Subfield: Algebra


Question: Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences

$$
\begin{aligned}
& P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\
& Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014)
\end{aligned}
$$

are permutations of each other.
Prove that there do not exist distinct block-similar polynomials of degree $n$.

Solution:
For convenience, we set $k=2015=2 \ell+1$.

For every polynomial $F(x)$ and every nonnegative integer $m$, define $\Sigma_{F}(m)=$ $\sum_{i=1}^{m} F(i)$; in particular, $\Sigma_{F}(0)=0$. It is well-known that for every nonnegative integer $d$ the $\operatorname{sum} \sum_{i=1}^{m} i^{d}$ is a polynomial in $m$ of degree $d+1$. Thus $\Sigma_{F}$ may also be regarded as a real polynomial of degree $\operatorname{deg} F+1$ (with the exception that if $F=0$, then $\Sigma_{F}=0$ as well). This allows us to consider the values of $\Sigma_{F}$ at all real points (where the initial definition does not apply).

Assume for the sake of contradiction that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Then both polynomials $\Sigma_{P-Q}(x)$ and $\Sigma_{P^{2}-Q^{2}}(x)$ have roots at the points $0, k, 2 k, \ldots, n k$. This motivates the following lemma, where we use the special polynomial

$$
T(x)=\prod_{i=0}^{n}(x-i k)
$$

Lemma. Assume that $F(x)$ is a nonzero polynomial such that $0, k, 2 k, \ldots, n k$ are among the roots of the polynomial $\Sigma_{F}(x)$. Then $\operatorname{deg} F \geqslant n$, and there exists a polynomial $G(x)$ such that $\operatorname{deg} G=\operatorname{deg} F-n$ and $F(x)=T(x) G(x)-T(x-1) G(x-1)$.

Proof. If $\operatorname{deg} F<n$, then $\Sigma_{F}(x)$ has at least $n+1$ roots, while its degree is less than $n+1$. Therefore, $\Sigma_{F}(x)=0$ and hence $F(x)=0$, which is impossible. Thus $\operatorname{deg} F \geqslant n$.

The lemma condition yields that $\Sigma_{F}(x)=T(x) G(x)$ for some polynomial $G(x)$ such that $\operatorname{deg} G=\operatorname{deg} \Sigma_{F}-(n+1)=\operatorname{deg} F-n$.

Now, let us define $F_{1}(x)=T(x) G(x)-T(x-1) G(x-1)$. Then for every positive integer $n$ we have

$$
\Sigma_{F_{1}}(n)=\sum_{i=1}^{n}(T(x) G(x)-T(x-1) G(x-1))=T(n) G(n)-T(0) G(0)=T(n) G(n)=\Sigma_{F}(n),
$$

so the polynomial $\Sigma_{F-F_{1}}(x)=\Sigma_{F}(x)-\Sigma_{F_{1}}(x)$ has infinitely many roots. This means that this polynomial is zero, which in turn yields $F(x)=F_{1}(x)$, as required.



First, we apply the lemma to the nonzero polynomial $R_{1}(x)=P(x)-Q(x)$. Since the degree of $R_{1}(x)$ is at most $n$, we conclude that it is exactly $n$. Moreover, $R_{1}(x)=\alpha \cdot(T(x)-T(x-1))$ for some nonzero constant $\alpha$.

Our next aim is to prove that the polynomial $S(x)=P(x)+Q(x)$ is constant. Assume the contrary. Then, notice that the polynomial $R_{2}(x)=P(x)^{2}-Q(x)^{2}=R_{1}(x) S(x)$ is also nonzero and satisfies the lemma condition. Since $n<\operatorname{deg} R_{1}+\operatorname{deg} S=\operatorname{deg} R_{2} \leqslant 2 n$, the lemma yields

$$
R_{2}(x)=T(x) G(x)-T(x-1) G(x-1)
$$

with some polynomial $G(x)$ with $0<\operatorname{deg} G \leqslant n$.

Since the polynomial $R_{1}(x)=\alpha(T(x)-T(x-1))$ divides the polynomial

$$
R_{2}(x)=T(x)(G(x)-G(x-1))+G(x-1)(T(x)-T(x-1)),
$$

we get $R_{1}(x) \mid T(x)(G(x)-G(x-1))$. On the other hand,

$$
\operatorname{gcd}\left(T(x), R_{1}(x)\right)=\operatorname{gcd}(T(x), T(x)-T(x-1))=\operatorname{gcd}(T(x), T(x-1))=1
$$

since both $T(x)$ and $T(x-1)$ are the products of linear polynomials, and their roots are distinct. Thus $R_{1}(x) \mid G(x)-G(x-1)$. However, this is impossible since $G(x)-G(x-1)$ is a nonzero polynomial of degree less than $n=\operatorname{deg} R_{1}$.

Thus, our assumption is wrong, and $S(x)$ is a constant polynomial, say $S(x)=\beta$. Notice that the polynomials $(2 P(x)-\beta) / \alpha$ and $(2 Q(x)-\beta) / \alpha$ are also block-similar and distinct. So we may replace the initial polynomials by these ones, thus obtaining two block-similar polynomials $P(x)$ and $Q(x)$ with $P(x)=-Q(x)=T(x)-T(x-1)$. It remains to show that this is impossible.

For every $i=1,2 \ldots, n$, the values $T(i k-k+1)$ and $T(i k-1)$ have the same sign. This means that the values $P(i k-k+1)=T(i k-k+1)$ and $P(i k)=-T(i k-1)$ have opposite signs, so $P(x)$ has a root in each of the $n$ segments $[i k-k+1, i k]$. Since $\operatorname{deg} P=n$, it must have exactly one root in each of them.

Thus, the sequence $P(1), P(2), \ldots, P(k)$ should change sign exactly once. On the other hand, since $P(x)$ and $-P(x)$ are block-similar, this sequence must have as many positive terms as negative ones. Since $k=2 \ell+1$ is odd, this shows that the middle term of the sequence above must be zero, so $P(\ell+1)=0$, or $T(\ell+1)=T(\ell)$. However, this is not true since

$$
|T(\ell+1)|=|\ell+1| \cdot|\ell| \cdot \prod_{i=2}^{n}|\ell+1-i k|<|\ell| \cdot|\ell+1| \cdot \prod_{i=2}^{n}|\ell-i k|=|T(\ell)|
$$

where the strict inequality holds because $n \geqslant 2$. We come to the final contradiction.
Assume again that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Let $R(x)=P(x)-Q(x)$ and $S(x)=P(x)+Q(x)$. For brevity, we also denote the segment $[(i-1) k+1, i k]$ by $I_{i}$, and the set $\{(i-1) k+1,(i-1) k+2, \ldots, i k\}$ of all integer points in $I_{i}$ by $Z_{i}$.

Step 1. We prove that $R(x)$ has exactly one root in each segment $I_{i}, i=1,2, \ldots, n$, and all these roots are simple.

Indeed, take any $i \in\{1,2, \ldots, n\}$ and choose some points $p^{-}, p^{+} \in Z_{i}$ so that

$$
P\left(p^{-}\right)=\min _{x \in Z_{i}} P(x) \text { and } \quad P\left(p^{+}\right)=\max _{x \in Z_{i}} P(x) \text {. }
$$

Since the sequences of values of $P$ and $Q$ in $Z_{i}$ are permutations of each other, we have $R\left(p^{-}\right)=P\left(p^{-}\right)-Q\left(p^{-}\right) \leqslant 0$ and $R\left(p^{+}\right)=P\left(p^{+}\right)-Q\left(p^{+}\right) \geqslant 0$. Since $R(x)$ is continuous, there exists at least one root of $R(x)$ between $p^{-}$and $p^{+}$- thus in $I_{i}$.

So, $R(x)$ has at least one root in each of the $n$ disjoint segments $I_{i}$ with $i=1,2, \ldots, n$. Since $R(x)$ is nonzero and its degree does not exceed $n$, it should have exactly one root in each of these segments, and all these roots are simple, as required.

Step 2. We prove that $S(x)$ is constant. 

We start with the following claim.

Claim. For every $i=1,2, \ldots, n$, the sequence of values $S((i-1) k+1), S((i-1) k+2), \ldots$, $S(i k)$ cannot be strictly increasing.

Proof. Fix any $i \in\{1,2, \ldots, n\}$. Due to the symmetry, we may assume that $P(i k) \leqslant Q(i k)$. Choose now $p^{-}$and $p^{+}$as in Step 1. If we had $P\left(p^{+}\right)=P\left(p^{-}\right)$, then $P$ would be constant on $Z_{i}$, so all the elements of $Z_{i}$ would be the roots of $R(x)$, which is not the case. In particular, we have $p^{+} \neq p^{-}$. If $p^{-}>p^{+}$, then $S\left(p^{-}\right)=P\left(p^{-}\right)+Q\left(p^{-}\right) \leqslant Q\left(p^{+}\right)+P\left(p^{+}\right)=S\left(p^{+}\right)$, so our claim holds.

We now show that the remaining case $p^{-}<p^{+}$is impossible. Assume first that $P\left(p^{+}\right)>$ $Q\left(p^{+}\right)$. Then, like in Step 1 , we have $R\left(p^{-}\right) \leqslant 0, R\left(p^{+}\right)>0$, and $R(i k) \leqslant 0$, so $R(x)$ has a root in each of the intervals $\left[p^{-}, p^{+}\right)$and $\left(p^{+}, i k\right]$. This contradicts the result of Step 1.

We are left only with the case $p^{-}<p^{+}$and $P\left(p^{+}\right)=Q\left(p^{+}\right)$(thus $p^{+}$is the unique root of $R(x)$ in $\left.I_{i}\right)$. If $p^{+}=i k$, then the values of $R(x)$ on $Z_{i} \backslash\{i k\}$ are all of the same sign, which is absurd since their sum is zero. Finally, if $p^{-}<p^{+}<i k$, then $R\left(p^{-}\right)$and $R(i k)$ are both negative. This means that $R(x)$ should have an even number of roots in $\left[p^{-}, i k\right]$, counted with multiplicity. This also contradicts the result of Step 1.

In a similar way, one may prove that for every $i=1,2, \ldots, n$, the sequence $S((i-1) k+1)$, $S((i-1) k+2), \ldots, S(i k)$ cannot be strictly decreasing. This means that the polynomial $\Delta S(x)=S(x)-S(x-1)$ attains at least one nonnegative value, as well as at least one nonpositive value, on the set $Z_{i}$ (and even on $Z_{i} \backslash\{(i-1) k+1\}$ ); so $\Delta S$ has a root in $I_{i}$.

Thus $\Delta S$ has at least $n$ roots; however, its degree is less than $n$, so $\Delta S$ should be identically zero. This shows that $S(x)$ is a constant, say $S(x) \equiv \beta$.

Step 3. Notice that the polynomials $P(x)-\beta / 2$ and $Q(x)-\beta / 2$ are also block-similar and distinct. So we may replace the initial polynomials by these ones, thus reaching $P(x)=-Q(x)$.

Then $R(x)=2 P(x)$, so $P(x)$ has exactly one root in each of the segments $I_{i}, i=1,2, \ldots, n$. On the other hand, $P(x)$ and $-P(x)$ should attain the same number of positive values on $Z_{i}$. Since $k$ is odd, this means that $Z_{i}$ contains exactly one root of $P(x)$; moreover, this root should be at the center of $Z_{i}$, because $P(x)$ has the same number of positive and negative values on $Z_{i}$.

Thus we have found all $n$ roots of $P(x)$, so

$$
P(x)=c \prod_{i=1}^{n}(x-i k+\ell) \quad \text { for some } c \in \mathbb{R} \backslash\{0\}
$$



where $\ell=(k-1) / 2$. It remains to notice that for every $t \in Z_{1} \backslash\{1\}$ we have

$$
|P(t)|=|c| \cdot|t-\ell-1| \cdot \prod_{i=2}^{n}|t-i k+\ell|<|c| \cdot \ell \cdot \prod_{i=2}^{n}|1-i k+\ell|=|P(1)|
$$

so $P(1) \neq-P(t)$ for all $t \in Z_{1}$. This shows that $P(x)$ is not block-similar to $-P(x)$. The final contradiction.