Show that the roots of the polynomial $x^4 - px^3 + qx^2 - pqx + 1 = 0$ satisfy a certain relationship

Here is the question:

If the roots of the equation $$ x^4 - px^3 + qx^2 - pqx + 1 = 0 $$ are $\alpha, \beta, \gamma,$ and $\delta$, show that $$ (\alpha + \beta + \gamma)(\alpha + \beta + \delta) (\alpha + \gamma + \delta)(\beta + \gamma + \delta) = 1. $$

Pretty much exhausted my resources.If there are more than one way of doing it please state and you can state some good books for this particular topic.

$$ 1+x^4+\text{qx}^2-\text{px}^3-\text{pqx} \equiv x^4+x^3 (-\alpha -\beta -\gamma -\delta )+x^2 (\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta )+x (-\alpha \beta \gamma -\alpha \beta \delta -\alpha \gamma \delta -\beta \gamma \delta )+\alpha \beta \gamma \delta $$

So, $$ \text{p} = \alpha+\beta +\gamma +\delta $$ $$ \text{q} = \alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta $$ $$ \text{p}\text{q}=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta $$ $$ 1= \alpha \beta \gamma \delta $$

Solution 1:

We know the factorisation:

$f(x) = x^4 - px^3 + qx^2 - pqx + 1 = (x - \alpha)(x - \beta)(x - \gamma)(x - \delta)$

Equating $x^3$ coefficients gives $\alpha + \beta + \gamma + \delta = p$.

Thus we can write the expression in question as:

$(p - \delta)(p - \gamma)(p - \beta)(p - \alpha) = f(p) = 1$.

Solution 2:

Without the trick in the answer by fretty (essentially recognising the expression asked for can be rewritten as $f(\alpha+\beta+\gamma+\delta)$) you can obtain the result through the approach you had originally chosen. For that it suffices to express $(\alpha+\beta+\gamma)(\alpha+\beta+\delta)(\alpha+\gamma+\delta)(\beta+\gamma+\delta)$, which is clearly a symmetric polynomial of $\alpha,\beta,\gamma,\delta$, in terms of the elementary symmetric polynomials (the right hand sides of your four equations; call them $e_1,e_2,e_3,e_4$ respectively).

Expansion of the product gives $3^4=81$ terms, which can be grouped into minimal (also called monomial) symmetric polynomials, the sum of all distinct permutations of a given monomial. I will denote these by $m_\lambda$ where $\lambda$ is the pattern of exponents; for instance $m_{2,1,1}$ is the sum of monomials like $\alpha^2\beta\gamma$ or $\alpha\gamma^2\delta$. With this notation one gets $$ (\alpha+\beta+\gamma) (\alpha+\beta+\delta) (\alpha+\gamma+\delta) (\beta+\gamma+\delta)= m_{3,1}+2m_{2,2}+4m_{2,1,1}+9m_{1,1,1,1}. $$ Now each minimal symmetric polynomial contains the leading term of a unique product of elementary symmetric polynomials, as explained here, and this leads to relations $$\begin{align} m_{3,1}&=e_2e_1^2 - 2m_{2,2} - 5m_{2,1,1} - 12m_{1,1,1,1}, \\ m_{2,2}&=e_2^2 - 2m_{2,1,1} - 6m_{1,1,1,1}, \\ m_{2,1,1}&=e_3e_1 - 4m_{1,1,1,1},\\ m_{1,1,1,1}&=e_4. \end{align} $$ Thus our equation becomes, after some more computation, $$ (\alpha+\beta+\gamma) (\alpha+\beta+\delta) (\alpha+\gamma+\delta) (\beta+\gamma+\delta)= e_2e_1^2 - e_3e_1 + e_4 = qp^2 - (pq)p + 1=1, $$ where your equations $p=e_1$, $q=e_2$, $pq=e_3$ and $1=e_4$ were substituted.

This is of course relatively laborious, but has the advantage of being a method that works for arbitrary symmetric polynomials of $\alpha,\beta,\gamma,\delta$.

The coefficients for the expression of products of elementary symmetric polynomials in terms of minimal ones are given by small combinatorial counting problems: the number of matrices with entries $0$ or $1$ with column sums given by the indices $i$ of the factors $e_i$, and row sums given by the indices of the $m_\lambda$. For instance the coefficient $5$ of $m_{2,1,1}$ in $e_2e_1^2$ equals the number of such matrices with row and column sums both given by the sequence $2,1,1$.