Let the degree-four polynomial be $k_4x^4 + k_3x^3 + k_2x^2 + k_1x + k_0$, with $k_4 \neq 0$. We look for quadratic trinomials $P(x) = \alpha x^2 + \beta x + \gamma$, $Q(x) = ax^2 + bx + c$, $R(x) = \alpha' x^2 + \beta' x + \gamma'$, $S(x) = a'x^2 + b'x + c'$. Computing $P(Q(x))$ and $R(S(x))$, and then by coefficient identification, we get a system of five equations in five unknowns $\alpha, \alpha', \beta, \beta', \gamma + \gamma'$, whose matrix (and extended matrix) is \[M = \left [ \begin{array} {c c c c c|c} a^2 & a'^2 & 0 & 0 & 0 & k_4 \\ 2ab & 2a'b' & 0 & 0 & 0 & k_3 \\ b^2 + 2ac & b'^2 + 2a'c' & a & a' & 0 & k_2 \\ 2bc & 2b'c' & b & b' & 0 & k_1 \\ c^2 & c'^2 & c & c' & 1 & k_0 \\ \end{array} \right ]\] By an immediate computation $\det M = 2aa'(ab' - a'b)^2$, which we wish be not-null, the same as $aa'$. Moreover, we need ensure $\alpha\alpha' \neq 0$. Thus, the first condition is over the field $\mathbb{K}$ where the coefficients of the polynomials live. We should have $2\neq 0$, thus the characteristic of $\mathbb{K}$ must not be $2$ (and we can se that otherwise, for $k_3\neq 0$, we cannot have a solution, thus the condition over the characteristic is also necessary). Then we need $b/a \neq b'/a'$. Since we will have $\alpha = a'(2k_4b' - k_3a')(ab' - a'b)$ and $\alpha' = a(2k_4b - k_3a)(ab' - a'b)$, we will need $b/a \neq k_3/2k_4$, and $b'/a' \neq k_3/2k_4$. These are all necessary conditions, which therefore impose $b/a$, $b'/a'$ and $k_3/2k_4$ to be distinct elements in $\mathbb{K}$. But any field of characteristic different from $2$ has at least three elements. We then obtain a solution (unique for each judicious choice of parameters $a,a',b,b'$ and $c,c'$). Moreover, surely instead of writing the system of five equations, one could solve the equations in turn, almost one by one; it is just more convenient and direct to proceed as above. Remark. A strange choice of a problem ... unless the author had a different approach in mind. Unfortunately, from my point of view, the very existence of the above solution makes of this problem a trite computational exercise.

Let $T(x)=mx^4+nx^3+px^2+qx+r$ ($m,n,p,q,r\in\Bbb{R}$). We choose $a,b,c,d,e,f\in\Bbb{R}$ such that $2ade=n$ and $2aef+2be=q$ (this is very simple; for example $a=d=1;e=\dfrac{n}{2};b=\dfrac{q-2aef}{2e}$ and $f$ random). Now we take $R(x)=ax^2+bx+x$ and $S(x)=dx^2+ex+f\Rightarrow$ there exist $\alpha,\beta,\gamma\in\Bbb{R}$ such that $R(S(x))=x^4\alpha+x^3n+x^2\beta+xq+\gamma$. Finally, we choose $Q(x)=x^2$ and $P(x)=(m-\alpha)x^2+(p-\beta)x+r-\gamma$.

mavropnevma wrote: Remark. A strange choice of a problem ... unless the author had a different approach in mind. Unfortunately, from my point of view, the very existence of the above solution makes of this problem a trite computational exercise. I think the point is almost arbitrary approach turns to be suitable. (You see, this is the first problem of a day...). My first thought was to add a constant to obtain a polinomial with positive values. Such a polinomial you can represent as a sum of two squares:)

My solution from the contest which is not the official one: Write the original polynomial as PQ with P, Q of degree 2 and use the identity $ \frac{ (P(x)+aQ(x) )^2 - ( P(x)-aQ(x))^2 } {4a}= P(x)Q(X)$

Both ideas mentioned in the last two posts are applicable when working with real polynomials (paul's is especially short and sweet, working also for complex polynomials - it needs factorization of the degree four polynomial). Of course, the problem setters omitted to specify that, and later circulated the information about the field of coefficients in the contest rooms. My updated post works in any field of characteristic different from $2$ (and also shows the conclusion otherways fails), via boring Linear Algebra techniques (which makes me dislike the problem).