INMO 2019

Since $p_{4} - p_{1} = 8$, and no prime is even, we observe that $\{p_{1}, p_{2}, p_{3}, p_{4}\}$ is a subset of $\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\}$. Moreover $p_{1}$ is larger than $3$. If $p_{1} \equiv 1 \pmod{3}$, then $p_{1} + 2$ and $p_{1} + 8$ are divisible by $3$. Hence we do not get $4$ primes in the set $\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\}$. Thus $p_{1} \equiv 2 \pmod{3}$ and $p_{1} + 4$ is not a prime. We get $p_{2} = p_{1} + 2$, $p_{3} = p_{1} + 6$, $p_{4} = p_{1} + 8$.

Consider the remainders of $p_{1}, p_{1} + 2, p_{1} + 6, p_{1} + 8$ when divided by $5$. If $p_{1} \equiv 2 \pmod{5}$, then $p_{1} + 8$ is divisible by $5$ and hence is not a prime. If $p_{1} \equiv 3 \pmod{5}$, then $p_{1} + 2$ is divisible by $5$. If $p_{1} \equiv 4 \pmod{5}$, then $p_{1} + 6$ is divisible by $5$. Hence the only possibility is $p_{1} \equiv 1 \pmod{5}$.

Thus we see that $p_{1} \equiv 1 \pmod{2}$, $p_{1} \equiv 2 \pmod{3}$ and $p_{1} \equiv 1 \pmod{5}$. We conclude that $p_{1} \equiv 11 \pmod{30}$.

Similarly $q_{1} \equiv 11 \pmod{30}$. It follows that $30$ divides $p_{1} - q_{1}$.

Observe that

This gives

We write this as

Using induction, we get

Observe that

Hence

Thus we obtain

Since $f_{0}(x), f_{1}(x)$ and $f_{2}(x)$ are polynomials with integer coefficients, induction again shows that $f_{n}(x)$ is a polynomial with integer coefficients.

Let $P$ be a good point. Let $l, m, n$ be respectively the number of parts the sides $BC, CA, AB$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices of the triangle $ABC$; otherwise we get some quadrilaterals.

Let $h_1$ be the distance of $P$ from $BC$. Then $h_1$ is the height for all the triangles with their bases on $BC$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $l x = a$. Similarly, taking $y$ and $z$ as the lengths of the bases of triangles on $CA$ and $AB$ respectively, we get $m y = b$ and $n z = c$. Let $h_2$ and $h_3$ be the distances of $P$ from $CA$ and $AB$ respectively. Then

where $\Delta$ denotes the area of the triangle $ABC$. These lead to

But

Thus we get

However, we also have

Adding these three relations,

Thus

We conclude that $l + m + n = 27$. Thus every good point $P$ determines a partition $(l, m, n)$ of $27$ such that there are $l, m, n$ equal segments respectively on $BC, CA, AB$.

Conversely, take any partition $(l, m, n)$ of $27$. Divide $BC, CA, AB$ respectively into $l, m, n$ equal parts. Define

Draw a line parallel to $BC$ at a distance $h_1$ from $BC$; draw another line parallel to $CA$ at a distance $h_2$ from $CA$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $ABC$. Then

Hence

This shows that the distance of $P$ from $AB$ is

Therefore each triangle with base on $CA$ has area $\frac{\Delta}{27}$. We conclude that all the triangles which partition $ABC$ have equal areas. Hence $P$ is a good point.

Thus the number of good points is equal to the number of positive integral solutions of the equation $l + m + n = 27$. This is equal to

Setting $y=0$ in the condition (ii), we get

for all $x$ (since $f(0) \neq 0$). Thus either $f(x)=0$ or $f(x)=f(0)$, for all $x \in \mathbb{Z}$. Now taking $x=y=0$ in (i), we see that $f(0)+f(0)^2=2 f(0)$. This shows that $f(0)=0$ or $f(0)=1$. Since $f(0) \neq 0$, we must have $f(0)=1$. We conclude that

This shows that the set of all possible values of $f(x)$ is $\{0,1\}$. This completes (a).

Let $S=\{n \in \mathbb{Z} \mid f(n) \neq 0\}$. Hence we must have $S=\{n \in \mathbb{Z} \mid f(n)=1\}$ by (a). Since $f(1)=0$, $1$ is not in $S$. And $f(0)=1$ implies that $0 \in S$. Take any $x \in \mathbb{Z}$ and $y \in S$. Using (ii), we get

This shows that $xy \in S$. If $x \in \mathbb{Z}$ and $y \in \mathbb{Z}$ are such that $xy \in S$, then (ii) gives

Thus $(f(x)-1)(f(y)-1)=0$. It follows that $f(x)=1$ or $f(y)=1$; i.e., either $x \in S$ or $y \in S$. We also observe from (ii) that $x \in S$ and $y \in S$ implies that $f(x-y)=1$ so that $x-y \in S$. Thus $S$ has the properties:

(A) $x \in \mathbb{Z}$ and $y \in S$ implies $xy \in S$;

(B) $x, y \in \mathbb{Z}$ and $xy \in S$ implies $x \in S$ or $y \in S$;

(C) $x, y \in S$ implies $x-y \in S$.

Now we know that $f(10) \neq 0$ and $f(2)=0$. Hence $f(10)=1$ and $10 \in S$; and $2 \notin S$. Writing $10=2 \times 5$ and using (B), we conclude that $5 \in S$ and $f(5)=1$. Hence $f(5k)=1$ for all $k \in \mathbb{Z}$ by (A).

Suppose $f(5k+l)=1$ for some $l$, $1 \leq l \leq 4$. Then $5k+l \in S$. Choose $u \in \mathbb{Z}$ such that $lu \equiv 1 \pmod{5}$. We have $(5k+l)u \in S$ by (A). Moreover, $lu=1+5m$ for some $m \in \mathbb{Z}$ and

This shows that $5(ku+m)+1 \in S$. However, we know that $5(ku+m) \in S$. By (C), $1 \in S$ which is a contradiction. We conclude that $5k+l \notin S$ for any $l$, $1 \leq l \leq 4$. Thus

Solution 1.

Suppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P > K O_{1} > K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given assumption, and therefore $P$ and $Q$ lie on the same side of the line joining $O_{1}$ and $O_{2}$. Since $K P = K Q$ it follows that $K$ lies on the radical axis of the given circles, which is the common tangent at $R$. Therefore $K P = K Q = K R$ and hence $K$ is the circumcenter of $\triangle P Q R$. On the other hand, $\triangle K Q O_{1}$ and $\triangle K R O_{1}$ are both right-angled triangles with $K Q = K R$ and $Q O_{1} = R O_{1}$, and hence the two triangles are congruent. Therefore $\widehat{Q K O_{1}} = \widehat{R K O_{1}}$, so $K O_{1}$, and hence $P K$ is perpendicular to $Q R$. Similarly, $Q K$ is perpendicular to $P R$, so it follows that $K$ is the orthocenter of $\triangle P Q R$. Hence we have that $\triangle P Q R$ is equilateral.

Solution 2.

We again rule out the possibility that $P$ and $Q$ are on the opposite side of the line joining $O_{1} O_{2}$, and assume that they are on the same side. Observe that $\triangle K P O_{2}$ is congruent to $\triangle K Q O_{1}$ (since $K P = K Q$). Therefore $O_{1} P = O_{2} Q = r$ (say). In $\triangle O_{1} O_{2} Q$, we have $\widehat{O_{1} Q O_{2}} = \pi / 2$ and $R$ is the midpoint of the hypotenuse, so $R Q = R O_{1} = r$. Therefore $\triangle O_{1} R Q$ is equilateral, so $\widehat{Q R O_{1}} = \pi / 3$. Similarly, $P R = r$ and $\widehat{P R O_{2}} = \pi / 3$, hence $\widehat{P R Q} = \pi / 3$. Since $P R = Q R$ it follows that $\triangle P Q R$ is equilateral.

Rewriting the given equation we have

The left hand side equals $(4 m+1)\left(m^{2}+3\right)$. Suppose that $\left(4 m+1, m^{2}+3\right)=1$. Then $\left(4 m+1, m^{2}+3\right)=\left(3 p^{n}, 1\right),\left(3, p^{n}\right),\left(p^{n}, 3\right)$ or $\left(1,3 p^{n}\right)$, a contradiction since $4 m+1, m^{2}+3 \geq 4$. Therefore $\left(4 m+1, m^{2}+3\right)>1$. Since $4 m+1$ is odd we have $\left(4 m+1, m^{2}+3\right)=\left(4 m+1,16 m^{2}+48\right)=(4 m+1,49)=7$ or $49$. This proves that $p=7$, and $4 m+1=3 \cdot 7^{k}$ or $7^{k}$ for some natural number $k$. If $(4 m+1,49)=7$ then we have $k=1$ and $4 m+1=21$ which does not lead to a solution. Therefore $\left(4 m+1, m^{2}+3\right)=49$. If $7^{3}$ divides $4 m+1$ then it does not divide $m^{2}+3$, so we get $m^{2}+3 \leq 3 \cdot 7^{2}<7^{3} \leq 4 m+1$. This implies $(m-2)^{2}<2$, so $m \leq 3$, which does not lead to a solution. Therefore we have $4 m+1=49$ which implies $m=12$ and $n=4$. Thus $(m, n, p)=(12,4,7)$ is the only solution.