Let $a,b,c,d$ be real roots of $x^4-px^3+qx^2-px+1=0$ $(a>b>c>d>0)$. Prove that $ad=bc=1$.

Question

Let $a,b,c,d$ be real roots of $x^4-px^3+qx^2-px+1 = 0\; (a>b>c>d>0).$ Prove that $ad=bc=1$.

My approach: $abcd=1$, and $$ a+b+c+d = \frac1a + \frac1b + \frac1c + \frac1d, $$ which can be shown using Vieta's theorem.

Rearranging, we get $$ (a+d) - \biggl(\frac1a + \frac1d\biggr) = (c+b) - \biggl(\frac1c + \frac1b\biggr). $$

Further we get $$ (a+d)(1-bc) = (b+c)(1-ad) $$ using $abcd=1$.

I am stuck here how do I show that $1-bc$ and $1-ad$ must be $0$?

Answer 1

I strongly suspect that there is a more elegant approach than the one that I will use. To understand my response, first see this answer.

Similar to the linked answer, the polynomial

$$f(x) = x^4 - px^3 + qx^2 - px + 1$$

is a symmetric $~4$-th degree polynomial. This is because the coefficients of the $~x^4~$ and $~x^0~$ terms are equal, and the coefficients of the $~x^3~$ and $~x^1~$ terms are also equal.

So, I will take the inelegant approach of actually computing the roots, subject to the constraint that the polynomial equation has $~4~$ distinct positive roots.

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Using the linked answer as a model, let

$$g(x) = x^2 - px + q - \frac{p}{x} + \frac{1}{x^2}. \tag1 $$

Clearly, the roots of $~g(x) = 0~$ are the same as the roots of $~f(x) = 0.$

Let $~w = x + \dfrac{1}{x} \implies $

$$w^2 = x^2 + 2 + \frac{1}{x^2} \implies (w^2 - 2) = x^2 + \frac{1}{x^2}.$$

Therefore, the equation $~g(x) = 0~$ can now be converted into the following quadratic equation in $~w~$:

$$(w^2 - 2) - pw + q = 0 \implies w^2 - pw + (q-2) = 0. \tag2 $$

This has roots

• $\displaystyle w_1 = \frac{1}{2} \left[ ~p + \sqrt{p^2 + 8 - 4q} ~\right]$

• $\displaystyle w_2 = \frac{1}{2} \left[ ~p - \sqrt{p^2 + 8 - 4q} ~\right]$

Further, given any root $~w,~$ the two corresponding roots of $~x~$ are generated by:

$$x + \frac{1}{x} = w \implies x^2 - wx + 1 = 0, \tag3 $$

Note that a value for $~x~$ will satisfy (1) above if and only if the constructed value for $~w~$ satisfies (2) above. Further the only values of $~x~$ that construct a value for $~w~$ that satisfies (2) above are those values of $~x~$ generated by plugging $~w_1~$ and $~w_2~$ into (3) above.

Therefore, the roots generated by plugging (for example) $~w_1~$ into (3) above must represent two of the distinct roots of $~g(x) = 0,~$ which has the same roots as $~f(x) = 0.$

At this point, the expression in (3) above makes it game over, because when you plug in $~w_1~$ into (1) above, the two roots must have product $~= 1.$

Since you know that plugging in $~w_1~$ into (3) above must generate two of the roots to the original equation $~f(x) = 0,~$ you can now deduce that there exists two distinct elements $~x_1, ~x_2~$ from the set $~\{a,b,c,d\}~$ such that $~x_1 \times x_2 = 1.~$

Further, if (for example) $~x_1 = a,~$ then clearly, $~x_2~$ must equal $~d,~$ and vice-versa. This is because $~a \times b \times c \times d = 1 ~$ and $~a > b > c > d > 0.$

Also, if neither $~x_1~$ nor $~x_2~$ equal either of $~a~$ or $~d,~$ then you have that $~b \times c = 1.$

Answer 2

If $a>0$ is a root then $a^4-pa^3+qa^2-pa+1=0$. By dividing this equation by $a^4$, we get $(\tfrac1a)^4-p(\tfrac1a)^3+q(\tfrac1a)^2-p(\tfrac1a)+1=0.$ So, $\frac1a$ is also a root. Since the roots are distinct and positive according to OP, we conclude that all the roots are $a,\frac1a, b,\frac1b$ for some positive real number $a,b.$

Without loss of generality, we may assume that $a>b>\frac1b.$ Where is $\frac1a ?:$ $$a>b>0\implies \frac1b>\frac1a.$$ Therefore, $a>b>c=\frac1b>d=\frac1a$.