Proposed by Liam Baker, South Africa

Note that $f$ is a solution if and only if $-f$ is. Since $f(1) \not=f(-1),$ we know that $f(0)\not=0$ and by the above we can assume that $f(0) \ge 1.$
If $f(0)=1,$ then choosing $m=m,n=0$ gives $f(m)^2 \mid f(m)-1.$ This implies that $f(m) \in \{-1,1\}.$ Every function for which $f(m)= \pm 1$ for every $m \in \mathbb Z$ and $f(1) \not = f(-1)$ satisfies the function equation.
If $f(0)=2$, we have $f(m)^2 \mid f(m)-2$ implying that $f(m) \in \pm \{1,2\}.$ Since $f(0)^2 \mid f(1)-f(-1),$ we have that one equals $2$ and the other one $-2$. Since $4 \mid f(1)^2 \mid f(m+1)-f(-m)$ and $4 \mid f(-1)^2 \mid f(m)-f(-m-1),$ we can prove by induction on the absolute value of $m$ that $f(m) = \pm 2$ for every $m \in \mathbb Z.$
If $f(0) \ge 3$, we have that $f(0)^2 \mid f(1)-f(-1)$ implies that one of $f(1),f(-1)$ is (strictly) larger than $f(0)$ in absolute value. But $f(1)^2 \mid f(1)-f(0)$ implies that this is not possible, since $0 \not=|{f(1)-f(0)} < 2|{f(1)} < f(1)^2$ in that case and similar for $f(-1).$
It is not hard to see that the two families of all solutions satisfy the equality.

1. Find out that when $f$ is a solution, then $-f$ is a solution. --- 0 point
2. Plug in $m=0$ to find that $f(n)$ divides $f(0)$ for any $n$. --- 1 point
3. Plug in $m=1,n=-1$ to find that $f(0)\ne 0$. --- 1 point
4. Plug in $m+n=0$ to find $|f(0)|\le 2$. --- 2 points
5. Find all solutions for $|f(0)|=1$, with proof. --- 1 point
6. Find all solutions for $|f(0)| = 2$, with proof. --- 2 points
7. Failure to verify some solutions (applicable when the proof is complete) --- -1 point
8. State the solutions without any proof (not additive). --- 1 point

Proposed by Kaarel Hänni, Estonia

Note that on the $i$th step, John places a total of $i\frac{n!}{i}=n!$ stones into the baskets. Hence, after all $n$ steps, there are $n! n$ stones in the baskets. The QM-AM inequality then gives \[\sum x_i^2\ge\frac{\left(\sum x_i\right)^2}{n!}=n! n^2,\] proving the first desired inequality. For the other inequality, note that $x_i=\sum_{j|i}j$, and hence \[\sum_i x_i^2=\sum_i \sum_{(j,k)\in [n]^2,\hspace{1mm}j|i,\hspace{1mm}k|i} jk.\] For fixed $j$ and $k$, we have $jk$ appearing in the index $i$ term of the first sum if and only if $j|i$ and $k|i$, which in turn happens if and only if $lcm(j,k)|i$. The number of such $i\in\{1,\dots, n!\}$ is exactly $\frac{n!}{lcm(j,k)}$. Hence, switching the order of summation to sum first by $(j,k)$, we get \[\sum_{i}x_i^2=\sum_{(j,k)\in [n]^2}jk \frac{n!}{lcm(j,k)}=n!\sum_{(j,k)\in [n]^2}\gcd(j,k).\] Now note that for some fixed $i\in [n]$ to be $\gcd(j,k)$, it is necessary (but not sufficient) that $i|j$ and $i|k$. Hence, for a fixed $i\in [n]$, the number of such pairs $(j,k)\in [n]^2$ is at most $\frac{n^2}{i^2}$, and so \[\sum_i x_i^2\le n!\sum_{i=1}^n i \frac{n^2}{i^2}=n! n^2 \sum_{i=1}^n \frac{1}{i}.\]

1. Showing $\sum_{i=1}^{n!} x_i^2 \ge n! n^2$. --- 2 points
• Showing $\sum_{i=1}^{n!} x_i = n! n$. --- 1 point
• Noticing that $\sum_{i=1}^{n!} x_i^2 = n! \sum_{a, b \le n} \gcd(a, b)$. --- 1 point
2. Showing $\sum_{i=1}^{n!} x_i^2 \le n! n^2 \sum_{i=1}^{n} \frac{1}{i} $. --- 5 points
• Noticing that $\sum_{i=1}^{n!} x_i^2 = n! \sum_{a, b \le n} \gcd(a, b)$. --- 1 points
• Showing $\sum_{a, b \le n} \gcd(a, b) = \sum_{d \le n} d \cdot \big|\{(a, b) \colon a, b \le n, \gcd(a, b) =d \}\big|$. --- 2 points
• Showing $\big|\{(a, b) \colon a, b \le n, \gcd(a, b) = d\}\big| \le \Big(\frac{n}{d}\Big)^2$ and concluding. --- 2 point

Subitems of 1 are not additive. Subitems of 2 are additive.

Proposed by Stijn Cambie, South Korea

We claim that there at most $18$ empty cells in an incomplete binoku that can be completed in a unique way. The binoku on the right contains $18$ zeros, each of which is the unique zero in either its row, its column, or its box (i.e. its $3\times 3$ subgrid), so the incomplete binoku obtained by emptying each of these 18 cells clearly has a unique solution (i.e. a unique binoku into which it can be completed). This shows that the upper bound can be attained.
To establish the upper bound, we begin by noticing that, for each empty cell in an incomplete binoku with a unique solution, the incomplete binoku with a single empty cell that is obtained from the solution by emptying that cell also has a unique solution. If the other cells in the row of this empty cell contain both zeros and ones, and the other cells in its column and the other cells in its box do so, too, then adding a $0$ or a $1$ to the empty cell both yield a binoku, contradicting uniqueness of the solution. Hence the other cells of either the row or the column or the box of the empty cell all contain the same number.
We call a cell of a binoku delible if it contains the only $0$ or the only $1$ in its row or column or box. Each empty cell in an incomplete binoku of which this binoku is the unique solution is therefore one of these delible cells. By definition, we can assign to each delible cell a row or a column or a box contaning it and in which all the other entries are different from the entry of that cell. This assignment is not necessarily unique (since a cell may, for example, contain the only $0$ both in its row and in its column), but the following holds:
Claim 1. No row, column, or box can be associated to more than one delible cell.
Proof of Claim 1 Suppose to the contrary that two delible cells are associated to the same row, column, or box. Without loss of generality, the first cell is $0$, so all the other cells (and in particular the second delible cell) in this row, column, or box are $1$. Because the second delible cell is~$1$, all the other cells in this row, column, or box are $0$, so any third cell is both $0$ and $1$, which is a contradiction.
This observation extends to yield the following:
Claim 2. If a row or column associated with a delible cell contains another delible cell, then the latter cannot be associated with a box.
Proof of Claim 2 Suppose to the contrary that it is associated with a box. The intersection of the box and row or column contains three cells, the delible cell associated to the box (which is $0$ without loss of generality) and two other cells (which are therefore $1$). Since they do not therefore all have the same value, the delible cell associated with the row or column must be among these two cells, so these two cells must have different values, which is a final contradiction.
A similar argument proves the following claim.
Claim 3. A box associated to a delible cell does not contain any other delible cell.
Finally, we observe the following claim.
Claim 4. Any box contains at most 3 delible cells associated to a row and at most 3 delible cells associated to a column.
Proof of Claim 4 If a box contained 4 delible cells associated to a row (or column), there would be two delible cells associated to the same row (or column), contradicting Claim 1.
Now let $b$, $c$, $r$ denote the number of boxes, columns, and rows of a binoku that are associated to a delible cell in this way. By Claim 1, $b+c+r$ is the number of delible cells and hence the largest possible number of empty cells in an incomplete binoku of which this binoku is the unique solution. Now each of the $r$ rows associated to a delible cell has $9$ cells and each of the $c$ columns associated to a delible cell contains $9-r$ cells not part of one these $r$ rows. Finally, the delible cell in each of the $b$ boxes is not in any of these $r$ rows or $c$ columns by Claim 2. There are 81 cells in total, so $$ 9r+(9-r)c+b\leq 81\quad\Longrightarrow\quad 81-(b+c+r)\geq 8(c+r)-cr\geq 8(c+r)-\dfrac{(c+r)^2}{4} $$ by the inequality between arithmetic and geometric means. Suppose to the contrary that $b+c+r\geq 19$. Then Equation (1) yields $f(c+r)\leq 248$, where $f(z)=32z-z^2$, which is increasing for $z\leq 16$ and decreasing for $z\geq 16$. Since $f(14)=f(18)=252$, this implies that $c+r\leq 13$ or $c+r>18$. Trivially, $c,r\leq 9$, so the latter is a contradiction. Thus $c+r\leq 13$ and hence $b\geq 6$ for $b+c+r\geq 13$.
Now suppose that any of the $9-b$ boxes not associated to a delible cell contains at most 4 delible cells. Each of these must be assigned to a row or column, so $$ c+r\leq 4(9-b)\quad\Longrightarrow\quad (b+c+r)+3b\leq 36. $$ If $b+c+r\geq 19$, this implies $3b\leq 17$, so $b\leq 5$, which is a contradiction. Hence there is a box $B$, not associated to a delible cell by Claim 3, that contains at least 5 delible cells. Hence, and without loss of generality, at least 3 of them are associated with rows. These rows are different by Claim 1 and they are shared by box $B$ with two other boxes, $B_1,B_2$. By Claim 2, $B_1,B_2$ are not associated with any delible cell either. In particular, this shows that $b\leq 9-3=6$. By the above, it follows that $b=6$ and hence $c+r=13$ for $b+c+r\geq 19$. Moreover, by Claim 3, the $b=6$ boxes associated to a delible cell do not contain any other delible cell. Finally, box $B$ contains at most $3+3=6$ delible cells by Claim 4, while boxes $B_1,B_2$ contain at most $3$ cells each associated to one of their columns by Claim 4 again, but do not contain any delible cells associated to one of their rows by Claim 1 because that row is already associated with a delible cell in box $B$. This proves that $c+r\leq 6+3+3=12$. This is the final contradiction showing that $b+c+r\leq 18$. This establishes the required bound and completes the solution.

Rubric

1. Construction --- 2 points
• Construction with at most $17$ empty cells. --- 0 point
• Construction with $18$ empty cells. --- 2 points
2. Proof of the upper bound --- 5 points
• Proving at least three of Claims 1, 2, 3 and 4 about the delible cells. --- 1 point
• Showing that $b$ in the solution is at least 6. --- 1 point
• Completing the proof. --- 3 points

Proposed by Leonardo Franchi, Italy

We claim that $P$, $Q$ and $I$ are aligned on the perpendicular line from $I$ to $BC$. In particular we are going to show that the line $IP$ is perpendicular to $BC$ and then thanks to the symmetry of the configuration the conclusion follows. Since I is incenter, we have by symmetry $ID=IE=IF$.
We now prove that $P,D$ and $E$ are colinear. In particular $\angle IPE=\angle ICA =\frac{\gamma}{2} $ since $PIEC$ is a cyclic quadrilateral. On the other hand $\angle IPD=\angle IFD$ since $FIDP$ is a cyclic quadrilateral. Now, since $\angle IFD=\angle IDF$, we get that $\angle IDF=\angle ADB-\angle FDB=\frac{\alpha}{2}+\gamma-(90-\frac{\beta}{2})=\frac{\gamma}{2}$ as we wanted.
Finally, since $\angle BDP=\angle EDC=90-\frac{\gamma}{2}$ and $IPD=\frac{\gamma}{2}$ the conclusion follows.

General rules for this problem

• We expect solutions to use either a synthetic or computational approach. Points from the two approaches cannot be added.
• Do not deduct points for configuration issues.
• Minor math mistakes generally lead to the deduction of 1 point, but should be dealt with on a case-by-case basis.
• Any equivalent approach should receive proper and proportionately judged equivalent marks.

Rubric for a synthetic approach

1. Claiming that $I$ lies on the line $PQ$. --- 1 point
2. Proving that $P,D,E$ are collinear. --- 4 point
• Claiming that $ID = IE = IF$. --- 1 point
• Claiming that $P,D,E$ are collinear. --- 1 point
3. Proving $IP$ is perpendicular to $BC$ --- 6 points

Items 2 and 3 are not additive.
Rubric for a computational approach

1. Any partial progress other than the items below (unless intermediate results are interpreted synthetically, in which case, see the synthetic rubric). --- 0 points
2. Complete solution with no mistakes, making clear that the student performed every step of the calculation. --- 7 points

Let $\alpha=\angle CAB$, $\beta=\angle ABC$ and $\gamma=\angle BCA$. After some angle chasing (see Figure~\ref{fig:P4_angles_1}) we find that $\angle FDI=\frac12\gamma$ and $\angle ICE=\frac12\gamma$. We also note that $IF=ID=IE$. From points of the arc $FDI$ in circle $FDI$ the segment $FI$ is seen at angle $\frac12\gamma$, and similarly from the points of the arc $ICE$ in circle $ICE$ the segment $EI$ is also seen at angle $\frac12\gamma$. Since the segments $EI$ and $FI$ are of the same length, the circles $FDI$ and $ICE$ are symmetric with respect to the angular bisector of angle $FIE$, in particular their second intersection point $P$ lies on the angular bisector. We can also prove analogously that $Q$ lies on the bisector of angle $FIE$. With some further angle chasing (see Figure~\ref{fig:P4_angles_1}) we find that this bisector is perpendicular to $BC$.
Remark: Another way of showing that the bisector of angle $EIF$ (which is also the perpendicular bisector of segment $EF$) is perpendicular to $BC$ is by noting that $BF=BD=\frac{c}{c+b}a$ and $CE=CD=\frac{b}{c+b}a$, hence $\frac{BF}{AB}=\frac{CE}{AC}$ and so $EF$ is parallel to $BC$.

1. Claiming that $P$ and $Q$ lie on the bisector of angle $FIE$ or claiming that they lie on the perpendicular bisector of segment $EF$. --- 1 point
2. Proving that $\angle FPI=\angle IPE$. --- 1 point
3. Claiming that $ID=IE=IF$. --- 1 point
4. Proving that the bisector of angle $FIE$ is perpendicular to $BC$ or proving that $EF$ is parallel to $BC$. --- 1 point

The above points are all additive.

$I$ is the center of $(DEF)$. Let $J$ be the second intersection of this circle with $BC$. Let $P'=ED \cap FJ$ and $Q'=EJ\cap FD$. A simple angle chasing proves that both $FBJI$ and $CEIJ$ are cyclic. A further angle chasing proves that $P'D=P'J$ and $Q'D=Q'J$. Thus $IP'Q'$ are collinear on a line, says $\ell$, which is perpendicular to $BC$. Finally, since $EDJF$ is cyclic, by consering the power of $P'$, one can show that $P'$ is on the radical axis of $(IED)$ and $(IFB)$, which is $IQ$, so $Q$ is on $\ell$. Similarly, $P$ is also on $\ell$, and we are done.

Proposed by Art Waeterschoot, Belgium

The largest number of red points is $1012$.
Proof of the upper bound. Let $N$ be the number of red points and let $v$ be a red point. Then $v$ is a vertex of at least $N-2$ triangles formed by red points whose interiors are disjoint. This means that there are at least $N-2$ blue points, so that $2N-2 \leq 2022$ and $N \leq 1012$.
Proof of the lower bound. Take the red points to be the vertices of a convex $1012$-gon. The lines that join the vertices divide the polygon into a number of regions. Choose a side $s$ of the polygon. For every vertex $v$ not on $s$, place a blue point in the region of the polygon that meets $v$ and that lies in the triangle formed by $s$ and $v$. This gives a total of $2022$ points. We claim that this construction is valid. Let $v_1$ and $v_2$ be the vertices of $s$, in clockwise order, and let $t_1, t_2, t_3$ be vertices of the polygon. We show that $\triangle t_1 t_2 t_3$ contains a blue point. If $s$ is a side of the triangle, we are done, by construction. Assume now, by changing the numbering if necessary, that $t_1$ and $t_2$ are not vertices of $s$, that $v_1, v_2, t_1, t_2$ lie in clockwise order and that $t_1, t_2, t_3$ lie in clockwise order. Then $t_3$ lies either strictly between $t_2$ and $v_2$ or strictly between $v_1$ and $t_1$. In the first case, $\triangle t_1 t_2 t_3$ contains the blue point in $\triangle t_2 v_1 v_2$. In the second case, $\triangle t_1 t_2 t_3$ contains the blue point in $\triangle t_1 v_1 v_2$.

1. Construction --- 5 points
• Example with $N = 1011$ and justification of the example. --- 1 point
• Example with $N = 1012$ only. --- 3 points
• Example with $N = 1012$ and justification of the example. --- 5 points
2. Proof of the upper bound --- 2 points
• Having Triangulation idea. --- 1 point

Proposed by Navid Safaei, Iran

Assume that $P(x)$ be a non-constant polynomial. Moreover, take $n$ large enough such that $P(n) \neq 0$. Consider the sequence $a_{m}=P^{(m)}(n)$. Let $m=q-n$ it follows that $q$ divides $P^{(m)}(n)-P^{(n)}(-n)$. The sequence $a_{m}$ is eventually periodic mod $q$. Let $T$ be the length of its fundamental period, we can assume that $T \leq q$. If $T=q$ then the set $\{P^{(r)}(n), \ldots, P^{(r+q)}(n)\}$ forms the complete residue system mod $q.$ If $T < q$ then $\gcd(T, q)=1$. Let $s$ be an arbitrary integer in $\{1, \ldots, q\}$ it follows that there is a positive integer $k$ such that $m+q k \equiv s(\bmod T)$. Setting $(m, n)=(m+q k, n)$ it follows that $q$ divides $$ P^{(m+q k)}(n)-P^{(n)}(m+q k) \equiv P^{(s)}(n)-P^{(n)}(m). $$ Hence, the sequence must be 1-periodic. That is, $$ P(a_{i}) \equiv a_{i}(\bmod q) \text { and } a_{i} \equiv a_{i+1}(\bmod q). $$ Therefore $T \in\{1, q\}$.
Note that if for all large enough $q$, we have $T=q$ then it follows that for all large enough $q$ the set $\{P(0), \ldots, P(q-1)\}$ is complete residue modulo $q$. If $P(x)$ is not linear, consider $P(x+1)-P(x)$ and use Schur's theorem to arrive at a contradiction.
Thus, there are infinitely many $q$ with $T=1$. Because the fundamental period mod $q$ starts before $q$, observe that $P^{(2q-n+1)}(n) \equiv P^{(2q-n)}(n)(\bmod q)$ for a fixed $n$ and infinitely many $q$.
Further, since $$ P^{(2q-n)}(n)-P^{(n)}(-n) \mid P(P^{(2q-n)}(n))-P(P^{(n)}(-n))=P^{(2q-n+1)}(n)-P^{(n+1)}(-n), $$ it follows that $q$ divides $P^{(2q-n+1)}(n)-P^{(n+1)}(-n)$. Since $$ P^{(2q-n+1)}(n) \equiv P^{(2q-n)}(n)(\bmod q), $$ it follows that $$ P^{(n+1)}(-n) \equiv P^{(n)}(-n)(\bmod q). $$ That is, for all large $l, l \geq n ; P^{(l)}(-n)$ is a root of $P(x) \equiv x(\bmod q)$. Hence, choose $q>|P^{(l)}(-n)|,|P^{(l+1)}(-n)|$ it follows that $$ P^{(l+1)}(-n)=P^{(l)}(-n). $$ Hence, $P^{(l)}(-n)$ is a root of $P(x)-x$. It follows that $P(x)-x$ must have $P^{(n)}(-n), P^{(n+1)}(-n), \ldots$ as its root.
It can be checked that $P(x)=c$ is a solution, and remember that $P(x)$ is not linear. In other cases, for sufficiently large $x$ it holds that $|P(x)|>|x|$. Taking $-n$ sufficiently small, it holds that $$ |-n| < |P(-n)|< |P^{(2)}(-n)|< \dots $$ therefore these are all distinct roots of $P(x)-x$. Impossible.

1. Consider the sequence $a_m=P^{(m)}(n)$ and prove it is eventually periodic mod $q$, of some period $T$. --- 1 point
2. Show that $T\in \{1,q\}$. --- 2 points
3. For $P(x)\neq x+1$, there exist infinitely many $q$ such that $T=1$. --- 2 points
4. Show that $P^{(n+1)}(-n) = P^{(n)}(-n)$, i.e. $P^{(n)}(-n)$ is a root of $P(x)-x$. Conclude that the polynomial $P(x)-x$, if nonlinear, has infinitely many \textbf{distinct} roots. --- 1 point
• Only prove one of the above statements. --- 0 points
5. Conclude or claim that $P(x)\equiv c$ and $P(x)\equiv x+1$ are the only solutions. --- 1 point