11.2. A trinomial of degree $p$ is a function of the form $f(x)=x^{p}+a x^{q}+1$, where $p, q-$ are natural numbers, $q<p$, and $a$ is an arbitrary real number (possibly equal to 0). Find all pairs of trinomials that give a trinomial of degree 15 when multiplied.

ANSWER: $\left(1+x^{5}\right)\left(1-x^{5}+x^{10}\right),\left(1-x^{3}+x^{9}\right)\left(1+x^{3}+x^{6}\right),\left(1-x^{6}+x^{9}\right)\left(1+x^{3}+x^{6}\right)$. CRITERIA

A grade of +/- was given for work in which a clear description of a certain final set of alternative options for the trinomial factors (with a complete proof of the non-feasibility of all discarded cases) was provided, and most of these alternative options were fully analyzed, but one of the options was considered with a calculation error and/or not completed or not considered at all, leading the author of the work to lose solutions.

A grade of +/2 was given for work in which a completely correct answer was found using additional assumptions that significantly narrow the search, but the validity of which is not proven in the work; typical examples of such assumptions: without proof, assume that the middle coefficients of the trinomials can only take the values $0, +1$, and $-1$, or without proof, assume that the middle exponents of the trinomial factors divide their leading exponents, etc.

A grade of -/+ was given for work in which explicit examples of the sought trinomials were provided, but the answer as a whole was incorrect (not all trinomials were found or several extra pairs were indicated, the product of which is not a trinomial), and this answer was obtained either based on additional assumptions that significantly narrow the search but are not proven in the work (as in the previous point), or by considering only a small part of a fairly large number of potential options arising in the work.

A grade of "-" was given in all other cases where something was written about this problem; in particular, "-" was given for all works containing one answer (even if correct) but not containing any explanation of its origin.
Solution. Let

$$
\left(x^{p}+a x^{q}+1\right)\left(x^{r}+b x^{s}+1\right)=x^{15}+c x^{t}+1
$$

In particular, the monomials of the highest degree on the right and left sides are equal: $x^{p+r}=x^{15}$. Since $p+r=15$, then $p \neq r$.

First, we investigate how many of the coefficients $a, b$ can be zero. If $a=b=0$, then due to $p \neq r$ the product $\left(x^{p}+1\right)\left(x^{r}+1\right)=x^{p+r}+x^{p}+x^{r}+1$ would not be a trinomial. Therefore, either $a$ or $b$ is not zero (the first case reduces to the second by changing the notation).

If $a=0$ and $b \neq 0$, then upon expanding the brackets in the equality $(*)$ we get $b x^{p+s}+x^{p}+x^{r}+b x^{s}=c x^{t}$. Note that if the degrees of the monomials $x^{p}$ and $x^{s}$ were not equal, then both of these monomials would enter the left side of the resulting equality with non-zero coefficients (because their degrees cannot equal the degrees of the other monomials in the left side), which would contradict the form of the right side. Therefore, $p=s$, and similarly $p+s=r$. In this case, the resulting equality takes the form $(1+b)\left(x^{p}+x^{2 p}\right)=c x^{t}$, which can only hold if $b=-1$ and $c=0$. Under these conditions, the equality $(*)$ takes the form $\left(x^{p}+1\right)\left(x^{2 p}-x^{p}+1\right)=x^{15}+1$, which holds for $p=5$.

Finally, suppose that $a$ and $b$ are non-zero. Upon expanding the brackets in the equality $(*)$ we get

$$
b x^{p+s}+x^{p}+a x^{q+r}+a b x^{q+s}+a x^{q}+x^{r}+b x^{s}=c x^{t}
$$

Note that if the degrees of the monomials $x^{q}$ and $x^{s}$ were not equal, then one of them (with the smaller degree) would enter the left side of the resulting equality with a non-zero coefficient (because its degree could not equal the degrees of the other monomials in the left side). The same can be said about the pair of monomials $a x^{q+r}$ and $b x^{p+s}$. Thus, either the monomials $x^{q}$ and $x^{s}$ have equal degrees, or the monomials $a x^{q+r}$ and $b x^{p+s}$ have equal degrees, and these cases exclude each other, since otherwise we would get $p=r$.

We assume that $q=s$ and $q+r>p+s$. The remaining case $q+r=p+s$ reduces to this case by substituting $x=1 / y$ in the equality $(*)$ and multiplying it by $y^{15}$, and therefore gives an answer that is obtained from the answer in the first case by the same substitution.

In the assumptions made, $r>p$, and the previous equality takes the form

$$
a x^{r+q}+b x^{p+q}+x^{r}+x^{p}+a b x^{2 q}+(a+b) x^{q}=c x^{t}
$$

Note that in the left side the degree of the monomial $x^{r+q}$ is greater than that of the others, so it cannot cancel out with them. Therefore, $a x^{r+q}=c x^{t}$, and $b x^{p+q}+x^{r}+x^{p}+a b x^{2 q}+(a+b) x^{q}=0$. Similarly, the degree of $x^{q}$ is less than that of the other monomials, so it cannot cancel out with them. Therefore, $a+b=0$, and $-a x^{p+q}+x^{r}+x^{p}-a^{2} x^{2 q}=0$. Note that if the degrees of the monomials $x^{p}$ and $x^{2 q}$ were not equal, then both of these monomials would enter the left side of the resulting equality with non-zero coefficients (because their degrees cannot equal the degrees of the other monomials in the left side), which would contradict the form of the right side. Therefore, $p=2 q$, and similarly, $p+q=r$. Under these conditions, the previous equality takes the form $(1-a) x^{r}+\left(1-a^{2}\right) x^{p}=0$, which holds only if $a=1$. Under these conditions, the equation $(*)$ takes the form $\left(x^{2 q}+x^{q}+1\right)\left(x^{3 q}-x^{q}+1\right)=x^{15}+x^{4 q}+1$, which holds for $q=3$. By substituting $x=1 / y$ and multiplying by $y^{15}$, the remaining answer is obtained from this equality.