Prove that polynomial of degree $4$ with real roots cannot have $\pm 1$ as coefficients (IITJEE)

Let $f(x)$ be any quartic polynomial with coefficients from $\{ -1, +1 \}$. Replacing $f(x)$ by $-f(x)$ if necessary, we can assume $f(x)$ is monic. i.e.

$$f(x) = x^4 + ax^3 + bx^2 + cx + d\quad\text{ with }\quad a,b,c,d \in \{ -1, +1 \}$$

If $f(x)$ has $4$ real roots $\lambda_1,\lambda_2,\lambda_3,\lambda_4$, then by Vieta's formula, we have

$$\sum_{i=1}^4 \lambda_i = -a, \sum_{1\le i < j\le 4} \lambda_i\lambda_j = b \quad\text{ and }\quad\prod_{i=1}^4 \lambda_i = d$$ Notice $$\sum_{i=1}^4 \lambda_i^2 = \left(\sum_{i=1}^4\lambda_i\right)^2 - 2\sum_{1\le i < j \le 4}\lambda_i\lambda_j = a^2 - 2b = 1 -2b$$

Since $\sum_{i=1}^4 \lambda_i^2 \ge 0$, we need $b = -1$. As a result, $$\sum_{i=1}^4 \lambda_i^2 = 3$$ By AM $\ge$ GM, this leads to

$$\frac34 = \frac14\sum_{i=1}^4 \lambda_i^2 \ge \left(\prod_{i=1}^4 \lambda_i^2\right)^{1/4} = (d^2)^{1/4} = 1$$ This is impossible and hence $f(x)$ cannot has 4 real roots.

It can be assumed WLOG that the leading coefficient is $\,+1\,$, so $\,P(x)=x^4\pm x^3\pm x^2\pm x\pm 1\,$.

• Then $\,P''(x)=12x^2 \pm 6x \pm 2\,$, and for the quadratic to have real roots it is necessary that the constant term be negative, so $\,P(x)=x^4\pm x^3 - x^2\pm x\pm 1\,$.

• $P(x)\,$ has all the roots real iff $\,x^4 P\left(\frac{1}{x}\right)\,$ has all real roots. By the same argument as above, the constant term of $\,P(x)\,$ must have opposite sign as the coefficient of $\,x^2\,$.

This leaves $4$ cases to check $\,P(x)=x^4\pm x^3 - x^2\pm x+1\,$.

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[ EDIT ]

• $P(x)\,$ has all the roots real iff $\,P\left(-x\right)\,$ has all real roots, so it is enough to consider the case where the coefficient of $\,x^3\,$ is $+1$.

This leaves $2$ cases to check $\,P(x)=x^4+ x^3 - x^2\pm x+1\,$.