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.

Click here to see all the problems posted so far. Keep checking everyday!

## Comments

Sort by:

TopNewestLet \(P(x)=\displaystyle \sum_{k=0}^n a_k x^k\) 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 · 2 years, 7 months ago

Log in to reply