# How to prove that my polynomial has distinct roots?

I want to prove that the polynomial

$$f_p(x) = x^{2p+2} - abx^{2p} - 2x^{p+1} +1$$

has distinct roots. Here $a$, $b$ are positive real numbers and $p>0$ is an odd integer. How can I prove that this polynomial has distinct roots for any arbitrary $a$,$b$ and $p$.

• Perhaps proving that $\gcd(f,f')=1$ is feasible. – lhf Apr 13 '16 at 12:57
• We tried to prove it with your way but didn't achieve. Could you please prove $gcd(f,f')=1$? – drxy Apr 13 '16 at 12:58
• Why do you have a coefficient $ab$? Why not just call it $c$? Is it a typo? – M. Vinay Apr 13 '16 at 13:14

Let $c$ denote $ab$. Note that \begin{equation*} f(x) = (x^{p+1}-1)^2 - cx^{2p} \end{equation*} and \begin{equation*} f'(x) = 2(p+1)x^p(x^{p+1}-1) - 2pcx^{2p-1} \end{equation*}

Then, $f(x) = 0 \iff$ \begin{equation*} c = \dfrac{(x^{p+1}-1)^2}{x^{2p}} = \varphi(x)\ (\text{say}) \end{equation*} and $f'(x) = 0 \iff$ \begin{align*} c & = \dfrac{(p+1)x^p(x^{p+1}-1)}{px^{2p-1}} = \dfrac{(p+1)x}{p}\dfrac{x^{p+1}-1}{x^p} \iff\\ c & = \left(\dfrac{p+1}{p}\right)x \sqrt{\varphi(x)}. \end{align*}

Thus, $f(x)$ and $f'(x)$ vanish for the same $x$ if and only if for some root $x$ of $f(x)$, \begin{align*} c & = \left(\dfrac{p+1}{p}\right)x \sqrt c \iff\\ x & = \dfrac{p\sqrt c}{p+1}. \end{align*}

Thus, for every $p$ and $c = ab$, if such an $x$ is a root, it is a multiple root.

Now, when does such a root $x$ exist? Let $x = t$ be one such. Then $c = \left(\dfrac{p+1}{p} \right)^2 t^2$. Then, since $f(t) = 0$, we have (from the original form of the equation): \begin{align*} t^{2p+2} - \left(\dfrac{p+1}{p}\right)^2 t^{2p + 2} - 2t^{p+1} + 1 = 0\\ -\dfrac{(2p + 1)}{p^2}t^{2p + 2} - 2t^{p+1} + 1 = 0\\ (2p + 1)t^{2(p + 1)} + 2p^2 t^{p + 1} - p^2 = 0. \end{align*}

When treated as a quadratic equation in $t^{p+1}$, the discriminant is \begin{equation*} 4p^4 + 4p^2(2p + 1) = 4p^2(p + 1)^2, \end{equation*} and therefore, the solutions are \begin{equation*} t^{p+1} = -p, \dfrac{p}{2p + 1}. \end{equation*} That is, \begin{equation*} t = (-p)^{\frac 1 {p + 1}}, \left(\dfrac p {2p + 1} \right)^{\frac 1 {p + 1}}. \end{equation*}

But substituting the same $c$ in $f'(t) = 0$, we get \begin{align*} & 2(p+1)t^p(t^{p+1}-1)-2p\left(\dfrac{p+1}{p}\right)^2t^{2p+1}=0\\ & p(t^{p+1}-1)-(p+1)t^{p+1}=0 \implies\\ & t = (-p)^{\frac{1}{p+1}} \end{align*}

Thus, only the first of the previous two solutions satisfies both equations.

Then, $c = \left( \dfrac{p+1}{p} \right)^2 t^2$ gives us \begin{equation*} \boxed{c= \dfrac{(p+1)^2}{(-p)^{\frac{2p}{p+1}}}}. \end{equation*}

Thus, the equation has multiple roots exactly when $c$ and $p$ are related as above.

Note that for odd values of $p$, $c$ will be a real number if and only if $p$ is of the form $4k + 1$, and then, $c < 0$. If, as stated in the question, $c = ab$ is a positive real number, the equation will have distinct roots.

Example

For $p = 1$, $f(x) = x^4 - cx^2 - 2x^2 + 1$ and $f'(x) = 4x^3 - 2cx - 4x$.

Then, $f(x) = 0$ and $f'(x) = 0$ imply that $c = \left(\dfrac{x^2 - 1}{x}\right)^2$ and $c = 2x\left(\dfrac{x^2-1}{x}\right)$ respectively. Thus, if $x$ is a multiple root, then $x = \dfrac{\sqrt c}{2}$.

Taking $t$ to be such a root, so that $c = 4t^2$, and substituting in $f(t) = 0$, we get \begin{align*} t^4 - 4t^4 - 2t^2 + 1 = 0\\ 3t^4 + 2t^2 - 1 = 0. \end{align*} Thus, $t^2 = -1, \dfrac 1 3$, of which only the first one satisfies $f'(t)=0$. Thus, $c = -4$.

For $c = 4$, $x^4 + 2x^2 + 1 = 0$ has roots $\pm i, \pm i$.

• I think an example would be useful. Perhaps p=1 or 3. – marty cohen Apr 13 '16 at 19:14
• @martycohen Added. Thanks, working out the example also brought out an error in the derivation, which I've fixed now. – M. Vinay Apr 14 '16 at 13:46
• Since $a$ and $b$ are positive real numbers, $c=ab$ must be positive. But in your answer $c=-4$. So does it mean there is no multiple root? – drxy Apr 14 '16 at 20:43
• @drxy Yes, for real $c > 0$, the equation has distinct roots (I suppose that's what you wish to prove). I've updated my answer with this information. – M. Vinay Apr 15 '16 at 2:32
• @M.Vinay thank you for invaluable response. If my polynomial turns into $f_p(x) = x^{2p+2} - abx^{2p} - (a+b)x^{p} -1$, where p is even, then have you got any idea that $f_p(x)$ has distinct roots? – drxy Apr 16 '16 at 9:37

The claim is false. Set $ab:=(\frac{729}{16})^\frac{1}{3}$ and $p:=2$

Then $f(x)=x^6-abx^4-2x^3+1$ has a double root at $x=-2^{\frac{1}{3}}$

• Sorry, I forgot that $p$ is odd. – drxy Apr 13 '16 at 18:15
• @drxy Even for odd $p$, it's possible to get multiple roots. – M. Vinay Apr 14 '16 at 13:48
• @M.Vinay If it is possible, can you give me some example for odd $p$. – drxy Apr 14 '16 at 14:04
• @drxy I have done so in my answer (both the general case describing exactly when [for what value of $c$] the polynomial will have multiple roots, as well as an example, with $p = 1$). – M. Vinay Apr 14 '16 at 14:05
• Oh, I have just saw it. So many thanks @M.Vinay – drxy Apr 14 '16 at 14:06

With the help of Mathematica the discriminant is given by:

$$\Delta = \left \{ \begin{array}{cc} (4c)^{p+1} \left[ p^{p} c^{(p+1)/2}+(p+1)^{p+1} \right]^{2} & \text{odd } p \\[5pt] (4c)^{p+1} [(p+1)^{2(p+1)}-p^{2p}c^{p+1}] & \text{even } p \end{array} \right. \\$$

$\Delta \neq 0 \,$ if $\, \left \{ \begin{array}{ll} \text{odd } p=4n-1 , & c\neq 0 \\[5pt] \text{odd } p=4n-3, & c^{(p+1)/2} \neq 0, \displaystyle -\frac{(p+1)^{p+1}}{p^{p}} \\[5pt] \text{even } p, & c^{p+1} \neq 0, \displaystyle \frac{(p+1)^{2(p+1)}}{p^{2p}} \end{array} \right.$

• Discriminant[x^(2n+2)- a x^(2n) - 2x^(n+1)+1 ,x] did not work for me... – lhf Apr 13 '16 at 14:14
• @lhf Table[{p, Discriminant[x^(2 p + 2) - c x^(2 p) - 2 x^(p + 1) + 1, x] // Simplify}, {p, 1, 10}] // TraditionalForm – Ng Chung Tak Apr 13 '16 at 14:28