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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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(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).

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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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  • $\begingroup$ Can you explain how did you know that there are $(p-1)/2$,please ? $\endgroup$
    – Frank
    Sep 28, 2012 at 10:20
  • $\begingroup$ @Mohammed, that's how many odd numbers there are less than $p$. $\endgroup$ Sep 28, 2012 at 13:19
  • $\begingroup$ How about $k$ ? $\endgroup$
    – Frank
    Sep 28, 2012 at 17:31
  • $\begingroup$ @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 $p$. $\endgroup$ Sep 28, 2012 at 22:35
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    $\begingroup$ @GerryMyerson, I think the OP is really asking, in effect, for the number of positive integers $k$ for which both $x^2-px+k$ and $x^2-px-k$ have integer roots. The more recent question that you marked as a duplicate of this one states the question somewhat more clearly. If I'm correct, the sought-for answer is $0$ (i.e., there are no postive $k$ for which both quadratics have integer roots), ultimately because $p\equiv-1$ mod $4$. $\endgroup$ May 19, 2017 at 10:43

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