The roots `alpha and beta` of a quadratic equation are the square of two consecutive natural numbers. The geometric mean of the two roots is 1 greater than the difference of the roots. If `alpha and beta` lies between the roots of `x^2-cx+ 1 = 0` then find the minimum integral value of c.

Answer

To solve the problem step-by-step, we will follow the clues given in the question.

### Step 1: Define the roots
Let the two consecutive natural numbers be \( a \) and \( a + 1 \). Therefore, the roots \( \alpha \) and \( \beta \) can be defined as:
\[ \alpha = a^2 \quad \text{and} \quad \beta = (a + 1)^2 \]

### Step 2: Calculate the geometric mean and the difference of the roots
The geometric mean of the roots is given by:
\[ \sqrt{\alpha \beta} = \sqrt{a^2 \cdot (a + 1)^2} = a(a + 1) \]
The difference of the roots is:
\[ \beta - \alpha = (a + 1)^2 - a^2 = (a^2 + 2a + 1) - a^2 = 2a + 1 \]

### Step 3: Set up the equation based on the given condition
According to the problem, the geometric mean is one greater than the difference of the roots:
\[ a(a + 1) = (2a + 1) + 1 \]
This simplifies to:
\[ a(a + 1) = 2a + 2 \]
Rearranging gives:
\[ a^2 + a - 2a - 2 = 0 \quad \Rightarrow \quad a^2 - a - 2 = 0 \]

### Step 4: Solve the quadratic equation
Factoring the quadratic equation:
\[ (a - 2)(a + 1) = 0 \]
This gives us the solutions:
\[ a = 2 \quad \text{or} \quad a = -1 \]
Since \( a \) must be a natural number, we take \( a = 2 \).

### Step 5: Find the values of \( \alpha \) and \( \beta \)
Substituting \( a = 2 \):
\[ \alpha = 2^2 = 4 \quad \text{and} \quad \beta = (2 + 1)^2 = 3^2 = 9 \]

### Step 6: Analyze the quadratic equation \( x^2 - cx + 1 = 0 \)
The roots of this equation are given by:
\[ x = \frac{c \pm \sqrt{c^2 - 4}}{2} \]
We need \( \alpha \) and \( \beta \) (4 and 9) to lie between the roots of this equation. Thus, we need:
\[ \frac{c - \sqrt{c^2 - 4}}{2} < 4 < 9 < \frac{c + \sqrt{c^2 - 4}}{2} \]

### Step 7: Set up inequalities
1. From \( \frac{c - \sqrt{c^2 - 4}}{2} < 4 \):
\[ c - \sqrt{c^2 - 4} < 8 \quad \Rightarrow \quad c - 8 < \sqrt{c^2 - 4} \]
Squaring both sides:
\[ (c - 8)^2 < c^2 - 4 \]
Expanding and simplifying gives:
\[ c^2 - 16c + 64 < c^2 - 4 \quad \Rightarrow \quad -16c + 68 < 0 \quad \Rightarrow \quad c > \frac{68}{16} = 4.25 \]

2. From \( 9 < \frac{c + \sqrt{c^2 - 4}}{2} \):
\[ 18 < c + \sqrt{c^2 - 4} \quad \Rightarrow \quad c + \sqrt{c^2 - 4} > 18 \]
Rearranging gives:
\[ \sqrt{c^2 - 4} > 18 - c \]
Squaring both sides:
\[ c^2 - 4 > (18 - c)^2 \]
Expanding and simplifying gives:
\[ c^2 - 4 > 324 - 36c + c^2 \quad \Rightarrow \quad 36c > 328 \quad \Rightarrow \quad c > \frac{328}{36} \approx 9.11 \]

### Step 8: Combine the inequalities
From the inequalities, we have:
\[ c > 4.25 \quad \text{and} \quad c > 9.11 \]
Thus, the minimum integral value of \( c \) is:
\[ c = 10 \]

### Final Answer
The minimum integral value of \( c \) is \( \boxed{10} \).