the number of the positive integer numbers $k$ that makes for the two quadratic equations $ \pm x^2 \pm px \pm k$ rational roots. 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. 
 A: (Assuming that the question is rather about $x^2-px+k$ and $x^2-px-k$ simultaneously having rational [=integer] solutions)
Let $a,p-a$ be the roots of $x^2-px-k$ and $b,p-b$ the roots of $x^2-px+k$. Then $a(p-a)=k=-b(p-b)$, i.e., 
$$\tag1 a^2+b^2=p(a+b).$$
Unless $p\mid a,b$, this means that $x^2+1\equiv 0\pmod p$ has a solution. But for $p\equiv -1\pmod 4$, no such solution exists. We conclude $p\mid a$, $p\mid b$, say $a=pu,b=pv$. Then $(1)$ becomes 
$$\tag2u^2+v^2=u+v.$$
For $x\in\Bbb Z$ we have $x^2\ge x$ with equality iff $x\in\{0,1\}$. 
Thus we concolude from $(2)$ that $a,b\in\{0,p\}$. At any rate, this implies that the  only possible value for $k$ si $k=a(p-a)=0$ (which is not positive).
A: $x^2+px+k=0$ 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. 
