Determine $x,y\in\mathbb{R}$ such that $t^4-2xt^2+y^2=0$ has $4$ real solutions in an arithmetic sequence.
I'm quite stuck with this problem right here because, although I get to an answer, it isn't the correct one.
If $t_1,t_2,t_3,t_4$ are the solutions of the equation and in an arithmetic sequence, that means $t_2 = t_1 + r,\ t_3 = t_1 + 2r,\ t_4 = t_1 + 3r$ and $t_1+t_4=t_2+t_3$. Therefore, using the first Viete sum, $t_1+t_4+t_2+t_3 = 0 \Rightarrow t_1 + t_4=t_2+t_3 = 0$.
From the second sum, $(t_1+t_4)(t_2+t_3) + t_1t_4+t_2t_3 = 2x \iff t_1t_4+t_2t_3 = 2x$.
$$\begin{cases} t_1\cdot t_4 = t_1^2+t_1r \\\ t_2\cdot t_3 = t_1^2+3t_1r+2r^2\end{cases} \Rightarrow t_1t_4+t_2t_3 = 2(t_1^2+2t_1r+r^2)=2x\\ \iff (t_1 + r)^2 = x \Rightarrow x \geq 0$$
By noting $t^2 = u$, the original equation becomes $u^2-2xu+y^2=0$. For there to be four solutions, the conditions are: t $\Delta > 0 \iff 4x^2-4y^2 > 0 \iff x^2 – y^2 > 0 \iff x^2 > y^2$ and $u_1$ and $u_2$ should be strictly greater than $0$, that is $\frac{2x\pm 2\sqrt{x^2-y^2}}{2}> 0$. Because the solution with plus is greater than the solution with minus, it suffices to say $x > \sqrt{x^2-y^2} \Rightarrow x > 0$
Returning to $(t_1+r)^2 = x$: $$t_2 = t_1 + r = \pm \sqrt{x} \Rightarrow t_3= t_1+2r = \mp \sqrt{x} \Rightarrow r = \mp 2\sqrt{x} \Rightarrow t_1 = \pm 3\sqrt{x}, \ t_4 = \mp 3 \sqrt{x}.$$
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If $r=-2\sqrt{x} \Rightarrow (-9x)(-x) = y^2 \iff 9x^2 = y^2 \iff 3x=|y|$ (from Viete's fourth sum)
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If $r=2\sqrt{x} \Rightarrow (-9x)(-x)=y^2 \iff 9x^2 = y^2 \iff 3x = |y|$
This, however, is the wrong answer according to the book, whose authors say that $x > 0, \ 3x=5|y|$ is the right conclusion. So, any ideas where I made a mistake or should further look for conditions? Any hints are much appreciated!
Best Answer
Let $t^4 - 2xt^2 + y^2 = 0$ be our equation, with roots $x_1, x_2, x_3, x_4 \in \mathbb{R}$ such that they are in arithmetic progression. Because of this, then there exists $\alpha, r \in \mathbb{R}$ such that $x_1 = \alpha - 3r, x_2 = \alpha -r, x_3 = \alpha + r, x_4 = \alpha + 3r$, and then our solutions are in an arithmetic progression of ratio $2r$.
Notice that $S_1 =x_1 + x_2 + x_3 + x_4 = 4\alpha$ by our assumption. By using the Viète sum, we obtain that $S_1 = 0$, therefore $\alpha = 0$, and our solutions become $x_1 = -3r, x_2 = -r, x_3 = r, x_4 = 3r$.
The Viète product of all solutions is $y^2$, therefore $9r^4 = y^2 \Leftrightarrow |y| = 3r^2$. If $r = 0$, then $x = y = 0$, and the solutions are in a constant arithmetic progression, $x_1 = x_2 = x_3 = x_4 = 0$. Suppose $r \neq 0$. Therefore, because $r$ is a solution, plugging it into our original equation, we obtain $r^4 - 2xr^2 + 9r^4 = 0 \Leftrightarrow x = 5r^2$. Finally, we obtain the following solutions:
$(x, y) \in \{(5r^2, 3r^2), (5r^2, -3r^2) : r \in \mathbb{R}\} = \{(5\beta, 3\beta), (5\beta, -3\beta) : \beta \geq 0\}$
Note: Whenever you have a polynomial with solutions either in an arithmetic or geometric progression, the idea is to consider the solutions in such a way so that when adding them (A.P.) or multiplying them (G.P.), you are only obtaining an equality in one term.