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What are the conditions for the existence of real solutions for the following equations:

$$\begin{align} x^2&=a\cdot y+b\\ y^2&=c\cdot x+d\end{align}$$

where $a,b,c,d $ are real numbers.

These represent two parabolas; how might we find out the conditions for the existence of $0,2,4$ real solutions of the equations?

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Don't forget the possibility of tangency. –  J. M. Dec 19 '11 at 7:23
    
Eliminating $y$ yields the (already) depressed quartic equation $x^4-2bx^2-a^2 cx+b^2-da^2$. This gives a nice overview of the possibilities for the roots of such a quartic. –  J. M. Dec 19 '11 at 7:30
    
That link is great, but the proofs are missing, I suppose? –  Abhinav Dec 19 '11 at 8:32
    
That's probably why I left it as a comment... –  J. M. Dec 19 '11 at 8:34
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2 Answers

Since these are parabolas, $a$ and $c$ must be nonzero. Let $x = a^{2/3} c^{1/3} X$ and $y = a^{1/3} c^{2/3} Y$. Under this scaling, the equations become $X^2 = Y + B$ and $Y^2 = X + D$ where $B = b a^{-4/3} c^{-2/3}$ and $D = d a^{-2/3} c^{-4/3}$. Now substituting $Y = B - X^2$ into $Y^2 = X + D$ we get the fourth-degree equation $X^4 - 2 B X^2 - X + B^2 - D = 0$. The discriminant of this, according to Maple, is $-256\,{B}^{3}+288\,B D -27+256\,{B}^{2}{D}^{2}-256\,{D} ^{3}$. The curve where the discriminant is $0$ separates the $BD$ plane into three regions like this:

We have: no real solution in the red region, one on the red-yellow boundary, two in the yellow region, three on the yellow-blue boundary (except at the sharp cusp $B=D=3/4$ where there are two), and four in the blue region.

enter image description here

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it seems quite interesting. Another analysis is done here: jstor.org/stable/2972804?seq=1, however without a proof?? anyone knows how to prove these results discussed on page 51? –  Abhinav Dec 19 '11 at 8:30
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Assume that $(x,y)$ is a point common to both parabolas. If we add the two equations together and complete the square we get the circle equation \begin{equation*} \left( x - (c/2) \right)^2 + \left(y - (a/2) \right)^2 = (a/2)^2 + (c/2)^2 + b + d \end{equation*} So one condition which is necessary for a solution is that the right-hand side be $\geq 0$. If we instead subtract the second equation from the first we get \begin{align*} (x-y)(x+y) &= ay - xc + b - d \newline &= \frac{a-c}{2}(x+y) - \frac{a+c}{2}(x-y) + b - d \end{align*} Let $w=(x+y)$, $z=(x-y)$, $k_1 = (a-c)/2$, $k_2=(a+c)/2$. Then \begin{equation*} wz = k_1w - k_2z + b-d \end{equation*} \begin{equation*} (w+k_2)(z-k_1) = (b-d) -k_1 k_2 \end{equation*} So the point $(w,z)$ lies on a certain hyperbola. Unfortunately I can't think of anything else that one can say with this line of reasoning, but I thought it was interesting so I decided to post this incomplete answer anyway.

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