Kvant Math Problem 1515

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 3m42s
Source on kvant.digital

Problem

It is known that $f(x)$, $g(x)$, and $h(x)$ are quadratic trinomials. Can the equation $f(g(h(x)))=0$ have roots 1, 2, 3, 4, 5, 6, 7, and 8?

S. I. Tokarev

Russian Mathematical Olympiad

Exploration

Let

$$F(x)=f(g(h(x))).$$

Since $f,g,h$ are quadratic trinomials, each has degree $2$, hence $F$ has degree $8$. If the roots $1,2,\dots,8$ are roots of $F$, then these are all roots of $F$, because an eighth degree polynomial cannot have more than eight distinct roots.

The equation $F(x)=0$ is solved by tracing preimages through the composition. Let the two roots of $f$ be $a$ and $b$. Then

$$g(h(x))\in{a,b}.$$

For a fixed value $c$, the equation $g(y)=c$ is quadratic in $y$, so it has at most two solutions. Hence $g(y)=a$ and $g(y)=b$ together produce at most four values $y$. For each such $y$, the equation $h(x)=y$ has at most two solutions. Thus $F(x)=0$ has at most $8$ roots, which matches the degree.

If $F$ actually has eight distinct roots, every intermediate quadratic equation must attain its maximal number of distinct solutions. This suggests a binary tree of preimages: two roots of $f$, each giving two distinct preimages under $g$, each of those giving two distinct preimages under $h$.

The question becomes whether the eight numbers $1,\dots,8$ can be arranged in such a tree.

A useful fact about a quadratic polynomial $q(x)$ is that whenever $q(u)=q(v)$ with $u\neq v$, the axis of symmetry equals

$$\frac{u+v}{2}.$$

Hence all pairs of distinct points with equal $q$-value have the same sum.

If the eight roots are $1,\dots,8$, then at the last stage, the roots must split into four pairs corresponding to equal values of $h$. All these pairs must have the same sum. Among $1,\dots,8$, the only way to partition all numbers into four pairs with a common sum is

$$(1,8),\ (2,7),\ (3,6),\ (4,5),$$

whose common sum is $9$.

The corresponding four values of $h$ must then split into two pairs having equal $g$-values. Since the arguments of $g$ occur in symmetric pairs about the axis of $h$, those four values themselves form an arithmetic pattern. Computing with a normalized quadratic should reveal whether they can be paired appropriately.

The delicate point is proving that the four values of $h$ obtained from the pairs above cannot be partitioned into two equal-sum pairs, as would be required for a quadratic $g$.

Problem Understanding

We are given quadratic trinomials $f(x)$, $g(x)$, and $h(x)$. The question is whether the equation

$$f(g(h(x)))=0$$

can have the eight numbers $1,2,3,4,5,6,7,8$ as roots.

This is a Type D problem: we must determine whether such a configuration exists. The core difficulty is that having eight distinct roots forces every quadratic stage of the composition to realize the maximal possible number of distinct preimages, imposing strong symmetry constraints on the roots.

The expected answer is that such a configuration does not exist. The reason is that equal values of a quadratic occur at points symmetric about its axis, forcing a rigid pairing structure that is incompatible with the numbers $1,\dots,8$.

Proof Architecture

First, prove that if $f(g(h(x)))$ has eight distinct roots, then the roots must be partitioned into four pairs on which $h$ takes equal values, and every such pair has the same sum.

Second, show that for the set ${1,2,\dots,8}$, the only partition into four pairs with a common sum is $(1,8),(2,7),(3,6),(4,5)$.

Third, normalize $h$ as $h(x)=A(x-\tfrac92)^2+B$, compute the four values corresponding to those pairs, and show that they are

$$B+\frac{49}{4}A,\quad B+\frac{25}{4}A,\quad B+\frac94A,\quad B+\frac14A.$$

Fourth, prove that if a quadratic $g$ takes equal values at two distinct arguments, then the sums of those arguments in each equal-value pair must be the same. Apply this to the four values above and show that no partition into two pairs has equal sums.

The hardest step is the last one, because it converts the geometric symmetry condition for $g$ into an arithmetic condition on the four values produced by $h$.

Solution

Assume that there exist quadratic trinomials $f,g,h$ such that

$$f(g(h(k)))=0$$

for $k=1,2,\dots,8$.

Since $f,g,h$ are quadratic, the polynomial

$$F(x)=f(g(h(x)))$$

has degree $8$. Because $1,2,\dots,8$ are eight distinct roots of $F$, they are all roots of $F$.

Let the roots of $f$ be $a$ and $b$. Then

$$F(x)=0$$

is equivalent to

$$g(h(x))=a$$

$$g(h(x))=b.$$

For each fixed constant $c$, the equation $g(y)=c$ is quadratic in $y$, hence has at most two solutions. Therefore the equations

$$g(y)=a,\qquad g(y)=b$$

produce at most four values of $y$.

For each such value $y$, the equation

$$h(x)=y$$

has at most two solutions. Consequently $F(x)=0$ has at most $8$ roots.

Since $F(x)$ actually has $8$ distinct roots, every one of these upper bounds is attained. Hence:

$f$ has two distinct roots $a,b$;
each of $g(y)=a$ and $g(y)=b$ has two distinct solutions;
for each resulting value $y$, the equation $h(x)=y$ has two distinct solutions.

Thus the eight roots $1,\dots,8$ are partitioned into four pairs, and within each pair the values of $h$ are equal.

Let

$$h(x)=A(x-\alpha)^2+\beta, \qquad A\neq0.$$

If $h(u)=h(v)$ and $u\neq v$, then

$$(u-\alpha)^2=(v-\alpha)^2,$$

hence

$$u+v=2\alpha.$$

Therefore every pair of roots corresponding to the same value of $h$ has the same sum.

The numbers $1,2,\dots,8$ must therefore be partitioned into four pairs having a common sum. The number $1$ must be paired with $8$, because otherwise the common sum would be less than $9$ or greater than $9$. After removing $1$ and $8$, the same argument gives the pairs

$$(2,7),\qquad (3,6),\qquad (4,5).$$

Hence the common sum is $9$, and

$$\alpha=\frac92.$$

Write

$$h(x)=A\left(x-\frac92\right)^2+B.$$

The four equal values of $h$ corresponding to the four pairs are

$$\begin{aligned} t_1&=h(1)=h(8)=B+\frac{49}{4}A,\ t_2&=h(2)=h(7)=B+\frac{25}{4}A,\ t_3&=h(3)=h(6)=B+\frac94A,\ t_4&=h(4)=h(5)=B+\frac14A. \end{aligned}$$

Since $F(x)$ has eight roots, the four numbers $t_1,t_2,t_3,t_4$ must be distinct and must be partitioned into two pairs on which $g$ takes equal values.

Let

$$g(y)=C(y-\gamma)^2+D.$$

If $g(r)=g(s)$ and $r\neq s$, then

$$(r-\gamma)^2=(s-\gamma)^2,$$

$$r+s=2\gamma.$$

Thus the two pairs of arguments giving equal values of $g$ must have the same sum.

We compute all possible pairings of $t_1,t_2,t_3,t_4$.

If the pairs are $(t_1,t_2)$ and $(t_3,t_4)$, then

$$t_1+t_2=2B+\frac{74}{4}A, \qquad t_3+t_4=2B+\frac{10}{4}A,$$

which are unequal.

If the pairs are $(t_1,t_3)$ and $(t_2,t_4)$, then

$$t_1+t_3=2B+\frac{58}{4}A, \qquad t_2+t_4=2B+\frac{26}{4}A,$$

which are unequal.

If the pairs are $(t_1,t_4)$ and $(t_2,t_3)$, then

$$t_1+t_4=2B+\frac{50}{4}A, \qquad t_2+t_3=2B+\frac{34}{4}A,$$

which are unequal.

No partition of ${t_1,t_2,t_3,t_4}$ into two pairs yields equal pair sums. Hence no quadratic polynomial $g$ can take equal values on two such pairs.

This contradiction shows that the assumed quadratic trinomials do not exist.

Therefore the equation $f(g(h(x)))=0$ cannot have roots $1,2,3,4,5,6,7,8$.

$$\boxed{\text{Such quadratic trinomials do not exist.}}$$

Verification of Key Steps

The first delicate step is the deduction that every upper bound is attained. The equation $g(y)=a$ has at most two solutions, and $g(y)=b$ has at most two solutions, so there are at most four intermediate values $y$. Each equation $h(x)=y$ contributes at most two roots. Hence $F(x)=0$ has at most $8$ roots. Since there are exactly eight distinct roots, neither stage can lose a solution. If even one equation $g(y)=c$ had only one solution, or one equation $h(x)=y$ had only one solution, the total number of roots would drop below $8$.

The second delicate step is the pairing of $1,\dots,8$. For a quadratic, equal values occur at points symmetric about the axis, so every pair has the same sum $2\alpha$. Since $1$ belongs to some pair and all numbers are between $1$ and $8$, the common sum must be $1+8=9$. This forces the unique partition

$$(1,8),(2,7),(3,6),(4,5).$$

The third delicate step is the impossibility for $g$. The four values produced by $h$ differ only by the coefficients

$$49,\ 25,\ 9,\ 1.$$

The three possible pairings give coefficient sums

$$74\ \text{and}\ 10,\qquad 58\ \text{and}\ 26,\qquad 50\ \text{and}\ 34.$$

None are equal. Since equal values of a quadratic require equal sums of the paired arguments, every pairing fails.

Alternative Approaches

Instead of writing $h(x)=A(x-\tfrac92)^2+B$, one may translate and scale so that the four values produced by $h$ are simply proportional to

$$49,\ 25,\ 9,\ 1.$$

The problem then becomes purely combinatorial. A quadratic $g$ identifies pairs of numbers symmetric about its axis, so among these four values there must exist a partition into two pairs with equal sums. Direct inspection of the three possible pairings shows that no such partition exists.

Another approach uses the graph of a quadratic. The roots $1,\dots,8$ force four horizontal chords of the parabola $y=h(x)$ centered on the same vertical axis. Their heights correspond to the four values above. The next quadratic $g$ would require those four heights to form two symmetric pairs about a second axis. The arithmetic of the heights prevents such a symmetry, yielding the same contradiction.