SAUDI ARABIAN MATHEMATICAL COMPETITIONS

We consider 3 cases.

1) Suppose that $\mathrm{deg}Q=1$. By looking at the leading coefficient, we have $Q(x)=\pm x+a$. If $Q(x)=x+a$, we have $$P(x)=(x-(a+1))(x-(a+2))\dots (x-(a+9)).$$ And if $Q(x)=-x+a$ we have $$P(x)=(a-1-x)(a-2-x)\dots (a-9-x).$$

2) Suppose that $\mathrm{deg}P=1$. By looking at the leading coefficient, we have $P(x)=\pm x+a$. In this case $$Q(x)=\pm (x-1)(x-2)\dots (x-9)-a.$$

3) Suppose $\mathrm{deg}P,\mathrm{deg}Q>1$. Then, the only option is $\mathrm{deg}P=\mathrm{deg}Q=3$. Since $1,2,\dots ,9$ are 9 roots of $P(Q(x))$, we have $P(x)$ have 3 roots $a,b,c$. For each root, for example $a$, the equation $Q(x)=a$ has 3 roots $a_1,a_2,a_3$ then $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3$ form a permutation of $1,2,\dots ,9$. By Vieta's formula, we have $$a_1+a_2+a_3=b_1+b_2+b_3\text{ and }{a}_1{a}_2+a_2{a}_3+a_3{a}_1=b_1{b}_2+b_2{b}_3+b_3{b}_1$$ so $a_1^2+a_2^2+a_3^2=b_1^2+b_2^2+b_3^2$. Since $x^2\equiv 0$ or $1(\mathrm{mod}4)$ so the number of odd numbers in $a_1,a_2,a_3$ is the same as the number of odd numbers in $b_1,b_2,b_3$. This implies that the number of odd numbers in $1,2,\dots ,9$ is divisible by 3, which is a contradiction.