need help with this equation!!

It's the homework the teacher gave me and I have tried many ways to do it. I think it's related to Delta somehow.
 
Hmmmm.... After looking at this for a bit I don't really like the problem. Supposedly you are supposed to somehow solve for p and x but I see no way to do it. Perhaps if you were taking number theory, but as you are referring to the discriminant of a quadratic equation simply by "delta" I'm guessing you are taking a High School Algebra class or something in college at a similar level.

Re-arranging your equation a bit:
[math]p^2 - p + 1 - x^3 = 0[/math]
[math]p^2 - p + (1 - x^3) = 0[/math]
Now set up the discriminant (delta) of the quadratic formula using a = 1, b = -1, and [math]c = 1 - x^3[/math]. The discriminant must be a perfect square for this to possibly work. So start picking values of x until you find one. Then find p from there and verify that it is indeed a prime number.

There appears to be only one solution, though I am unable to prove it.

-Dan
 
Dan is right to work it to a quadratic which can then be solved.

The first(?) solution (only solution?) is that the prime number p is 19 and the integer x is 7.
 
Theoretically we can say that [MATH] p (p-1) = (x-1) ( x^2 + x + 1)[/MATH]. It is obvious that x <p, so p divides [MATH]x^2 + x + 1[/MATH]. So
[MATH] p (p-1) = (x-1) k p [/MATH] and [MATH] p-1 = k (x-1) [/MATH]. We can therefore draw [MATH]x = \frac{p-1}{k} + 1[/MATH] . And so
[MATH] ( \frac{p-1}{k} + 1)^3=x^3=p^2-p+1[/MATH] which gives [MATH]p^2+p(-2+3k-k^3)+3k^2-3k+1=0[/MATH]. So p divides [MATH]3k^2-3k+1[/MATH] . If we take [MATH]p=3k^2-3k+1[/MATH] and we replace p by this value we find [MATH]-k^3+3k=0[/MATH]. k = 0 is not possible. So k = 3 and then p = 19 and x = 7.
 
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