# $32^{\text {nd }}$ Indian National Mathematical Olympiad-2017 

## Problems and Solutions

1. In the given figure, $A B C D$ is a square paper. It is folded along $E F$ such that $A$ goes to a point $A^{\prime} \neq C, B$ on the side $B C$ and $D$ goes to $D^{\prime}$. The line $A^{\prime} D^{\prime}$ cuts $C D$ in $G$. Show that the inradius of the triangle $G C A^{\prime}$ is the sum of the inradii of the triangles $G D^{\prime} F$ and $A^{\prime} B E$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf078063ce163dc741bfg-1.jpg?height=268&width=255&top_left_y=538&top_left_x=1309)

Solution: Observe that the triangles $G C A^{\prime}$ and $A^{\prime} B E$ are similar to the triangle $G D^{\prime} F$. If $G F=u, G D^{\prime}=v$ and $D^{\prime} F=w$, then we have

$$
A^{\prime} G=p u, C G=p v, A^{\prime} C=p w, \quad A^{\prime} E=q u, B E=q w, A^{\prime} B=q v
$$

If $r$ is the inradius of $\triangle G D^{\prime} F$, then $p r$ and $q r$ are respectively the inradii of triangles $G C A^{\prime}$ and $A^{\prime} B E$. We have to show that $p r=r+q r$. We also observe that

$$
A E=E A^{\prime}, \quad D F=F D^{\prime}
$$

Therefore

$$
p w+q v=q w+q u=w+u+p v=v+p u
$$

The last two equalities give $(p-1)(u-v)=w$. The first two equalities give $(p-q) w=q(u-v)$. Hence

$$
\frac{p-q}{q}=\frac{u-v}{w}=\frac{1}{p-1}
$$

This simplifies to $p(p-q-1)=0$. Since $p \neq 0$, we get $p=q+1$. This implies that $p r=q r+r$.

2. Suppose $n \geq 0$ is an integer and all the roots of $x^{3}+\alpha x+4-\left(2 \times 2016^{n}\right)=0$ are integers. Find all possible values of $\alpha$.

Solution 1: Let $u, v, w$ be the roots of $x^{3}+\alpha x+4-\left(2 \times 2016^{n}\right)=0$. Then $u+v+w=0$ and $u v w=-4+\left(2 \times 2016^{n}\right)$. Therefore we obtain

$$
u v(u+v)=4-\left(2 \times 2016^{n}\right)
$$

Suppose $n \geq 1$. Then we see that $u v(u+v) \equiv 4\left(\bmod 2016^{n}\right)$. Therefore $u v(u+v) \equiv 1$ $(\bmod 3)$ and $u v(u+v) \equiv 1(\bmod 9)$. This implies that $u \equiv 2(\bmod 3)$ and $v \equiv 2(\bmod 3)$. This shows that modulo 9 the pair $(u, v)$ could be any one of the following:

$$
(2,2),(2,5),(2,8),(5,2),(5,5),(5,8),(8,2),(8,5),(8,8)
$$

In each case it is easy to check that $u v(u+v) \not \equiv 4(\bmod 9)$. Hence $n=0$ and $u v(u+v)=2$. It follows that $(u, v)=(1,1),(1,-2)$ or $(-2,1)$. Thus

$$
\alpha=u v+v w+w u=u v-(u+v)^{2}=-3
$$

for every pair $(u, v)$.

Solution 2: Let $a, b, c \in \mathbb{Z}$ be the roots of the given equation for some $n \in \mathbb{N}_{0}$. By Vieta Theorem, we know that

$$
\begin{gathered}
a+b+c=0 \\
a b+b c+c a=\alpha \\
a b c=2 \times 2016^{n}-4
\end{gathered}
$$

If possible, let us have $n \geq 1$. Since $7 \mid 2016$, we have that

$$
7|a b c+4 \Longrightarrow 7| 3(a b c+4) \Longrightarrow 7|3 a b c+12 \Longrightarrow 7| 3 a b c+5
$$

Since we have $a+b+c=0$, we get that $3 a b c=a^{3}+b^{3}+c^{3}$. Substituting this in the earlier expression, we get that

$$
a^{3}+b^{3}+c^{3}+5 \equiv 0 \quad(\bmod 7)
$$

Consider below, a table calculating the residues of cubes modulo 7 .

| $x$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x^{3}$ | 0 | 1 | 1 | -1 | 1 | -1 | -1 |

Hence, we know that if $x \in \mathbb{N}$, then we have $x^{3} \equiv 0,1,-1(\bmod 7)$. Since $a^{3}+b^{3}+c^{3} \equiv 2$ $(\bmod 7)$, we see that we must have one of the numbers divisible by 7 and the other two numbers, when cubed, must leave 1 as remainder modulo 7. Without of generality, let us assume that

$$
a \equiv 0 \quad(\bmod 7), \quad b^{3}, c^{3} \equiv 1 \quad(\bmod 7)
$$

Hence, we have $b, c \equiv 1,2,4(\bmod 7)$. We will consider all possible values of $b+c$ modulo 7 . Since the expression is symmetric in $b, c$, modulo 7 , we will consider $b \leq c$.

| $b$ | 1 | 1 | 1 | 2 | 2 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $c$ | 1 | 2 | 4 | 2 | 4 | 4 |
| $b+c$ | 2 | 3 | 5 | 4 | 6 | 1 |

We see that, in all the above cases, we get $7 \nmid\langle b+c$. But this is a contradiction, since 7$| a+b+c$ and $7 \mid a$ together imply that $7 \mid b+c$. Hence, we cannot have $n \geq 1$. Hence, the only possible value is $n=0$. Substituting this value in the original equation, the equation becomes

$$
x^{3}+\alpha x+2=0
$$

Solving the equations $a+b+c=0$ and $a b c=-2$ in integers, we see that the only possible solutions $(a, b, c)$ are permutations of $(1,1,-2)$. In case of any permutation, $\alpha=-3$. Substituting this value of $\alpha$ back in the equation, we see that we indeed, get integer roots. Hence, the only possible value for $\alpha$ is -3 .

3. Find the number of triples $(x, a, b)$ where $x$ is a real number and $a, b$ belong to the set $\{1,2,3,4,5,6,7,8,9\}$ such that

$$
x^{2}-a\{x\}+b=0
$$

where $\{x\}$ denotes the fractional part of the real number $x$. (For example $\{1.1\}=0.1=$ $\{-0.9\}$.

Solution: Let us write $x=n+f$ where $n=[x]$ and $f=\{x\}$. Then

$$
f^{2}+(2 n-a) f+n^{2}+b=0
$$

Observe that the product of the roots of (1) is $n^{2}+b \geq 1$. If this equation has to have a solution $0 \leq f<1$, the larger root of (1) is greater 1 . We conclude that the equation (1) has a real root less than 1 only if $P(1)<0$ where $P(y)=y^{2}+(2 n-a) y+n^{2}+2 b$. This gives

$$
1+2 n-a+n^{2}+2 b<0
$$

Therefore we have $(n+1)^{2}+b<a$. If $n \geq 2$, then $(n+1)^{2}+b \geq 10>a$. Hence $n \leq 1$. If $n \leq-4$, then again $(n+1)^{2}+b \geq 10>a$. Thus we have the range for $n:-3,-2,-1,0,1$. If $n=-3$ or $n=1$, we have $(n+1)^{2}=4$. Thus we must have $4+b<a$. If $a=9$, we must have $b=4,3,2,1$ giving 4 values. For $a=8$, we must have $b=3,2,1$ giving 3 values. Similarly, for $a=7$ we get 2 values of $b$ and $a=6$ leads to 1 value of $b$. In each case we get a real value of $f<1$ and this leads to a solution for $x$. Thus we get totally $2(4+3+2+1)=20$ values of the triple $(x, a, b)$.

For $n=-2$ and $n=0$, we have $(n+1)^{2}=1$. Hence we require $1+b<a$. We again count pairs $(a, b)$ such that $a-b>1$. For $a=9$, we get 7 values of $b$; for $a=8$ we get 6 values of $b$ and so on. Thus we get $2(7+6+5+4+3+2+1)=56$ values for the triple $(x, a, b)$.

Suppose $n=-1$ so that $(n+1)^{2}=0$. In this case we require $b<a$. We get $8+7+6+5+$ $4+3+2+1=36$ values for the triple $(x, a, b)$.

Thus the total number of triples $(x, a, b)$ is $20+56+36=112$.

4. Let $A B C D E$ be a convex pentagon in which $\angle A=\angle B=\angle C=\angle D=120^{\circ}$ and whose side lengths are 5 consecutive integers in some order. Find all possible values of $A B+B C+C D$.

Solution 1: Let $A B=a, B C=b$, and $C D=c$. By symmetry, we may assume that $c<a$. We show that $D E=a+b$ and $E A=b+c$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf078063ce163dc741bfg-3.jpg?height=393&width=333&top_left_y=1533&top_left_x=885)

Draw a line parallel to $B C$ through $D$. Extend $E A$ to meet this line at $F$. Draw a line parallel to $C D$ through $B$ and let it intersect $D F$ in $G$. Let $A B$ intersect $D F$ in $H$. We have $\angle F D E=60^{\circ}$ and $\angle E=60^{\circ}$. Hence $E F D$ is an equilateral triangle. Similarly $A F H$ and $B G H$ are also equilateral triangles. Hence $H G=G B=c$. Moreover, $D G=b$. Therefore $H D=b+c$. But $H D=A E$ since $F H=F A$ and $F D=F E$. Also $A H=a-B H=$ $a-B G=a-c$. Hence $E D=E F=E A+A F=b+c+A H=(b+c)+(a-c)=b+a$.

We have five possibilities:

(1) $b<c<a<b+c<a+b$;

(2) $c<b<a<b+c<a+b$;

(3) $c<a<b<b+c<a+b$;

(4) $b<c<b+c<a<a+b$;

(5) $c<b<b+c<a<a+b$.

In (1), we see that $c<a<b+c$ are three consecutive integers provided $b=2$. Hence we get $c=3$ and $a=4$. In this case $b+c=5$ and $a+b=6$ so that we have five consecutive integrs $2,3,4,5,6$ as side lengths. In (2), $b<a<b+c$ form three consecutive integrs only when $c=2$. Hence $b=3, a=4$. But then $b+c=5$ and $a+b=7$. Thus the side lengths are $2,3,4,6,7$ which are not consecutive integers. In case (3), $b<b+c$ are two consecutive integrs so that $c=1$. Hence $a=2$ and $b=3$. We get $b+c=4$ and $a+b=5$ so that the consecutive integers $1,2,3,4,5$ form the side lengths. In case (4), we have $c<b+c$ as two consecutive integers and hence $b=1$. Therefore $c=2, b+c=3, a=4$ and $a+b=5$ which is admissible. Finally, in case (5) we have $b<b+c$ as two consecutive integers, so that $c=1$. Thus $b=2, b+c=3, a=4$ and $a+b=6$. We do not get consecutive integers.

Therefore the only possibilities are $(a, b, c)=(4,2,3),(2,3,1)$ and $(4,1,2)$. This shows that $a+b+c=9,6$ or 7 . Thus there are three possible sums $A B+B C+C A$, namely, 6,7 or 9 .

Solution 2: As in the earlier solution, $E D=d=a+b$ and $E A=e=b+c$. Let the sides be $x-2, x-1, x, x+1, x+2$. Then $x \geq 3$. We also have $x+2 \geq x-1+x-2$ so that $x \leq 5$. Thus $x=3,4$ or 5 . If $x=5$, the sides are $\{3,4,5,6,7\}$ and here we do not have two pairs which add to a number in the set. Hence $x=3$ or 4 and we get the sets as $\{1,2,3,4,5\}$ or $\{2,3,4,5,6\}$. With the set $\{1,2,3,4,5\}$ we get

$$
(a, b, c, d, e)=(2,3,1,5,4),(4,1,2,5,3)
$$

From the set $\{2,3,4,5,6\}$, we get $(a, b, c, d, e)=(4,2,3,6,5)$. Thus we see that $a+b+c=6,7$ or 9 .

Solution 3: We use the same notations and we get $d=a+b$ and $e=b+c$. If $a \geq 5$, we see that $d-b \geq 5$. But the maximum difference in a set of 5 consecutive integers is 4 . Hence $a \leq 4$. Similarly, we see $b \leq 4$ and $c \leq 4$. Thus we see that $a+b+c \leq 2+3+4=9$. But $a+b+c \geq 1+2+3=6$. It follows that $a+b+c=6,7,8$ or 9 . If we take $(a, b, c, d, e)=(1,3,2,4,5)$, we get $a+b+c=6$. Similarly, $(a, b, c, d, e)=(2,1,4,3,5)$ gives $a+b+c=7$, For $a+b+c=8$, the only we we can get $1+3+4=8$. Here we cannot accommodate 2 and consecutiveness is lost. For 9 , we can have $(a, b, c, d, e)=(3,2,4,5,6)$ and $a+b+c=9$.

5. Let $A B C$ be a triangle with $\angle A=90^{\circ}$ and $A B<A C$. Let $A D$ be the altitude from $A$ on to $B C$. Let $P, Q$ and $I$ denote respectively the incentres of triangles $A B D, A C D$ and $A B C$. Prove that $A I$ is perpendicular to $P Q$ and $A I=P Q$.

Solution: Draw $P S \| B C$ and $Q S \| A D$. Then $P S Q$ is a right-angled triangle with $\angle P S Q=90^{\circ}$. Observe that $P S=r_{1}+r_{2}$ and $S Q=r_{2}-r_{1}$, where $r_{1}$ and $r_{2}$ are the inradii of triangles $A B D$ and $A C D$, respectively. We observe that triangles $D A B$ and $D C A$ are similar to triangle $A C B$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf078063ce163dc741bfg-4.jpg?height=333&width=637&top_left_y=1991&top_left_x=1161)

Hence

$$
r_{1}=\frac{c}{a} r, \quad r_{2}=\frac{b}{a} r
$$

where $r$ is the inradius of triangle $A B C$. Thus we get

$$
\frac{P S}{S Q}=\frac{r_{2}+r_{1}}{r_{2}-r_{1}}=\frac{b+c}{b-c}
$$

On the otherhand $A D=h=b c / a$. We also have $B E=c a /(b+c)$ and

$$
B D^{2}=c^{2}-h^{2}=c^{2}-\frac{b^{2} c^{2}}{a^{2}}=\frac{c^{4}}{a^{2}}
$$

Hence $B D=c^{2} / a$. Therefore

$$
D E=B E-B D=\frac{c a}{b+c}-\frac{c^{2}}{a}=\frac{c b(b-c)}{a(b+c)}
$$

Thus we get

$$
\frac{A D}{D E}=\frac{b+c}{b-c}=\frac{P S}{S Q}
$$

Since $\angle A D E=90^{\circ}=\angle P S Q$, we conclude that $\triangle A D E \sim \triangle P S Q$. Since $A D \perp P S$, it follows that $A E \perp P Q$.

We also observe that

$$
P Q^{2}=P S^{2}+S Q^{2}=\left(r_{2}+r_{1}\right)^{2}+\left(r_{2}-r_{1}\right)^{2}=2\left(r_{1}^{2}+r_{2}^{2}\right)
$$

However

$$
r_{1}^{2}+r_{2}^{2}=\frac{c^{2}+b^{2}}{a^{2}} r^{2}=r^{2}
$$

Hence $P Q=\sqrt{2} r$. We also observe that $A I=r \operatorname{cosec}(A / 2)=r \operatorname{cosec}\left(45^{\circ}\right)=\sqrt{2} r$. Thus $P Q=A I$.

Solution 2: In the figure, we have made the construction as mentioned in the hint. Since $P, Q$ are the incentres of $\triangle A B D, \triangle A C D, D P, D Q$ are the internal angle bisectors of $\angle A D B, \angle A D C$ respectively. Since $A D$ is the altitude on the hypotenuse $B C$ in $\triangle A B C$, we have that $\angle P D Q=45^{\circ}+45^{\circ}=90^{\circ}$. It also implies that

$$
\triangle A B C \sim \triangle D B A \sim \triangle D A C
$$

This implies that all corresponding length in the above mentioned triangles have the same ratio.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf078063ce163dc741bfg-6.jpg?height=323&width=491&top_left_y=397&top_left_x=947)

In particular,

$$
\begin{aligned}
& \frac{A I}{B C}=\frac{D P}{A B}=\frac{D Q}{A C} \\
\Longrightarrow \quad & \frac{A I^{2}}{B C^{2}}=\frac{D P^{2}}{A B^{2}}=\frac{D Q^{2}}{A C^{2}}=\frac{D P^{2}+D Q^{2}}{A B^{2}+A C^{2}} \\
\Longrightarrow \quad & \frac{A I^{2}}{B C^{2}}=\frac{P Q^{2}}{B C^{2}}, \quad \text { by Pythagoras Theorem in } \triangle A B C, \triangle P D Q \\
\Longrightarrow \quad & A I=P Q
\end{aligned}
$$

as required.

For the second, part, we note that from the above relations, we have $\triangle A B C \sim \triangle D P Q$. Let us take $\angle A C B=\theta$. Then, we get

$$
\begin{aligned}
\angle P S D & =180^{\circ}-(\angle S P D+\angle S D P) \\
& =180^{\circ}-\left(90^{\circ}-\theta+45^{\circ}\right) \\
& =45^{\circ}+\theta
\end{aligned}
$$

This gives us that

$$
\begin{aligned}
\angle A R S & =180^{\circ}-(\angle A S R+\angle S A R) \\
& =180^{\circ}-(\angle P S D+\angle S A C-\angle I A C) \\
& =180^{\circ}-\left(45^{\circ}+\theta+90^{\circ}-\theta-45^{\circ}\right) \\
& =90^{\circ}
\end{aligned}
$$

as required. Hence, we get that $A I=P Q$ and $A I \perp P Q$.

Solution 3: We know that the angle bisector of $\angle B$ passes through $P, I$ which implies that $B, P, I$ are collinear. Similarly, $C, Q, I$ are also collinear. Since $I$ is the incentre of $\triangle A B C$, we know that

$$
\angle P I Q=\angle B I C=90^{\circ}+\frac{\angle A}{2}=135^{\circ}
$$

Join $A P, A Q$. We know that $\angle B A P=\frac{1}{2} \angle B A D=\frac{1}{2} \angle C$. Also, $\angle A B P=\frac{1}{2} \angle B$. Hence by Exterior Angle Theorem in $\triangle A B P$, we get that

$$
\angle A P I=\angle A B P+\angle B A P=\frac{1}{2}(\angle B+\angle C)=45^{\circ}
$$

Similarly in $\triangle A D C$, we get that $\angle A Q I=45^{\circ}$. Also, we have

$$
\angle P A I=\angle B A I-\angle B A P=45^{\circ}-\frac{\angle C}{2}=\frac{\angle B}{2}
$$

Similarly, we get $\angle Q A I=\frac{\angle C}{2}$.

Now applying Sine Rule in $\triangle A P I$, we get

$$
\frac{I P}{\sin \angle P A I}=\frac{A I}{\sin \angle A P I} \Longrightarrow I P=\sqrt{2} A I \sin \frac{B}{2}
$$

Similarly, applying Sine Rule in $\triangle A Q I$, we get

$$
\frac{I Q}{\sin \angle P A I}=\frac{A I}{\sin \angle A Q I} \Longrightarrow I Q=\sqrt{2} A I \sin \frac{C}{2}
$$

Applying Cosine Rule in $\triangle P I Q$ gives us that

$$
\begin{aligned}
P Q^{2} & =I P^{2}+I Q^{2}-2 \cdot I P \cdot I Q \cos \angle P I Q \\
& =2 A I^{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\sqrt{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)
\end{aligned}
$$

We will prove that $\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\sqrt{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)=\frac{1}{2}$. In any $\triangle X Y Z$, we have that

$$
\sum_{c y c} \sin ^{2} \frac{X}{2}=1-2 \prod \sin \frac{X}{2}
$$

Using this in $\triangle A B C$, and using the fact that $\angle A=90^{\circ}$, we get

$$
\begin{aligned}
& \sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
\Longrightarrow \quad & \frac{1}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-\sqrt{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
\Longrightarrow \quad & \left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\sqrt{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)=\frac{1}{2}
\end{aligned}
$$

which was to be proved. Hence we get $P Q=A I$.

The second part of the problem can be obtained by angle-chasing as outlined in Solution 2 .

Solution 4: Observe that $\angle A P B=\angle A Q C=135^{\circ}$. Thus $\angle A P I=\angle A Q I=45^{\circ}$ (since $B-P-I$ and $C-Q-I)$. Note $\angle P A Q=1 / 2 \angle A=45^{\circ}$. Let $X=B I \cap A Q$ and $Y=C I \cap A P$. Therefore $\angle A X P=180-\angle A P I-\angle P A Q=90^{\circ}$. Similarly $\angle A Y Q=90^{\circ}$. Hence $I$ is the orthocentre of triangle $P A Q$. Therefore $A I$ is perpendicular to $P Q$. Also $A I=2 R_{P A Q} \cos 45^{\circ}=2 R_{P A Q} \sin 45^{\circ}=P Q$.

6. Let $n \geq 1$ be an integer and consider the sum

$$
x=\sum_{k \geq 0}\binom{n}{2 k} 2^{n-2 k} 3^{k}=\binom{n}{0} 2^{n}+\binom{n}{2} 2^{n-2} \cdot 3+\binom{n}{4} 2^{n-4} \cdot 3^{2}+\cdots
$$

Show that $2 x-1,2 x, 2 x+1$ form the sides of a triangle whose area and inradius are also integers.

Solution: Consider the binomial expansion of $(2+\sqrt{3})^{n}$. It is easy to check that

$$
(2+\sqrt{3})^{n}=x+y \sqrt{3}
$$

where $y$ is also an integer. We also have

$$
(2-\sqrt{3})^{n}=x-y \sqrt{3}
$$

Multiplying these two relations, we obtain $x^{2}-3 y^{2}=1$.

Since all the terms of the expansion of $(2+\sqrt{3})^{n}$ are positive, we see that

$$
2 x=(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}=2\left(2^{n}+\binom{n}{2} 2^{n-2} \cdot 3+\cdots\right) \geq 4
$$

Thus $x \geq 2$. Hence $2 x+1<2 x+(2 x-1)$ and therefore $2 x-1,2 x, 2 x+1$ are the sides of a triangle. By Heron's formula we have

$$
\Delta^{2}=3 x(x+1)(x)(x-1)=3 x^{2}\left(x^{2}-1\right)=9 x^{2} y^{2}
$$

Hence $\Delta=3 x y$ which is an integer. Finally, its inradius is

$$
\frac{\text { area }}{\text { perimeter }}=\frac{3 x y}{3 x}=y
$$

which is also an integer.

Solution 2: We will first show that the numbers $2 x_{n}-1,2 x_{n}, 2 x_{n}+1$ form the sides of a triangle. To show that, it suffices to prove that $2 x_{n}-1+2 x_{n}>2 x_{n}+1$. If possible, let the converse hold. Then, we see that we must have $4 x_{n}-1 \leq 2 x_{n}+1$, which implies that $x_{n} \leq 1$. But we see that even for the smallest value of $n=1$, we have that $x_{n}>1$. Hence, the numbers are indeed sides of a triangle.

Let $\Delta_{n}, r_{n}, s_{n}$ denote respectively, the area, inradius and semiperimeter of the triangle with sides $2 x_{n}-1,2 x_{n}, 2 x_{n}+1$. By Heron's Formula for the area of a triangle, we see that

$$
\Delta_{n}=\sqrt{3 x_{n}\left(x_{n}-1\right) x_{n}\left(x_{n}+1\right)}=x_{n} \sqrt{3\left(x_{n}^{2}-1\right)}
$$

If possible, let $\Delta_{n}$ be an integer for all $n \in \mathbb{N}$. We see that due to the presence of the first term $\binom{n}{0} 2^{n}$, we have $3 \nmid x_{n}, \forall n \in \mathbb{N}$. Hence, we get that $3 \mid x_{n}^{2}-1$. Hence, we can write $x_{n}^{2}-1$ as $3 m$ for some $m \in \mathbb{N}$. Then, we can also write

$$
\Delta_{n}=3 x_{n} \sqrt{m}
$$

Note that we have assumed that $\Delta_{n}$ is an integer. Hence, we see that we must have $m$ to be a perfect square. Consequently, we get that

$$
r_{n}=\frac{\Delta_{n}}{s_{n}}=\frac{\Delta_{n}}{3 x_{n}}=\sqrt{m} \in \mathbb{Z}
$$

Hence, it only remains to show that $\Delta_{n} \in \mathbb{Z}, \forall n \in \mathbb{N}$. In other words, it suffices to show that $3\left(x_{n}^{2}-1\right)$ is a perfect square for all $n \in \mathbb{N}$.

We see that we can write $x_{n}$ as

$$
\begin{aligned}
x_{n} & =\frac{1}{2}\left(2 \sum_{k \geq 0}\binom{n}{2 k} 2^{n-2 k} 3^{k}\right) \\
& =\frac{1}{2}\left((2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right) \\
3 x_{n}^{2}-3 & =\frac{3}{4}\left((2+\sqrt{3})^{2 n}+(2-\sqrt{3})^{2 n}+2(2+\sqrt{3})^{n}(2-\sqrt{3})^{n}\right)-3 \\
& =\frac{3}{4}\left((2+\sqrt{3})^{2 n}+(2-\sqrt{3})^{2 n}-2(2+\sqrt{3})^{n}(2-\sqrt{3})^{n}\right) \\
& =\left(\frac{\sqrt{3}}{2}\left((2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right)\right)^{2}
\end{aligned}
$$

We are left to show that the quantity obtained in the above equation is an integer. But we see that if we define

$$
a_{n}=\frac{\sqrt{3}}{2}\left((2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right), \quad \forall n \in \mathbb{N}
$$

the sequence $\left\langle a_{k}\right\rangle_{k=1}^{\infty}$ thus obtained is exactly the solution for the recursion given by

$$
a_{n+2}=4 a_{n+1}-a_{n}, \quad \forall n \in \mathbb{N}, \quad a_{1}=3, a_{2}=12
$$

Hence, clearly, each $a_{n}$ is obviously an integer, thus completing the proof. $\qquad$