Are there positive integers $m,n$ such that there exist at least $2012$ positive integers $x$ such that both $m-x^2$ and $n-x^2$ are perfect squares?

[i]David Yang.[/i]
To determine if there exist positive integers \( m \) and \( n \) such that there are at least 2012 positive integers \( x \) for which both \( m - x^2 \) and \( n - x^2 \) are perfect squares, we need to analyze the conditions under which \( m - x^2 = a^2 \) and \( n - x^2 = b^2 \) for some integers \( a \) and \( b \).

1. **Express the conditions mathematically:**
   \[
   m - x^2 = a^2 \quad \text{and} \quad n - x^2 = b^2
   \]
   Rearrange these equations to:
   \[
   m = x^2 + a^2 \quad \text{and} \quad n = x^2 + b^2
   \]

2. **Subtract the two equations:**
   \[
   m - n = (x^2 + a^2) - (x^2 + b^2) = a^2 - b^2
   \]
   This simplifies to:
   \[
   m - n = (a - b)(a + b)
   \]

3. **Analyze the equation \( m - n = (a - b)(a + b) \):**
   For \( m - n \) to be expressed as a product of two factors in at least 2012 different ways, \( m - n \) must have at least 2012 pairs of factors \((a - b)\) and \((a + b)\).

4. **Consider the number of divisors:**
   The number of ways to express \( m - n \) as a product of two factors is related to the number of divisors of \( m - n \). For \( m - n \) to have at least 2012 pairs of factors, it must have at least \( 2 \times 2012 = 4024 \) divisors (since each pair \((a - b, a + b)\) corresponds to two divisors).

5. **Find a suitable \( m - n \):**
   To find such an \( m - n \), we need a number with at least 4024 divisors. The number of divisors of a number \( k \) is given by the product of one more than each of the exponents in its prime factorization. For example, if \( k = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r} \), then the number of divisors of \( k \) is \((e_1 + 1)(e_2 + 1) \cdots (e_r + 1)\).

6. **Construct an example:**
   Consider \( k = 2^{10} \times 3^5 \times 5^4 \). The number of divisors of \( k \) is:
   \[
   (10 + 1)(5 + 1)(4 + 1) = 11 \times 6 \times 5 = 330
   \]
   This is not sufficient. We need a larger number. Let's try \( k = 2^{10} \times 3^5 \times 5^4 \times 7^3 \):
   \[
   (10 + 1)(5 + 1)(4 + 1)(3 + 1) = 11 \times 6 \times 5 \times 4 = 1320
   \]
   Still not sufficient. We need to increase the exponents further.

7. **Find a larger \( k \):**
   Consider \( k = 2^{10} \times 3^5 \times 5^4 \times 7^3 \times 11^2 \):
   \[
   (10 + 1)(5 + 1)(4 + 1)(3 + 1)(2 + 1) = 11 \times 6 \times 5 \times 4 \times 3 = 3960
   \]
   This is sufficient as it has more than 4024 divisors.

8. **Conclusion:**
   Therefore, we can choose \( m \) and \( n \) such that \( m - n = 2^{10} \times 3^5 \times 5^4 \times 7^3 \times 11^2 \). This ensures that there are at least 2012 positive integers \( x \) such that both \( m - x^2 \) and \( n - x^2 \) are perfect squares.

The final answer is \( \boxed{ m } \) and \( n \) such that \( m - n = 2^{10} \times 3^5 \times 5^4 \times 7^3 \times 11^2 \).