Parametric equation

Question

Please help with this problem:

Parameter p is a given integer. Find all integer solutions of the equation:

(p+1/p)⋅(x−1/x)+(p−1/p)⋅(x+1/x)=4px+5+1/p

I appreciate any help you can provide.

Answer 1

(p + 1/p)(x - 1/x) + (p - 1/p)(x + 1/x) = 4px + 5 + 1/p

Due to each of x and p being in a denominator, we must have x ≠ 0 and p ≠ 0.

So we can multiply both sides by px (because we would not be multiplying by 0 and creating an issue).

We get:

[(p + 1/p)(x - 1/x) + (p - 1/p)(x + 1/x)]px = (4px + 5 + 1/p)px

(p + 1/p)(x - 1/x)px + (p - 1/p)(x + 1/x)px = (4px + 5 + 1/p)px

p(p + 1/p)x(x - 1/x) + p(p - 1/p)x(x + 1/x) = 4pxpx + 5px + (1/p)px

(p² + 1)(x² - 1) + (p² - 1)(x² + 1) = 4p²x² + 5px + x

p²x² + x² - p² - 1 + p²x² - x² + p² - 1 = 4p²x² + 5px + x

2p²x² - 2 = 4p²x² + 5px + x

2p²x² + 5px + x + 2 = 0

2p²x² + (5p + 1)x + 2 = 0

This is now a quadratic equation in the variable x.

The discriminant of this quadratic equation is given by:

(5p + 1)² - 4 * 2p² * 2 = 25p² + 10p + 1 - 16p² = 9p² + 10p + 1 = (9p + 1)(p + 1)

This discriminant is greater than or equal to 0 provided that p ≤ - 1 or p ≥ -1/9.

So, provided that p ≤ - 1 or p ≥ -1/9, the real solution(s) of the quadratic equation is (are) given by:

x = [-(5p + 1) + (9p² + 10p + 1)^0.5] / 4p²

x = [-(5p + 1) - (9p² + 10p + 1)^0.5] / 4p²

At this point, I graphed both of these on Desmos and it seems to me that the only integer solution x = 1 happens when p = -1.

You can plug those values in to verify that this is indeed a solution.

However, in all honesty, I might be missing other possible solutions.

Answer 2

(p² + 1)(x² - 1) + (p² - 1)(x² + 1) = 4p²x² + 5px + x; 2p²x² - 2 = 4p²x² + 5px + x; 2p²x² + x(5p + 1) + 2 = 0

D = (5p + 1)² - 16p² = 9p² + 10p + 1 = (p + 1)(9p + 1); D = 0; p = -1,

x = [- 5p - 1 ±√(9p² + 10p + 1)]/(4p²); x = 1.

9p² + 10p + 1 = n²; p = (- 5 ±√16 + 9n²)/9

When n = 1; p = 0(not in domain of p) or p = - 10/9(not in domain of P)

Answer: x = 1 if p = - 1