Determine $x,y\in\mathbb{R}$ such that $t^4-2xt^2+y^2=0$ has $4$ real solutions in an arithmetic progression

Question

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 $(t_1+r)^2 = x$: $$t_2 = t_1 + r = \pm \sqrt{x} \Rightarrow t_3= t_1+2r = \mp \sqrt{x} \Rightarrow r = \mp 2\sqrt{x} \Rightarrow t_1 = \pm 3\sqrt{x}, \ t_4 = \mp 3 \sqrt{x}.$$

1. If $r=-2\sqrt{x} \Rightarrow (-9x)(-x) = y^2 \iff 9x^2 = y^2 \iff 3x=|y|$ (from Viete's fourth sum)

2. 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!

Answer 1

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.

Answer 2

Another way

WLOG the roots be $$a\pm3d, a\pm d$$

$$\implies a+3d+a+d+a-d+a-3d=0\implies a=0$$ and $d^2\ge0$ as $d$ has to be real

$$\implies t^4-2xt^2+y^2=(t-d)(t+d)(t-3d)(t+3d)=\cdots=t^4-10d^2t^2+9d^4$$

$$\implies2x=10d^2,y^2=9d^2$$

Compare the values of $d^2$

Answer 3

[without applying the Viete relations]

Since $ \ t^4 - 2xt^2 + y^2 \ = \ 0 \ $ is a biquadratic equation, we can determine that $ \ t^2 \ = \ x \ \pm \ \sqrt{x^2 - y^2} \ \ . \ $ In order for there to be the four distinct real roots that the arithmetic progression requires, we must have $ \ x \ > \ 0 \ \ , \ \ x^2 \ > \ y^2 \ \ $ and $ \ y \ \neq \ 0 \ \ . \ $ Where the roots are real, the order of the progression is then $$ -\left( \ \sqrt{x \ + \ \sqrt{x^2 - y^2}} \ \right) \ \ , \ \ -\left( \ \sqrt{x \ - \ \sqrt{x^2 - y^2}} \ \right) \ \ , $$ $$ \left( \ \sqrt{x \ - \ \sqrt{x^2 - y^2}} \ \right) \ \ , \ \ \left( \ \sqrt{x \ + \ \sqrt{x^2 - y^2}} \ \right) \ \ , \ \ $$ as $ \ x \ > \ \sqrt{x^2 - y^2} \ \ . $

Arithmetic progression then implies that $$ d \ \ = \ \ \left( \ \sqrt{x \ - \ \sqrt{x^2 - y^2}} \ \right) \ - \ \left[ \ -\left( \ \sqrt{x \ - \ \sqrt{x^2 - y^2}} \ \right) \ \right] $$ $$ = \ \ 2 \left( \ \sqrt{x \ - \ \sqrt{x^2 - y^2}} \ \right) $$ and $$ 3d \ \ = \ \ \left( \ \sqrt{x \ + \ \sqrt{x^2 - y^2}} \ \right) \ - \ \left[ \ -\left( \ \sqrt{x \ + \ \sqrt{x^2 - y^2}} \ \right) \ \right] $$ $$ = \ \ 2 \left( \ \sqrt{x \ + \ \sqrt{x^2 - y^2}} \ \right) \ \ . $$ Consequently, $$ 9d^2 \ \ = \ \ 4 · \left( \ x \ + \ \sqrt{x^2 - y^2} \ \right) \ \ = \ \ 9·\left[ \ 4 · \left( \ x \ - \ \sqrt{x^2 - y^2} \ \right) \ \right] $$ $$ \Rightarrow \ \ 32x \ \ = \ \ 40·\sqrt{x^2 - y^2} \ \ \Rightarrow \ \ 16x^2 \ \ = \ \ 25·(x^2 - y^2) $$ $$ \Rightarrow \ \ 25y^2 \ \ = \ \ 9x^2 \ \ . $$

Under the conditions mentioned in the first paragraph, we may have either $ \ x \ > \ y \ > \ 0 \ $ or $ \ x \ > \ 0 \ > \ y \ \ . \ $ Writing our result in terms of $ \ |y| \ $ will make $ \ x \ $ positive, so we have $ \ 5·|y| \ = \ 3x \ \ , \ \ y \ \neq \ 0 \ \ . $

Because the constant term in the equation is $ \ y^2 \ $ is positive, the sign of $ \ y \ $ is "invisible", so there are always two implied solutions for any choice of (positive) $ \ x \ \ . \ $ As an example, for $ \ x \ = \ 10 \ \ , \ \ y \ = \ \pm 6 \ \ , \ $ the roots of $ \ t^4 - 20t^2 + 36 \ = \ 0 \ $ are $ \ -3 \sqrt2 \ \ , \ \ - \sqrt2 \ \ , \ \ \sqrt2 \ \ , \ \ 3 \sqrt2 \ \ . \ $ EDIT (3/31) -- (Incidentally, "flipping" the sign of $ \ x \ $ still produces a set of roots in arithmetic progression, but now along the imaginary axis: $ \ t^4 + 20t^2 + 36 \ = \ 0 \ $ has the roots $ \ -i·3 \sqrt2 \ \ , \ \ - i·\sqrt2 \ \ , \ \ i·\sqrt2 \ \ , \ \ i·3 \sqrt2 \ \ . \ $ This is reasonable to expect since $ \ (it)^4 - 2x·(it)^2 + y^2 \ = \ t^4 + 2xt^2 + y^2 \ \ . ) $