Problem A. 937. (May 2026)
A. 937. Let \(\displaystyle P \in \mathbb{C}[x_1,\ldots,x_n]\) be an irreducible complex polynomial of degree at least 2. Assume that there exists an integer \(\displaystyle M>1\) such that \(\displaystyle P(x_1,\dots,x_n)\mid P(x_1^M,\dots,x_n^M)\). Prove that there exist a nonzero complex number \(\displaystyle c\), a complex root of unity \(\displaystyle \zeta\) and non-negative integers \(\displaystyle a_1\), \(\displaystyle \ldots\), \(\displaystyle a_n\), \(\displaystyle b_1\), \(\displaystyle \dots\), \(\displaystyle b_n\) such that \(\displaystyle P(x_1,\dots,x_n)=c\left(x_1^{a_1}\cdots x_n^{a_n}-\zeta\,x_1^{b_1}\cdots x_n^{b_n}\right)\).
Proposed by Navid Safaei, Tehran
(7 pont)
Deadline expired on June 10, 2026.
Solution. If polynomial \(\displaystyle P\) has a variable on which it does not depend, let's omit this variable. If \(\displaystyle P\) can be written as \(\displaystyle Q(x_1^{k_1},...,x_n^{k_n})\), then polynomial \(\displaystyle Q\) also satisfis the condition of the problem, and proving that \(\displaystyle Q\) can be written in the given form, the same will hold for \(\displaystyle P\). If at least one of the exponents \(\displaystyle k_i\) is bigger than 1, then the sum of the exponents of the non-zero monomials of \(\displaystyle Q\) is smaller then the sum of the exponents in \(\displaystyle P\) (using the fact that each variable appears in at least one non-zero monomial), thus after finitely many such steps this process will terminate. Thus we can assume that \(\displaystyle P\) cannot be rewritten in the given form.
Let \(\displaystyle \varepsilon\) be a primitive root of unity such that \(\displaystyle \varepsilon^M=1\). EUsing the condition of the problem we get that \(\displaystyle P(\varepsilon^ix_1,x_2,...,x_n)|P(x_1^M,x_2^M,...,x_n^M)\). We will prove that polynomials \(\displaystyle P_i(x_1,...,x_n)=P(\varepsilon^ix_1,x_2,...,x_n)\) (for\(\displaystyle i=0,1,...,M-1\)) are irreducible and not associates of each other.
Let's prove the irreducibility first: if \(\displaystyle P(zx_1,x_2,...,x_n)=A(x_1,...x_n)B(x_1,...,x_n)\) for some complex constant \(\displaystyle z\), then \(\displaystyle P(x_1,...,x_n)=A(z^{-1}x_1,x_2,...,x_n)B(z^{-1}x_1,x_2,...,x_n)\), which contradicts the fact that \(\displaystyle P\) is irreducible.
Now let's prove that the polynomials are not associates of each other. Let's assume that for some non-zero and distinct complex constants \(\displaystyle z_1\) and \(\displaystyle z_2\) and complex constant \(\displaystyle C\) \(\displaystyle P(z_1x_1,x_2,...,x_n)=CP(z_2x_1,x_2,...,x_n)\). Then \(\displaystyle P(x_1,...,x_n)=CP(z_1^{-1}z_2x_1,x_2,...,x_n)\). Since \(\displaystyle P\) is irreducible and not constant, it contains a non-zero monomial in which variable \(\displaystyle x_1\) does not appear: since this monomial is the same on both sides, we get that \(\displaystyle C=1\). Now considering a non-zero monomial containing \(\displaystyle x_1\) we get that for every \(\displaystyle i>0\) which can be the exponent of \(\displaystyle x_1\) in a non-zero monomial of \(\displaystyle P\), \(\displaystyle (z_1^{-1}z_2)^i=0\). This implies that \(\displaystyle z_1^{-1}z_2\) is a root of unity different from 1, and if it's a primitive \(\displaystyle m^{\text{th}}\) root of unity, then \(\displaystyle m|i\), therefore \(\displaystyle P(x_1,...,x_n)=Q(x_1^m,x_2,...,x_n)\), and we've already ruled out this possibility in the beginning of our argument. Thus \(\displaystyle P_0P_1...P_{M-1}|P(x_1^M,...,x_n^M)\), and since they have the same degree, \(\displaystyle P(x_1^M,...,x_n^M)=CP_0P_1...P_{M-1}\) with some complex constant \(\displaystyle C\).
We can repeat this argument with the other variables. Since the fundamental theorem of arithmetic is (also) valid for multivariate polynomial with complex coefficient, we will get the factors from before possible in a different order and with different constant factors. Thus for every \(\displaystyle 1\le k\le n\) we can find \(\displaystyle 0\le \ell_k\le M-1\) and complex constant \(\displaystyle C_k\) such that \(\displaystyle P(x_1,...,\varepsilon x_k,...,x_n)=C_kP(\varepsilon^{\ell_k}x_1,x_2,...,x_n)\). Looking at a non-zero monomial \(\displaystyle x_1^{i_1}...x_n^{i_n}\) we get that \(\displaystyle \varepsilon^{i_k}-C_k\varepsilon^{{\ell_k}i_1}=0\). Therefore \(\displaystyle C_k\) is a power of \(\displaystyle \varepsilon\): \(\displaystyle C_k=\varepsilon^{m_k}\) for some integer \(\displaystyle m_k\), and thus \(\displaystyle i_k\equiv m_k+\ell_ki_1\) mod \(\displaystyle M\).
Now consider all pairs of exponents \(\displaystyle (i_1,i_k)\) which appear as the exponents of \(\displaystyle x_1\) and \(\displaystyle x_k\) in a non-zero monomial of \(\displaystyle P\): we've showed above that \(\displaystyle i_k\equiv m_k+\ell_ki_1\) modulo \(\displaystyle M\) is true for all such pairs. This means that these pairs are collinear modulo \(\displaystyle M\). We will prove the following lemma:
Lemma: If lattice points \(\displaystyle (x_1,y_1)\)...., \(\displaystyle (x_m,y_m)\in \mathbb{Z}^2\) are collinear modulo \(\displaystyle M\), where \(\displaystyle |x_i|,|y_i|\le K\) and \(\displaystyle M>6K^2\), then they are also collinear in \(\displaystyle \mathbb{Z}^2\).
Bizonyítás: Points \(\displaystyle (x_{1},y_{1})\), \(\displaystyle (x_2,y_2)\), \(\displaystyle (x_3,y_3)\) being collinear in \(\displaystyle \mathbb{Z}^2\) is equivalent to
\(\displaystyle (x_2y_3-x_3y_2)+(x_3y_1-x_1y_3)+(x_1y_2-x_2y_1) = 0. \)
However, since the absolute value of the sum is at most \(\displaystyle 6K^2\) and it is also divisible by \(\displaystyle M\), the sum is indeed 0.
We want to apply this lemma. By iterating the condition \(\displaystyle P(x_1,...,x_n)|P(x_1^M,...,x_n^M)\) we get that \(\displaystyle P(x_1,...,x_n)|P(x_1^{M^t},...,x_n^{M^t})\), and choosing \(\displaystyle t\) to be large enough the conditions of the lemma will be satisfied. Therefore there exists rational numbers \(\displaystyle r_k\) and \(\displaystyle s_k\) such that \(\displaystyle i_k=r_ki_1+s_k\) (if \(\displaystyle i_1\) is constant, then \(\displaystyle P\) is not irreducible since \(\displaystyle x_1^{i_1}\) can be factored out) is true for every pair of exponents \(\displaystyle (i_1,i_k)\) appearing in a non-zero monomial of \(\displaystyle P\). If \(\displaystyle r_k=a/b\), where \(\displaystyle a\) and \(\displaystyle b\) are relatively primes, then \(\displaystyle b|i_1\), which can only be true for \(\displaystyle b=1\), therefore \(\displaystyle r_k,s_k\in \mathbb{Z}\). So the non-zero monomials can be written as \(\displaystyle (x_1x_2^{r_2}...x_n^{r_n})^{i_1}x_2^{s_2}...x_n^{s_n}\). Observe that the integer \(\displaystyle s_k\) cannot be negative, since there is a non-zero monomial in which \(\displaystyle i_1=0\), and thus the corresponding exponent \(\displaystyle i_k\) must be equal to \(\displaystyle s_k\).
Let \(\displaystyle X=x_1x_2^{r_2}...x_n^{r_n}\) and \(\displaystyle C=x_2^{s_2}...x_n^{s_n}\). Polynomial \(\displaystyle P\) can be rewritten as the polynomial of \(\displaystyle X\) as \(\displaystyle C\sum c_iX^i\) with suitable complex coefficients \(\displaystyle c_i\). If polynomial \(\displaystyle \sum c_iX^i\) is at least degree 2, then it can be written as a product of linear factors by the fundamental theorem of algebra. Suppose that
\(\displaystyle \sum c_iX^i=D(X-d_1)(X-d_2)...(X-d_m) \)
(where \(\displaystyle m\) is the degree of \(\displaystyle \sum c_iX^i\)). Then \(\displaystyle P\) can be written as
\(\displaystyle Dx_2^{s_2}...x_n^{s_n}(x_1x_2^{r_2}...x_n^{r_n}-d_1)(x_1x_2^{r_2}...x_n^{r_n}-d_2)...(x_1x_2^{r_2}...x_n^{r_n}-d_m). \)
We know that \(\displaystyle r_km+s_k\ge 0\), thus \(\displaystyle s_k\ge -r_km\), therefore \(\displaystyle s_k=r'_km+s'_k\), where \(\displaystyle r'_k\ge \max(-r_k,0)\) and \(\displaystyle s'_k\ge 0\) (we took advantage of the fact that \(\displaystyle s_k\ge 0\)). Then the expression above can be written as
\(\displaystyle Dx_2^{s'_2}...x_n^{s'_n}(x_1x_2^{r_2+r'_2}...x_n^{r_n+r'_n}-d_1x_2^{r'_2}...x_n^{r'_n})(x_1x_2^{r_2+r'_2}...x_n^{r_n+r'_n}-d_2x_2^{r'_2}...x_n^{r'_n})...(x_1x_2^{r_2+r'_2}...x_n^{r_n+r'_n}-d_mx_2^{r'_2}...x_n^{r'_n}), \)
and this implies that \(\displaystyle P\) is not irreducible. This shows that the exponent of variable \(\displaystyle x_1\) can only be 0 or 1 in the non-zero terms of the polynomial, and this holds for every other variable, too. Since the degree of \(\displaystyle P\) is at least two, it cannot contain a single monomial, because a single monomial of degree at least 2 is reducible. Our argument also shows that the number of terms cannot be bigger than the number of different exponents of \(\displaystyle x_1\) (since we've showed that the exponent of \(\displaystyle x_1\) uniquely defines all the other exponents in a non-zero monomial), so it must be exactly two terms, which cannot share a common variable, because then the polynomial would be reducible (the common variable can be factored out). So after rearranging the variable we get that \(\displaystyle P=ax_1...x_i-bx_{i+1}...x_n\). Now substituting \(\displaystyle x_2=...=x_n=1\) we get that polynomial \(\displaystyle ax_1-b\) divides polynomial \(\displaystyle ax_1^M-b\). This implies that \(\displaystyle b/a\) is the root of the second polynomial, and this leads to \(\displaystyle (b/a)^{M-1}=1\), so \(\displaystyle b/a\) is a root of unity, and our proof is finished.
Statistics:
5 students sent a solution. 3 points: 1 student. 2 points: 1 student. 1 point: 1 student. 0 points: 1 student. irregular: 1 script.
Problems in Mathematics of KöMaL, May 2026