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

Caleb Stanford
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Irtiza
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    Try Viete's Formulae. – Teoc Jan 02 '15 at 07:35
  • http://math.stackexchange.com/questions/1082398/find-coefficient-in-quartic-given-product-of-roots – Bumblebee Jan 02 '15 at 07:41
  • Tried that.Some other technique is required here. – Irtiza Jan 02 '15 at 07:41
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    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 02 '15 at 07:46
  • @MathN00b Vieta! I know it's named after François Viète but the Latinised form of his name is more popular. – Nick Jan 02 '15 at 08:00
  • @Nick It is merely a matter of notation, it does not matter. – Teoc Jan 02 '15 at 08:04
  • @MathN00b: I know nothing really matters but Viète's formula is $$\lim_{n \to \infty} \prod_{i=1}^{n} \frac{a_1}{2} = \frac{2}{\pi}\ ;\ a_n = \sqrt{2 + a_{n-1}},\ a_1 = \sqrt{2}$$ The matter of notation matters here does matter and I don't care what you have to say about interpretation through circumstance. Spell names the way they're supposed to spelt, please. – Nick Jan 02 '15 at 08:13
  • @Irtiza: This isn't really a "topic", you can't learn everything from a book. Sometimes you have to use what you know and think. I didn't find the answer by memorizing some solution in a book. – fretty Jan 02 '15 at 09:17

2 Answers2

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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:

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

fretty
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  • can you explain your last step – Irtiza Jan 02 '15 at 09:02
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    @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 02 '15 at 09:11
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    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 02 '15 at 09:15
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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 $2,1,1$.

Marc van Leeuwen
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  • I was going to go down this route but at the last minute saw that the expression simply was $f(p)$. – fretty Jan 02 '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 02 '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 02 '15 at 12:13