^4 + 1}{ab}$.

5. **Analyze the value of $k$:**
   - Since $k \mid (b - a)^4 + 1$, for every odd prime dividing $k$, there must exist a number with order 8, so it must be $1 \mod 8$.
   - The smallest prime $1 \mod 8$ is 17, which is greater than 16. Therefore, $k$ is a power of 2.
   - Since $(b - a)^4 \not\equiv 3 \pmod 4$, we get $k = 1, 2$.

6. **Consider $k = 2$:**
   - If $k = 2$, then $2 \mid (b - a)^4 + 1$, so $a \not\equiv b \pmod 2$.
   - This implies $ab \equiv 0 \pmod 2$, so $4 \mid (b - a)^4 + 1$, which is a contradiction.

7. **Consider $k = 1$:**
   - If $k = 1$, then $ab = (b - a)^4 + 1$.
   - Assume without loss of generality $b \geq a$, and let $b = a + n$.
   - Then $a(a + n) = n^4 + 1$, so $a^2 + an - (n^4 + 1) = 0$.
   - The discriminant of this equation must be a perfect square, so $n^2 + 4n^4 + 4$ must be a perfect square.

8. **Analyze the discriminant:**
   - For $n \geq 2$, we have $4n^4 + n^2 + 4 > 4n^4$ and $4n^4 + n^2 + 4 < (2n^2 + 1)^2 = 4n^4 + 4n^2 + 1$.
   - Therefore, it cannot be a square for $n \geq 2$.

9. **Consider $n = 0$ and $n = 1$:**
   - If $n = 0$, then $a^2 = 1$, so $a = 1$ and $b = 1$.
   - If $n = 1$, then $a^2 + a - 2 = 0$, so $a = 1$ and $b = 2$.

Therefore, the only solutions are $\boxed{(1, 1)}$, $\boxed{(1, 2)}$, and $\boxed{(2, 1)}$.