INMO 2017

Suppose $0 \in A$. Then $a^{2} = a^{2} + 0 \times b$ is rational and $a b = 0^{2} + a b$ is also rational for all $a, b$ in $A$, $a \neq 0$, $b \neq 0$, $a \neq b$. Hence $a = a_{1} \sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \neq 0$, we have

which is a rational number. Hence we may assume $0$ is not in $A$. If there is a number $a$ in $A$ such that $-a$ is also in $A$, then again we can get the conclusion as follows. Consider two other elements $c, d$ in $A$. Then $c^{2} + d a$ is rational and $c^{2} - d a$ is also rational. It follows that $c^{2}$ is rational and $d a$ is rational. Similarly, $d^{2}$ and $c a$ are also rationals. Thus $d / c = (d a) / (c a)$ is rational. Note that we can vary $d$ over $A$ with $d \neq c$ and $d \neq a$. Again $c^{2}$ is rational implies that $c = c_{1} \sqrt{M}$ for some rational $c_{1}$ and natural number $M$. We observe that $c \sqrt{M} = c_{1} M$ is rational, and

so that $a \sqrt{M}$ is a rational number. Similarly is the case with $-a \sqrt{M}$. For any other element $d$,

is a rational number. Thus we may now assume that $0$ is not in $A$ and $a + b \neq 0$ for any $a, b$ in $A$. Let $a, b, c, d$ be four distinct elements of $A$. We may assume $|a| > |b|$. Then $d^{2} + a b$ and $d^{2} + b c$ are rational numbers and so is their difference $a b - b c$. Writing $a^{2} + a b = a^{2} + b c + (a b - b c)$, and using the facts $a^{2} + b c$, $a b - b c$ are rationals, we conclude that $a^{2} + a b$ is also a rational number. Similarly, $b^{2} + a b$ is also a rational number. Consider

Note that $a^{2} + a b > 0$. Thus $q$ is a rational number and $a = b q$. This gives $a^{2} + a b = b^{2}(q^{2} + q)$. Let us take $b^{2}(q^{2} + q) = l$. Then

where $x$ and $y$ are natural numbers. Take $M = x y$. Then $|b| \sqrt{M} = x$ is a rational number. Finally, for any $c$ in $A$, we have

is also a rational number.

Consider the lattice points (points with integer coordinates) on the lines $y=0$ and $y=1$, other than $(0,0)$ and $(0,1)$. If one of them, say $A=(p, 1)$, is coloured green, then we have a right-angled triangle with $(0,0)$, $(0,1)$ and $A$ as vertices, all having different colours. (See Figures 1 and 2.)

If not, the lattice points on $y=0$ and $y=1$ are all red or blue. We consider three different cases.

Case 1. Suppose a point $B=(c, 0)$ is blue. Consider a green point $D=(p, q)$ in the plane. Suppose $p \neq 0$. If its projection $(p, 0)$ on the $x$-axis is red, then $(p, q)$, $(p, 0)$ and $(c, 0)$ are the vertices of a required type of right-angled triangle. If $(p, 0)$ is blue, then we can consider the triangle whose vertices are $(0,0)$, $(p, 0)$ and $(p, q)$. If $p=0$, then the points $D$, $(0,0)$ and $(c, 0)$ will work. (Figure 3.)

Case 2. A point $D=(c, 1)$, on the line $y=1$, is red. A similar argument works in this case.

Fig-4

Case 3. Suppose all the lattice points on the line $y=0$ are red and all on the line $y=1$ are blue points. Consider a green point $E=(p, q)$, where $q \neq 0$ and $q \neq 1$. (See Figure 4.) Consider an isosceles right-angled triangle $E K M$ with $\angle E=90^{\circ}$ such that the hypotenuse $K M$ is a part of the $x$-axis. Let $E M$ intersect $y=$ in $L$. Then $K$ is a red point and $L$ is a blue point. Hence $E K L$ is a desired triangle.

Let $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.

Part (i) We may write

Define

Then $Q(x)$ is also a polynomial with integer coefficients and

is a polynomial in $x^{2}$.

Part (ii) We write again

where

Note that $A(x), B(x)$ and $C(x)$ are polynomials with integer coefficients and each of these is a polynomial in $x^{3}$. We may introduce

where $\omega$ is an imaginary cube-root of unity. Then

since $\omega^{3}=1$ and $\omega+\omega^{2}=-1$. Taking $R(x)=S(x) T(x)$, we obtain

which is a polynomial in $x^{3}$. This follows from the identity

Alternately, $R(x)$ may be directly defined by

Let $A = 2\alpha$. Then $\angle A_1AC = \angle BAA_1 = \alpha$. Thus

We also have $\angle B_1CQ = \angle AA_1B_1 = \beta$, say. It follows that triangles $MA_1B_1$ and $QCB_1$ are similar and hence

Similarly, triangles $ACM$ and $C_1A_1M$ are similar and we get

Using the point $P$, we get similar ratios:

Thus,

and

However, triangles $C_1BM$ and $B_1CM$ are similar, which gives

Putting this in the last expression, we get

We conclude that $PQ$ is parallel to $BC$.

Suppose $n = p q r$, where $p < q$ are primes and $r > 1$. Then $p \geq 2$, $q \geq 3$ and $r \geq 2$, not necessarily a prime. Thus we have

Observe that $p, q, p r, q r$ are all distinct. Hence their product divides $(n-2)!$. Thus $n^{2} = p^{2} q^{2} r^{2}$ divides $(n-2)!$ in this case. We conclude that either $n = p q$ where $p, q$ are distinct primes or $n = p^{k}$ for some prime $p$.

Case 1. Suppose $n = p q$ for some primes $p, q$, where $2 < p < q$. Then $p \geq 3$ and $q \geq 5$. In this case

Thus $p, q, 2p, 2q$ are all distinct numbers in the set $\{1, 2, 3, \ldots, n-2\}$. We see that $n^{2} = p^{2} q^{2}$ divides $(n-2)!$. We conclude that $n = 2q$ for some prime $q \geq 3$. Note that $n-2 = 2q-2 < 2q$ in this case so that $n^{2}$ does not divide $(n-2)!$.

Case 2. Suppose $n = p^{k}$ for some prime $p$. We observe that $p, 2p, 3p, \ldots, (p^{k-1}-1)p$ all lie in the set $\{1, 2, 3, \ldots, n-2\}$. If $p^{k-1}-1 \geq 2k$, then there are at least $2k$ multiples of $p$ in the set $\{1, 2, 3, \ldots, n-2\}$. Hence $n^{2} = p^{2k}$ divides $(n-2)!$. Thus $p^{k-1}-1 < 2k$.

If $k \geq 5$, then $p^{k-1}-1 \geq 2^{k-1}-1 \geq 2k$, which may be proved by an easy induction. Hence $k \leq 4$. If $k=1$, we get $n=p$, a prime. If $k=2$, then $p-1<4$ so that $p=2$ or $3$; we get $n=2^{2}=4$ or $n=3^{2}=9$. For $k=3$, we have $p^{2}-1<6$ giving $p=2$; $n=2^{3}=8$ in this case. Finally, $k=4$ gives $p^{3}-1<8$. Again $p=2$ and $n=2^{4}=16$. However $n^{2}=2^{8}$ divides $14!$ and hence is not a solution.

Thus $n = p, 2p$ for some prime $p$ or $n = 8, 9$. It is easy to verify that these satisfy the conditions of the problem.

Since $x y z \neq 0$, we can divide the second relation by the first. Observe that

holds for any $x, y$. Thus we get

However, for any real numbers $x, y$, we have

Since $x^{2} y^{2} z^{2}=|x y||y z||z x|$, we get

This is possible only if

hold simultaneously. However $|x y|= \pm x y$. If $x^{2}-x y+y^{2}=-x y$, then $x^{2}+y^{2}=0$ giving $x=y=0$. Since we are looking for nonzero $x, y, z$, we conclude that $x^{2}-x y+y^{2}=x y$ which is same as $x=y$. Using the other two relations, we also get $y=z$ and $z=x$. The first equation now gives $27 x^{6}=x^{3}$. This gives $x^{3}=1 / 27$ (since $x \neq 0$ ), or $x=1 / 3$. We thus have $x=y=z=1 / 3$. These also satisfy the second relation, as may be verified.