The integral values of x for which `x^2 +17x+71` is perfect square of a rational number are a and b, then `|a-b|`=

Answer

To solve the problem, we need to find the integral values of \( x \) for which \( x^2 + 17x + 71 \) is a perfect square of a rational number. Let's denote this perfect square by \( y^2 \). Therefore, we can rewrite the equation as:

\[ x^2 + 17x + 71 = y^2 \]

Rearranging gives us:

\[ x^2 + 17x + (71 - y^2) = 0 \]

This is a quadratic equation in \( x \). For \( x \) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac = 17^2 - 4 \cdot 1 \cdot (71 - y^2) \]

Calculating the discriminant:

\[ D = 289 - 4(71 - y^2) = 289 - 284 + 4y^2 = 5 + 4y^2 \]

For \( D \) to be a perfect square, we can set:

\[ 5 + 4y^2 = k^2 \]

for some integer \( k \). Rearranging gives us:

\[ k^2 - 4y^2 = 5 \]

This can be factored using the difference of squares:

\[ (k - 2y)(k + 2y) = 5 \]

Next, we need to consider the factor pairs of 5, which are \( (1, 5) \), \( (-1, -5) \), \( (5, 1) \), and \( (-5, -1) \).

### Case 1: \( k - 2y = 1 \) and \( k + 2y = 5 \)

Adding these equations:

\[ 2k = 6 \implies k = 3 \]

Substituting \( k \) back to find \( y \):

\[ 3 - 2y = 1 \implies 2y = 2 \implies y = 1 \]

### Case 2: \( k - 2y = 5 \) and \( k + 2y = 1 \)

Adding these equations:

\[ 2k = 6 \implies k = 3 \]

Substituting \( k \) back to find \( y \):

\[ 3 - 2y = 5 \implies 2y = -2 \implies y = -1 \]

### Case 3: \( k - 2y = -1 \) and \( k + 2y = -5 \)

Adding these equations:

\[ 2k = -6 \implies k = -3 \]

Substituting \( k \) back to find \( y \):

\[ -3 - 2y = -1 \implies 2y = -2 \implies y = -1 \]

### Case 4: \( k - 2y = -5 \) and \( k + 2y = -1 \)

Adding these equations:

\[ 2k = -6 \implies k = -3 \]

Substituting \( k \) back to find \( y \):

\[ -3 - 2y = -5 \implies 2y = 2 \implies y = 1 \]

Now we have two distinct values for \( y \): \( y = 1 \) and \( y = -1 \).

### Finding \( x \)

Substituting \( y = 1 \) into the discriminant:

\[ D = 5 + 4(1^2) = 9 \quad (\text{which is } 3^2) \]

Calculating \( x \):

\[ x = \frac{-17 \pm \sqrt{9}}{2} = \frac{-17 \pm 3}{2} \]

Calculating the two values:

1. \( x = \frac{-14}{2} = -7 \)
2. \( x = \frac{-20}{2} = -10 \)

Now substituting \( y = -1 \):

\[ D = 5 + 4(-1)^2 = 9 \quad (\text{which is } 3^2) \]

Calculating \( x \):

\[ x = \frac{-17 \pm \sqrt{9}}{2} = \frac{-17 \pm 3}{2} \]

This gives the same values:

1. \( x = -7 \)
2. \( x = -10 \)

### Final Calculation

Let \( a = -10 \) and \( b = -7 \). We need to find \( |a - b| \):

\[ |a - b| = |-10 - (-7)| = |-10 + 7| = |-3| = 3 \]

Thus, the final answer is:

\[ \boxed{3} \]