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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?

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    The quantification is not clear. Are $z$ and $f$ previously-given parameters, or what?2012-09-29
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    @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.2012-09-29
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    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.2012-09-29
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    "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?2012-10-02

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