Proof Problem Of The Day - Co-efficient Restriction!

Let $P(x)$ be a polynomial such that its co-efficients are equal to $\pm1$ and its roots are all real.

Prove that $\text{deg}(P)\leq 3$

Bonus: Find all such polynomials.

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#Algebra #Polynomials #Vieta'sFormula #DegreeOfAPolynomial #AM-GMInequality

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Comments

Let $P(x)=\displaystyle \sum_{k=0}^n a_k x^k$ so that $\deg P=n$ and let it's roots be $r_i\in\mathbb{R}$ for $1\le i\le n$ . WLOG let $a_n=1$ . Applying Vieta's we get

$\sum_{\text{cyc}} r_i^2=\left(\sum_{\text{cyc}}r_i\right)^2-2\sum_{\text{cyc}} r_ir_j=3.$

Notice that we must have $\displaystyle \sum_{\text{cyc}} r_ir_j=-1$ because otherwise the sum of squares become negative. Finally applying Vieta's once again along with AM-GM inequality gives

$n=n\sqrt[n]{\prod_{\text{cyc}} r_i^2}\le \sum_{\text{cyc}} r_i^2=3$

which is $\deg P\le 3$ as desired. $\square$

Constant: No solutions.

Linear: $P(x)=\pm x\pm 1$ .

Quadratic: For $b,c\in\{1,-1\}$ consider $P(x)=x^2+bx+c$ so discriminant $b^2-4c$ , We must have $1=b^2\ge 4c$ so $c=-1$ . Hence $P(x)=\pm\left(x^2\pm x-1\right)$ .

Cubic: For $b,c,d\in\{1,-1\}$ consider $P(x)=x^3+bx^2+cx+d$ so discriminant

$b^2c^2-4c^3-4b^3d-27d^2+18bcd=1-4c^3-4b^3d-27+18bcd\ge 0.$

This rearranges to $18bcd\ge 4c^3+4b^3d+26$ . Now notice that the left side is $\pm 18$ . The right side is $\pm 4\pm 4+26$ . From this we deduce that both sides must be equal to $18$ . This implies $c=-1$ and $(b,d)=(1,-1),(-1,1)$ . Therefore the solutions are $P(x)=\pm\left(x^3\pm x^2-x\mp 1\right)$ . Note that since the discriminant is zero, there's a repeated root of these cubics.

Jubayer Nirjhor - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$