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If we assume that $p=2^{24036583}-1$ is the greatest prime number until now .How to find the number of the positive integer numbers $k$ that makes for the two quadratic equations $ \pm x^2 \pm px \pm k$ rational roots.

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$x^2+px+k=0$ will have rational (indeed, integer) roots if and only if $p^2-4k$ is a square, $p^2-4k=q^2$. Let's write this as $p^2-q^2=4k$. This works for every odd number $q$ less than $p$, so there are $(p-1)/2$ such numbers $k$.

I don't know what you mean by "the two quadratic equations." The equation $x^2+px+k=0$ is the same as $-x^2-px-q=0$. If you want to choose the signs independently, then you get four (pairs of inequivalent) equations, not two, but they can be handled by the same methods as the one I did.

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Can you explain how did you know that there are $(p-1)/2$,please ? – Frank Sep 28 '12 at 10:20
@Mohammed, that's how many odd numbers there are less than $p$. – Gerry Myerson Sep 28 '12 at 13:19
How about $k$ ? – Frank Sep 28 '12 at 17:31
@Mohammed, $p^2-q^2=4k$ gives you a value of $k$ for every odd number $q$ less than $p$. There are $(p-1)/2$ odd numbers less than $p$. – Gerry Myerson Sep 28 '12 at 22:35

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