Let $n$ be a positive integer. 
- Find, in terms of $n$, the number of pairs $(x,y)$ of positive integers that are solutions of the equation : $$x^2-y^2=10^2.30^{2n}$$
- Prove further that this number is never a square 
1. We start by rewriting the given equation \( x^2 - y^2 = 10^2 \cdot 30^{2n} \) using the difference of squares formula:
   \[
   x^2 - y^2 = (x - y)(x + y) = 10^2 \cdot 30^{2n}
   \]
   This implies that \( (x - y) \) and \( (x + y) \) are factors of \( 10^2 \cdot 30^{2n} \).

2. Next, we factorize \( 10^2 \cdot 30^{2n} \):
   \[
   10^2 \cdot 30^{2n} = 2^2 \cdot 5^2 \cdot (2 \cdot 3 \cdot 5)^{2n} = 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2}
   \]

3. We need to count the number of pairs \((a, b)\) such that \( a = x - y \) and \( b = x + y \) are factors of \( 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2} \) and satisfy \( ab = 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2} \).

4. The number of ways to choose \( a \) and \( b \) such that \( ab = 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2} \) is given by the number of divisors of \( 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2} \). The number of divisors of a number \( p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is \( (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \).

5. Therefore, the number of divisors of \( 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2} \) is:
   \[
   (2n+2+1)(2n+1)(2n+2+1) = (2n+3)(2n+1)(2n+3)
   \]

6. Since each pair \((a, b)\) corresponds to a unique pair \((x, y)\), the number of pairs \((x, y)\) is:
   \[
   \frac{(2n+3)(2n+1)(2n+3)}{2} = \frac{(2n+3)^2(2n+1)}{2}
   \]

7. Simplifying, we get:
   \[
   \frac{(2n+3)^2(2n+1)}{2} = \frac{4n^2 + 12n + 9}{2} \cdot (2n+1) = 2n^3 + 7n^2 + 7n + 3
   \]

8. To prove that this number is never a square, we assume for contradiction that \( 2n^3 + 7n^2 + 7n + 3 \) is a perfect square. Let \( k^2 = 2n^3 + 7n^2 + 7n + 3 \).

9. Consider the polynomial \( f(n) = 2n^3 + 7n^2 + 7n + 3 \). We need to show that \( f(n) \) cannot be a perfect square for any integer \( n \).

10. We analyze the expression modulo small primes. For instance, modulo 3:
    \[
    f(n) \equiv 2n^3 + n^2 + n + 3 \pmod{3}
    \]
    Simplifying, we get:
    \[
    f(n) \equiv 2n^3 + n^2 + n \pmod{3}
    \]
    We check the values for \( n = 0, 1, 2 \):
    \[
    f(0) \equiv 0 \pmod{3}, \quad f(1) \equiv 2 + 1 + 1 = 4 \equiv 1 \pmod{3}, \quad f(2) \equiv 16 + 4 + 2 = 22 \equiv 1 \pmod{3}
    \]
    Since \( f(n) \equiv 0 \pmod{3} \) only for \( n \equiv 0 \pmod{3} \), we need to check further.

11. We can also check modulo 4:
    \[
    f(n) \equiv 2n^3 + 3n^2 + 3n + 3 \pmod{4}
    \]
    Simplifying, we get:
    \[
    f(n) \equiv 2n^3 + 3n^2 + 3n + 3 \pmod{4}
    \]
    We check the values for \( n = 0, 1, 2, 3 \):
    \[
    f(0) \equiv 3 \pmod{4}, \quad f(1) \equiv 2 + 3 + 3 + 3 = 11 \equiv 3 \pmod{4}, \quad f(2) \equiv 16 + 12 + 6 + 3 = 37 \equiv 1 \pmod{4}, \quad f(3) \equiv 54 + 27 + 9 + 3 = 93 \equiv 1 \pmod{4}
    \]
    Since \( f(n) \equiv 3 \pmod{4} \) for \( n \equiv 0, 1 \pmod{4} \), it cannot be a perfect square.

12. Therefore, \( 2n^3 + 7n^2 + 7n + 3 \) is never a perfect square.

\(\blacksquare\)

The final answer is \( \boxed{ 2n^3 + 7n^2 + 7n + 3 } \)