[Edit] : Incorrect ...

If $\mu,\rho$ are the roots of $ax^2+bx+c=0$, then ${1\over\mu},{1\over\rho}$ are the roots of $cx^2+bx+a=0$. WLOG assume $\mu\in\mathbb{Z}$. If ${1\over\mu}$ is also an integer, then $\mu=\pm 1$. If $\mu=1$, there's nothing to prove, so assume $\mu=-1$. Then by Vieta, $b=(1-\rho)a$ and $c=-\rho a$. Thus the respective discriminants of the equations $bx^2+cx+a=0$ and $cx^2+ax+b=0$ are $D_1=c^2-4ab=a^2(\rho^2+4\rho-4)$ and $D_2=a^2-4bc=a^2(-4\rho^2+4\rho+1)$. Let's discuss their sign: $D_1$ is non-negative for $\rho\leqslant -2-2\sqrt{2}\lor\rho\geqslant -2+2\sqrt{2}$, and $D_2$ is non-negative for ${1-\sqrt{2}\over 2}\leqslant\rho\leqslant{1+\sqrt{2}\over 2}$. Thus, if the equations are to have real solutions, then $2(\sqrt{2}-1)\leqslant\rho\leqslant{1+\sqrt{2}\over 2}$. The roots of $ax^2+cx+b=0\iff ax^2-\rho ax+(1-\rho)a=0$ are ${\rho\pm\sqrt{\rho^2+4\rho-4}\over 2}$. By the initial condition, there must exist an integer $v$ such that ${\rho\pm\sqrt{\rho^2+4\rho-4}\over 2}=v\implies\rho^2+4\rho-4=(2v-\rho)^2$, which yields $\rho={v^2+1\over v+1}=v-1+{2\over v+1}$. If $v\leqslant -2$, then $\rho$ is negative and falls out of the favorable interval. If $v\geqslant 2$, then $\rho\geqslant{5\over 3}$ and also falls out of the favorable interval. Hence $v\in\{0,1\}$, yielding $\rho=1$ and the claim is again proven. So assume $\mu\neq\pm 1$. Then we must have ${1\over\rho}\in\mathbb{Z}$, hence the roots of $ax^2+bx+c=0$ are of the form $m,{1\over n}$ for some integers $m,n\neq\pm 1$. Similarly for the equations $bx^2+cx+a=0$ and $cx^2+ax+b=0$. Hence by Vieta we can write $-{b\over a}=m+{1\over n}$ $-{c\over b}=p+{1\over q}$ $-{a\over c}=r+{1\over s}$ where $p,q,r,s$ are some integers different from $\pm 1$. If the absolute values of $m,n,p,q,r,s$ are all at least $2$, then the three equations yield $\left|{b\over a}\right|>1, \left|{c\over b}\right|>1,\left|{a\over c}\right|>1$, which after multiplying out yields $1>1$. Therefore the absolute value of at least one number in question must be $1$, and then it makes no difference if the actual number is $1$ or $-1$, as shown in the first part of the proof. Hence there exists at least one permutation of $a,b,c$ - denote it by $\alpha,\beta,\gamma$ - such that $g(x)=\alpha x^2+\beta x+\gamma$ vanishes at $x=1$. But as $0=g(1)=\alpha+\beta+\gamma=a+b+c=f(1)$, the statement is proven.

If three quadratic trinomials : $ax^2+bx+c=0,\ bx^2+cx+a=0,\ cx^2+ax+b=0$ have a real common root $\alpha$. Since $\alpha$ satisfies those equations, then, we have $a\alpha ^ 2+b\alpha +c=0,\ b\alpha ^2 +c\alpha +a=0,\ c\alpha ^ 2 +a\alpha +b=0$. Summing them up gives $(a+b+c)(\alpha ^ 2+\alpha +1)=0$. Since $\alpha$ is real, we have $\alpha ^ 2+\alpha+1>0$, so the necessary condition is $a+b+c=0$. Conversely if this condition holds, $f(x)=ax^2+bx+c=0$, since $abc\neq 0$, the solutions are $x=1,\ \frac{c}{a}$, analogously from other equations, we obtain real solutions $x=1,\ \frac{a}{b}$ or $x=1,\ \frac{b}{c}$, thus $ x=1$ is the common root, which shows the sufficient condition. Therefore $f(1)=a+b+c=0$. We should consider other cases. Last Edited

kunny wrote: We will find the condition for which three quadratic trinomials : $ax^2+bx+c=0,\ bx^2+cx+a=0,\ cx^2+ax+b=0$ have a real common root $\alpha$. Since $\alpha$ satisfies those equations, we have $a\alpha ^ 2+b\alpha +c=0,\ b\alpha ^2 +c\alpha +a=0,\ c\alpha ^ 2 +a\alpha +b=0$. Summing them up gives $(a+b+c)(\alpha ^ 2+\alpha +1)=0$. Since $\alpha$ is real, we have $\alpha ^ 2+\alpha+1>0$, so the necessary condion is $a+b+c=0$. Conversely if this condition holds, $f(x)=ax^2+bx-(a+b)=0\Longleftrightarrow (x-1)(ax-b)=0$, since $abc\neq 0$, yielding real solutions $x=1,\ \frac{b}{a}$, analogously from other equations, we obtain real solutions $x=1,\ \frac{c}{b}$ or $x=1,\ \frac{a}{c}$, thus $ x=1$ is the common root, which shows the sufficient condition. Therefore $f(1)=a+b+c=0$. I don't think "some of them have a common root" is the same as "all of them have an integer root".

You are right. My solution was not perfect. I should have said, if three equations have real roots, ... Anyway, I made some typos in my post above, I have just edited it.