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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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    "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

2 Answers 2

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$ 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 $f$.

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    If that's what he meant, he certainly wasn't clear about it.2012-10-02
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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 $y$.