Does anyone know how to find integer solutions of the quadratic equation

$$y^2+y+z=f$$

where $z$ is a fixed odd prime or $1$ and $f$ is a fixed odd prime greater than $3$?

This problem arose from the Diophantine equation $A+B=C$ where $A,B,C$ are natural numbers with no common factor. The managers of this site asked me to make my questions harder for this reason I will restate the above. Does anyone know if the quadratic equation $x^2-2x-[a^5+b^5]=0$ has infinite integer solutions?

-
The quantification is not clear. Are $z$ and $f$ previously-given parameters, or what? –  Lubin Sep 29 '12 at 19:08
@Lubin.only one parameter is given. In sort the question asks to find the pair of primes such as that.f-z=y[y+1].We know the difference of two primes can be expressed as the difference of two squares.So the above is saying there are infinite pairs of primes which their difference is expressed as above. –  Vassilis Parassidis Sep 29 '12 at 19:20
The difference of two primes always has two factors which are the following.2[2x-2y+1] or 4[x-y] which are obtained from the difference of their squares. –  Vassilis Parassidis Sep 29 '12 at 22:23
"We know the difference of two primes can be expressed as the difference of two squares." 11 and 5 are prime, and their difference is 6. How do you propose to express 6 as a difference of two squares? –  Gerry Myerson Oct 2 '12 at 0:43

Why dont you express the equation in the form $y^2+y+(z-f)=0$ and use the discriminant $b^2-4ac$ where $a=1$, $b=1$ and $c= z-f$. $c$ could be the difference of two primes or check the sequence. find the possible values of $y$.
$$y^2 + y + z = f$$ $$y^2 + y + z - f = 0$$ $$ay^2 + by + c = 0$$ where $a=1$, $b=1$, and $c=z-f$. So $$y = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-1\pm\sqrt{1-4(z-f)}}{2}$$ In its method of solution, it's no different from any other quadratic equation.
Whether the solutions are integers depends of course on $z$ and $f$.
My guess is that what OP is asking is, given a prime $f$, how do you find $y$ and prime $z$ such that $y^2+y+z=f$. –  Gerry Myerson Oct 2 '12 at 6:34