Let $a,x,y$ be positive integer such that $a>100,x>100,y>100$ and $y^2-1=a^2(x^2-1)$  . Find the minimum value of $\frac{a}{x}$.
Given the equation \( y^2 - 1 = a^2(x^2 - 1) \), we need to find the minimum value of \( \frac{a}{x} \) under the conditions \( a > 100 \), \( x > 100 \), and \( y > 100 \).

1. **Rewrite the given equation:**
   \[
   y^2 - 1 = a^2(x^2 - 1)
   \]
   This can be rearranged as:
   \[
   y^2 = a^2x^2 - a^2 + 1
   \]

2. **Assume \( \frac{a}{x} < 2 \):**
   If \( \frac{a}{x} < 2 \), then \( a < 2x \). We need to check if this assumption leads to a contradiction.

3. **Consider the expression \( (ax - 1)^2 + 2ax - a^2 \):**
   \[
   (ax - 1)^2 + 2ax - a^2 = y^2
   \]
   Expanding \( (ax - 1)^2 \):
   \[
   (ax - 1)^2 = a^2x^2 - 2ax + 1
   \]
   Adding \( 2ax - a^2 \):
   \[
   a^2x^2 - 2ax + 1 + 2ax - a^2 = a^2x^2 - a^2 + 1
   \]
   Thus, we have:
   \[
   y^2 = a^2x^2 - a^2 + 1
   \]

4. **Check the inequality \( 2ax - a^2 > 0 \):**
   Since \( 2ax - a^2 > 0 \), we have:
   \[
   2ax > a^2 \implies 2x > a \implies \frac{a}{x} < 2
   \]
   This is consistent with our assumption.

5. **Verify the minimum value of \( \frac{a}{x} \):**
   To find the minimum value, we need to check if \( \frac{a}{x} = 2 \) is achievable. If \( \frac{a}{x} = 2 \), then \( a = 2x \).

6. **Substitute \( a = 2x \) into the original equation:**
   \[
   y^2 - 1 = (2x)^2(x^2 - 1)
   \]
   Simplifying:
   \[
   y^2 - 1 = 4x^2(x^2 - 1) = 4x^4 - 4x^2
   \]
   Thus:
   \[
   y^2 = 4x^4 - 4x^2 + 1
   \]
   We need to check if \( y \) is an integer. Let \( y = 2x^2 - 1 \):
   \[
   y^2 = (2x^2 - 1)^2 = 4x^4 - 4x^2 + 1
   \]
   This matches the equation, so \( y = 2x^2 - 1 \) is an integer.

Therefore, the minimum value of \( \frac{a}{x} \) is 2, which can be achieved by choosing \( (x, y, a) = (n, 2n^2 - 1, 2n) \) where \( n > 100 \).

The final answer is \(\boxed{2}\)