# 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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Note by Mursalin Habib
3 years, 11 months ago

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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.

- 3 years, 11 months ago