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

• Try Viete's Formulae. – Cyclohexanol. Jan 2 '15 at 7:35
• math.stackexchange.com/questions/1082398/… – Bumblebee Jan 2 '15 at 7:41
• Tried that.Some other technique is required here. – Irtiza Jan 2 '15 at 7:41
• This is a standard problem using Viete's formulae, as said above...see my answer. Of course you have to think a bit, it isn't a case of just using the formulae blindly. – fretty Jan 2 '15 at 7:46
• @MathN00b Vieta! I know it's named after François Viète but the Latinised form of his name is more popular. – Nick Jan 2 '15 at 8:00

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:

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

• can you explain your last step – Irtiza Jan 2 '15 at 9:02
• @Irtiza If you substitute $p$ for $x$ in the original quartic, the first and second terms sum to zero as do the third and fourth terms leaving you with $1$. – Mike Jan 2 '15 at 9:11
• The expression turns into $(p-\alpha)(p-\beta)(p-\gamma)(p-\delta)$. But what is this? It is exactly $f(p)$...so plug in $x=p$ into the polynomial and what do you get... – fretty Jan 2 '15 at 9:15

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 $2,1,1$.

• I was going to go down this route but at the last minute saw that the expression simply was $f(p)$. – fretty Jan 2 '15 at 11:53
• @fretty: Certainly your method is much more painless. Maybe there is an advantage of having two methods at hand: my solution involves the number $9$ of derangements of $4$ elements as coefficient of $m_{1,1,1,1}$ in the given expression, and comparing two approaches (which must give the same result) might produce an equation for that number of derangements. Of course this requires more work, and is likely to produce a known expression. – Marc van Leeuwen Jan 2 '15 at 12:06
• Well of course the method used this question has an obvious generalization. If we let $f(x) = x^n - a_{n-1}x^{n-1}+... + a_1 x + a_0$ be a degree $n$ polynomial with roots $\alpha_i$ and $S_i = \sum_{j\neq i}\alpha_j$ then $\prod_i S_i = f(a_{n-1})$. Maybe this will shed some light. – fretty Jan 2 '15 at 12:13