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If

$x^5+ax^4+ bx^3+cx^2+dx+e$ where $a,b,c,d,e \in {\bf R}$

and

- $\ker \phi = (a_1, …, a_n)$ for a ring homomorphism $\phi: R \to R$
- Content of a polynomial
- Understanding proof by infinite descent, Fermat's Last Theorem.
- Replacing in equation introduces more solutions
- Derivation of binomial coefficient in binomial theorem.
- Irreducibility of a polynomial with degree $4$

$2a^2< 5b$

then the polynomial has at least one non-real root.

We have $-a = x_1 + \dots + x_5$ and $b = x_1 x_2 + x_1 x_3 + \dots + x_4 x_5$.

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- Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree
- Elementary proof of the irreducibility of $T^4 - a T - 1$ in $\mathbf{Q}$ when $a\in\mathbf{Z}-\{0\}$
- Can we prove that all zeros of entire function cos(x) are real from the Taylor series expansion of cos(x)?
- Unique expression of a polynomial under quotient mapping?

We will prove the slightly stronger statement

Claim:Let

$$

p(x) = x^n + ax^{n-1} + bx^{n-2} + \cdots + c

$$

be a polynomial with real coefficients. If $n=2,3,4,5$ and $2a^2 < 5b$ then $p$ has at least one non-real zero.If $n \geq 6$ then $p$ may have all real zeros regardless of whether $2a^2 < 5b$ or $2a^2 \geq 5b$. For example, the polynomial

$$

\left(x – \sqrt{\frac{2}{n(n-1)}}\right)^n = x^n – \sqrt{\frac{2n}{n-1}} x^{n-1} + x^{n-2} + \cdots + \left(\frac{2}{n(n-1)}\right)^n

$$

has all real zeros with $2a^2 < 5b$.

(In fact $p$ has at least two real zeros since they will come in complex conjugate pairs.)

As noted in the question, if $x_1,\ldots,x_n$ are the zeros of $p$ then

$$

a = -\sum_{k=1}^{n}x_k \qquad \text{and} \qquad b = \sum_{1 \leq j < k \leq n} x_j x_k,

$$

so that

$$

\begin{align}

2a^2 &= 2\sum_{k=1}^{n} x_k^2 + 4\sum_{1 \leq j < k \leq n} x_j x_k \\

&= 2\sum_{k=1}^{n} x_k^2 + 4b

\end{align}

$$

or, rearranging,

$$

2a^2 – 5b = 2\sum_{k=1}^{n} x_k^2 – \sum_{1 \leq j < k \leq n} x_j x_k.

$$

To prove your statement it therefore suffices to show that

$$

2\sum_{k=1}^{n} x_k^2 \geq \sum_{1 \leq j < k \leq n} x_j x_k \tag{1}

$$

for $x_1,\ldots,x_n \in \mathbb R$ and $n = 5$.

It’s clearly true if

$$

\sum_{1 \leq j < k \leq n} x_j x_k \leq 0,

$$

so suppose that

$$

\sum_{1 \leq j < k \leq n} x_j x_k > 0

$$

and set

$$

y_\ell = \left( \sum_{1 \leq j < k \leq n} x_j x_k \right)^{-1/2} x_\ell, \qquad \ell = 1,\ldots,n.

$$

Then

$$

\sum_{k=1}^{n} x_k^2 = \left(\sum_{1 \leq j < k \leq n} x_j x_k \right) \sum_{k=1}^{n} y_k^2,

$$

so that $(1)$ becomes

$$

2 \sum_{k=1}^{n} y_k^2 \geq 1. \tag{2}

$$

We also note that

$$

\sum_{1 \leq j < k \leq n} y_j y_k = 1.

$$

Our mode of attack to prove the new inequality $(2)$ will be to use Lagrange multipliers to minimize $\sum_{k=1}^{n} y_k^2$ subject to this constraint. To this end, define

$$

f(y_1,\ldots,y_n,\lambda) = \sum_{k=1}^{n} y_k^2 + \lambda\left(\sum_{1 \leq j < k \leq n} y_j y_k – 1\right).

$$

We calculate

$$

f_{y_k}(y_1,\ldots,y_n,\lambda) = 2y_k + \lambda \sum_{j\neq k} y_j = (2-\lambda) y_k + \lambda \sum_{j=1}^{n} y_j, \qquad k=1,\ldots,n,

$$

and thus we need to solve the $n+1$ equations

$$

\begin{align}

0 &= (2-\lambda) y_k + \lambda \sum_{j=1}^{n} y_j, \qquad k=1,\ldots,n, \\

1 &= \sum_{1 \leq j < k \leq n} y_j y_k.

\end{align}

$$

If $\lambda \neq 2$ then

$$

y_k = \frac{\lambda}{\lambda-2} \sum_{j=1}^{n} y_j

$$

and so $y_1 = y_2 = \cdots = y_n$, and if $y_k \neq 0$ then this yields

$$

\lambda = \frac{2}{1-n}.

$$

Further, the equation $1 = \sum_{1 \leq j < k \leq n} y_j y_k$ becomes

$$

y_k = \sqrt{\frac{2}{n(n-1)}}.

$$

We will omit checking that this is a local minimum of the problem. Finally, toward $(2)$ we calculate

$$

2\sum_{k=1}^{n} y_k^2 = \frac{4}{n-1} \geq 1

$$

for $n = 2,3,4,5$, from which the claim, and the result in the question, follows. In particular, taking $x_\ell = y_\ell$ furnishes the counterexample at the end of the claim.

Here is a simple proof.

Let’s prove this by contraposition: if $x_1, x_2, x_3, x_4, x_5$ are all real, then $2a^2 \geq 5b$ should hold.

You have a formula for $-a$ and $b$ already; from these you can easily calculate that

$$2a^2-5b = \left( 2x_1^2 + 2x_2^2 + 2x_3^2 + 2x_4^2 + 2x_5^2 \right) – x_1x_2 – x_1x_3 – \dots – x_4x_5.$$

Double it once more, and it starts to remind us of something:

$$2(2a^2-5b) = \color{tomato}{\left( 4x_1^2 + 4x_2^2 + 4x_3^2 + 4x_4^2 + 4x_5^2 \right)} \color{steelblue}{- 2x_1x_2 – 2x_1x_3 – \dots – 2x_4x_5}.$$

We have a bunch of $x_i^2 + x_j^2$ and a bunch of $-2x_ix_j$, which looks like the expansion of $(x_i – x_j)^2$, doesn’t it? Using this clever insight, we can rewrite this expression into a sum of squares. For each term subtracted in the blue half, we combine it the squares it needs from the red half. Since each $x_i \in \{x_1, x_2, x_3, x_4, x_5\}$ occurs *exactly four times* in the blue half, all of our squares in the red half nicely vanish. All we are left with is

\begin{align*}

2(2a^2-5b) &= \left(\color{tomato}{x_1^2} \color{steelblue}{-2x_1 x_2} \color{tomato}{+x_2^2}\right) + \left(\color{tomato}{x_1^2} \color{steelblue}{-2x_1 x_3} \color{tomato}{+x_3^2}\right) + \dots

\\ &= (x_1-x_2)^2 + (x_1-x_3)^2 + \dots + (x_4-x_5)^2

\end{align*}

(If you’re not convinced, try expanding this back.)

Since we’ve written $2(2a^2-5b)$ as a sum of squares, clearly it is $\geq 0$. We can conclude that $2a^2 \geq 5b$.

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