# Prove that the cubic has at least one of its four coefficients equals to or less than $-2$.

Suppose that $P(x)$ is a polynomial of degree $3$ with integer coefficients and that $P(1)=0,P(2)=0$. Prove that at least one of its four coefficients is equal to or less than $−2$.

We can write such a polynomial as $(x-1)(x-2)(ax+b) = a x^3-3 a x^2+2 a x+b x^2-3 b x+2 b$. What is the next step to prove that at least one of its four coefficients is equal to or less than $−2$?

• We have $ax^3+x^2(b-3a)+x(2a-3b)+2b$. Then I will show by contradiction that at least one of its coefficients is equal to or less than $-2$. Assume on the contrary the opposite. Then $b \geq -1, a \geq -1$. Then what? – user19405892 Dec 6 '15 at 15:55
• We have $ax^3+x^2(b-3a)+x(2a-3b)+2b$. If $a$ is negative we have $ax^3+x^2(b+3a)+x(-2a-3b)+2b$ for $a \geq 1$. We then see that $a$ can't be negative otherwise $x(2a-3b)$ will have a coefficient that is less than or equal to $-2$. Therefore, $a \geq 1$. We also have that $2b > -2$ and $b \geq 1$. – user19405892 Dec 6 '15 at 16:08
• @turkeyhundt $a$ is an integer. – rogerl Dec 6 '15 at 16:09
• We have $b-3a > -2; 2a-3b > -2$ and from that we get $1 \leq a < \dfrac{8}{7}$ implying that $a = 1$. Our polynomial then becomes $x^3+x^2(b-3)+x(2-3b)+2b$ and from which we obtain $1 < b < 4/3$, a contradiction. Therefore, at least one of its four coefficients is equal to or less than $-2$. – user19405892 Dec 6 '15 at 16:15
• @rogerl Ha. It pays to read the original question. Thx. – turkeyhundt Dec 6 '15 at 16:20

Necessarily there is a third real root $x_0$. Then one has $$f(x)=x^3-(3-x_0)x^2+(2+3x_0)x-2x_0$$ We can verify that the system of inequalities $$-3+x_0>-2\iff x_0>1$$ and $$-2x_0>-2\iff x_0<1$$ is incompatible so one of the two considered coefficients must be $\le -2$
Let $P(x)=-(ax^3+bx^2+cx+d)$, where the negative sign out front allows us to reformulate the condition as one of the integer coefficients $a$, $b$, $c$, or $d$ is at least $2$. Now suppose the condition is not met, so $a,b,c,d\le1$.
The assumption $P(1)=0$ implies $a+b+c+d=0$. There are four possibilities:
$$\{a,b,c,d\}= \begin{cases} \{1,1,1,-3\}\\ \{1,1,-1,-1\}\\ \{1,0,0,-1\}\\ \{1,1,0,-2\}\\ \end{cases}$$
The first two cases can be dismissed because if all four coefficients are odd, then $P(2)$ is odd. The third case can be dismissed because no two powers of $2$ are equal, and the fourth case can be dismissed because no two powers of $2$ sum to another power of $2$.
Note, this proof applies, with a tiny extra argument, to any polynomial with at most $4$ nonzero (integer) coefficients. (The extra argument is to factor out and ignore the largest power of $x$ dividing the polynomial.)