INMO problems + Summary of 2021

Welcome back! I tried the previous year INMO problems. And yes, I did a lot of them, here are a few!

Here are the problems! ( they are kept in random order)

[2008 P2]: Find all triples $ \left(p,x,y\right)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.

[some INMO]:Find all $m,n\in\mathbb N$ and primes $p\geq 5$ satisfying

$$m(4m^2+m+12)=3(p^n-1).$$

[2015 P3] Find all real functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2+yf(x))=xf(x+y)$.

[2005 P6]: Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ such that\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]for all $x, y, z \in \mathbb{R}$.

[2012 P6]Let $f : \mathbb{Z} \to \mathbb{Z}$ be a function satisfying $f(0) \ne 0$, $f(1) = 0$ and

$(i) f(xy) + f(x)f(y) = f(x) + f(y)$

$(ii)\left(f(x-y) - f(0)\right ) f(x)f(y) = 0 $ for all $x,y \in \mathbb{Z}$, simultaneously.

$(a)$ Find the set of all possible values of the function $f$.

$(b)$ If $f(10) \ne 0$ and $f(2) = 0$, find the set of all integers $n$ such that $f(n) \ne 0$.

[2011 P4]: Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.

[2011 P4]Find all functions $f:\mathbb{R}\to \mathbb R$ satisfying

$$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$$ For all $x,y\in\mathbb R$.

[2014 P2]: Let $n$ be a natural number. Prove that,

$$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor $$

is even.

[2012 P2]:Let $p_1<p_2<p_3<p_4$ and $q_1<q_2<q_3<q_4$ be two sets of prime numbers, such that $p_4 - p_1 = 8$ and $q_4 - q_1= 8$. Suppose $p_1 > 5$ and $q_1>5$. Prove that $30$ divides $p_1 - q_1$.

[2011 P2]Call a natural number $n$ faithful if there exist natural numbers $a<b<c$ such that $a|b,$ and $b|c$ and $n=a+b+c

$(i)$ Show that all but a finite number of natural numbers are faithful.

$(ii)$ Find the sum of all natural numbers which are not faithful.

[2017 P1]: In the given figure, $ABCD$ is a square sheet of paper. It is folded along $EF$ such that $A$ goes to a point $A'$ different from $B$ and $C$, on the side $BC$ and $D$ goes to $D'$. The line $A'D'$ cuts $CD$ in $G$. Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$.

[2020 P6]:A stromino is a $3 \times 1$ rectangle. Show that a $5 \times 5$ board divided into twenty-five $1 \times 1$ squares cannot be covered by $16$ strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.)

[2020 p1]: Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.

[2016 P4]:Suppose $2016$ points of the circumference of a circle are coloured red and the remaining points are coloured blue . Given any natural number $n\ge 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.

[2018 P6] Let $\mathbb N$ denote set of all natural numbers and let $f:\mathbb{N}\to\mathbb{N}$ be a function such that

$\text{(a)} f(mn)=f(m).f(n)$ for all $m,n \in\mathbb{N}$;

$\text{(b)} m+n$ divides $f(m)+f(n)$ for all $m,n\in \mathbb N$.

Prove that, there exists an odd natural number $k$ such that $f(n)= n^k$ for all $n$ in $\mathbb{N}$.

[2017 P6] Let $n\ge 1$ be an integer and consider the sum$$x=\sum_{k\ge 0} \dbinom{n}{2k} 2^{n-2k}3^k=\dbinom{n}{0}2^n+\dbinom{n}{2}2^{n-2}\cdot{}3+\dbinom{n}{4}2^{n-k}\cdot{}3^2 + \cdots{}.$$Show that $2x-1,2x,2x+1$ form the sides of a triangle whose area and inradius are also integers.

[2021 P1]Suppose $r \ge 2$ is an integer, and let $m_1, n_1, m_2, n_2, \dots, m_r, n_r$ be $2r$ integers such that$$\left|m_in_j-m_jn_i\right|=1$$for any two integers $i$ and $j$ satisfying $1 \le i<j \le r$. Determine the maximum possible value of $r$.

[2006 P2] Prove that for every positive integer $n$ there exists a unique ordered pair $(a,b)$ of positive integers such that

$$ n = \frac{1}{2}(a + b - 1)(a + b - 2) + a . $$

The solutions:

[2008 P2]: Find all triples $ \left(p,x,y\right)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.

Solution: If $p$ is two $\implies y $ is even. So $4|y^4$ and clearly $x\ge 3.$ Then divide the equation by $4.$ We get $2|\text{RHS}$ but $2\nmid \text{ LHS}$

So $p$ is an odd prime. Note that $$p^x=y^4+4=(y^2-2y+2)(y^2+2y+2)\implies y^2-2y+1|y^2+2y+1\implies y^2-2y+2|4y$$

$$y^2-2y+2\le 4y\implies y=2,1$$

Verrifying, we get $y=1$ satisfying. Giving us $(p,z,y)=\boxed{(5,1,1)}$ as the only solution.

[some INMO]:Find all $m,n\in\mathbb N$ and primes $p\geq 5$ satisfying

$$m(4m^2+m+12)=3(p^n-1).$$

Solution:  $$m(4m^2+m+12)=3(p^n-1)\implies \frac{(m^2+3)(4m+1)}{3}=p^n$$Note that$$\gcd(4m+1,m^2+3)=\gcd(4m^2+m,4m^2+12\gcd(m-12,4m^2+12)$$$$=\gcd(4m-48,4m+1)|49\implies p=7.$$

Now we have $m^2+3, 4m+1=p^a,3\cdot p^b$ and gcd is $49.$ Hence one of $a,b$ is $2.$

Case1: $m^2+3=49$ not possible

Case2: $m^2+3=49\cdot 3\implies m=12\implies 4\cdot 12+1=49\implies n=4$

Case3: $4m+1=49\implies m=12$

Case4: $4m+1=49\cdot 3$ not possible.

We can do similar  for gcd is $7.$

So the only solution is $\boxed{(p,m,n)=(7,12,4)}$

[2015 P3] Find all real functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2+yf(x))=xf(x+y)$.

Solution: Let $P(x,y)$ denote the assertion.

$$P(x,0)\implies f(x^2)=xf(x)=-xf(-x)\implies f(x)=-f(-x)$$

So $$f(0)=0$$

Now, we will show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 0$

$$P(t,0)\implies f(t^2)=0$$

$$P(t,t-a)\implies 0=f(t^2+(t-a)f(t))=tf(a),\forall a$$

This would imply that $f$ is $0$ constant function or $t=0.$ So assuming that $f$ is not $0,$ we get $f$ injective at $0.$

$$P(a,-a)\implies f(a^2-af(a))=f(a-a)=0\implies a^2-af(a)=0\implies f(a)=a$$

So the solutions are $\boxed{ f=0, f(a)=a}$

[2005 P6]: Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ such that\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]for all $x, y, z \in \mathbb{R}$.

Solution: Let $P(x,y)$ denote the assertion.

Then $$P(x,0,0)\implies f(x^2)=xf(x)\implies f(x)=-f(-x)$$

And $$f(x^2+y^2)=f(x^2)+f(y^2)\implies f(a+b)=f(a)+f(b), a,b \in \mathbb{R}^{+}$$

Also $f(0)=0.$ Now we show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 1.$ So  $$P(0,1,t)\implies 0=f(f(t))=tf(1)\implies f(1)=0\text{ or } f=0$$

Assuming $f$ is non constant, we get $f(1)=0.$ Then $$P(0,a,1)\implies 0=f(a)\forall a.$$

Not possible, hence $f$ is injective at $0.$

Moreover, $$P(0,1,x)\implies f(f(x))=xf(1)$$

Note that $$f(f(x))=xf(1)\implies f \text { is bijective }$$

Actually, we don't need the following claim, but I found it too good to not add.

Claim: $f(1)=1$ 

Proof: $$P(0,y,1)\implies f(yf(1))=f(y)\implies yf(1)=y\implies f(1)=1$$

Alternate proof: Using subjectivity, $$\exists k, f(k)=1\implies P(0,k,k)\implies 1=f(kf(k))=k\implies k=1$$

Now, fix $x.$ We carry $y$ and choose $z$ such that  $$f(x^2+yf(z)=xf(x)+zf(y)=0$$ 

Then $$x^2+yf(z)=0, xf(x)+zf(y)=0$$

Take $y=x\implies z=-x$ in $xf(x)+zf(y)=0.$ So, we get $$x^2+xf(-x)=0\implies f(-x)=-x\implies f(x)=x$$

Hence the solutions are $\boxed{ f=0,f(x)=x}$

[2012 P6]Let $f : \mathbb{Z} \to \mathbb{Z}$ be a function satisfying $f(0) \ne 0$, $f(1) = 0$ and

$(i) f(xy) + f(x)f(y) = f(x) + f(y)$

$(ii)\left(f(x-y) - f(0)\right ) f(x)f(y) = 0 $ for all $x,y \in \mathbb{Z}$, simultaneously.

$(a)$ Find the set of all possible values of the function $f$.

$(b)$ If $f(10) \ne 0$ and $f(2) = 0$, find the set of all integers $n$ such that $f(n) \ne 0$.

Solution: This problem was actually pretty fun to do!

Let $P(x,y)$ be the assertion.

Note that when $f(x)\ne 0$ then by (ii) with $P(x,0),$ we get $f(x)=0.$

Note that $P(0,0)$ in (i) gives us, $f(0)=1.$

So the possible values of $f$ are $1,0.$

Note that 

$\bullet f(x)=0,f(y)=0\implies f(xy)=0$

$\bullet f(x)=1,f(y)=0\implies f(xy)=1$

$\bullet f(x)=1,f(y)=1\implies f(xy)=1$

$\bullet f(x)=1,f(y)=1\implies f(x-y)=1$

Since $f(10)=1,f(2)=0\implies f(5)=1.$

So all multiples of $5$ are mapped to $1$ i.e $f(5k)=1.$

Now if there is a number $a$ such that $\gcd(a,5)=1,$ we get that all multiples of $a$ are mapped to $1.$

Since the gcd is $1.$ By bezout's theorem, we get that $\exists m,n \text{ such that } am+5n=1.$ Also, clearly one of $m,n$ is negative and other is positive. 

Case 1: $n$ is negative

Since $$f(ma)=1, f(5(-n))=1 \implies f(1)=f(ma-5(-n))=0$$ Not possible. 

Case 2: $m$ is negative

Since $$f((-m)a)=1, f(5n)=1 \implies f(1)=f(-ma+5n)=0$$ Not possible. 

Hence, only numbers divisible by $5$ are mapped to $1.$ Rest all the numbers are mapped to $0.$

[2011 P4]: Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.

Proof: Note that whenever we choose $2$ points, we form an edge and there are $3$ edges parallel to our edge every time.

Now suppose we have $4$ points chosen and these $4$ points do not form a trapezoid.

So we have $6$ non-parallel edges. Let the vertices of the quadrilateral be called " Quadratic vertices". Now, each of the six edges is parallel to three other edges, which are not part of the quadrilateral's edge ( If they would have, then we would have got a trapezoid).

Hence each of the quadratic vertices can not be connected to three other vertexes out of the $5$ vertexes left.

For example, in the image we have $E$ can not be connected to $D, A, G.$

Now, we interpret it in GT terms. Define a $bad$ edge, an edge that is parallel to one of the quadratic edges. Now consider the graph of all the $bad $ edges. Note that this graph is a bipartite graph with one vertice set as the quadratic vertices set ( which has $4$ vertices ) and the other set as non-quadratic vertice set ( which has $5$ vertices)

Note that $6\times 3=18$ edges are parallel to one of the quadrilateral edges and hence are bad edges. Suppose there exist a non-quadratic vertex $X$ such that we don't get a trapezoid. Hence, the degree of $X$ is $0.$ So we can safely delete $X.$ So the non-quadratic vertices set has $4$ vertices and the quadratic vertices set has $4$ vertices. But the maximum edges, we can get is $4\times 4=16$ as the graph is bipartite.

A contradiction as we have $18$ edges in the graph.

[2014 P2]: Let $n$ be a natural number. Prove that,

$$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor $$

is even.

Claim: $$ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor =d(1)+d(2)+\dots+d(n)$$

Proof: Define $$a_n=\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor .$$

Hence, $$a_{n}-a_{n-1}=\sum \left\lfloor \frac{n}{k} \right\rfloor-\left\lfloor \frac{n-1}{k} \right\rfloor=d(n)$$

As $\left\lfloor \frac{n}{k} \right\rfloor\ge\left\lfloor \frac{n-1}{k} \right\rfloor, \left\lfloor \frac{n}{k} \right\rfloor-\left\lfloor \frac{n-1}{k} \right\rfloor=1 \text { iff } k|n $

By telescoping, we get $$a_n=\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor =d(1)+d(2)+\dots+d(n).$$

Main proof: Note that $d(x)$ is odd iff $x$ is an perfect square.

We use induction. For $n=1,2$ we have $$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor $$

 even. Suppose it's true for $n=l$ and we will show it's true for $n=l+1.$

$$\left\lfloor \frac{l}{1} \right\rfloor+ \left\lfloor \frac{l}{2} \right\rfloor + \cdots + \left\lfloor \frac{l}{l} \right\rfloor + \left\lfloor \sqrt{l} \right\rfloor \text { is even }$$

So we have two cases.

Case 1: $l+1$ is perfect square

Then, we have $d(l+1)$ odd and $\left\lfloor \sqrt{l+1} \right\rfloor-\left\lfloor \sqrt{l} \right\rfloor=1.$

Case 2: $l+1$ is not perfect square

Then, we have $d(l+1)$ even  and $\left\lfloor \sqrt{l+1} \right\rfloor-\left\lfloor \sqrt{l} \right\rfloor=0.$

In both cases, 

So $$a_{l+1}=a_{l}+d(l+1)-\left\lfloor \sqrt{l+1} \right\rfloor-\left\lfloor \sqrt{l} \right\rfloor\text { is even. }$$

[2011 P4]Find all functions $f:\mathbb{R}\to \mathbb R$ satisfying

$$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$$ For all $x,y\in\mathbb R$.

Solution: Let $P(x,y)$ denote the assertion.

$$P(0,0)=f(0)^2=4f(0)^2\implies f(0)=0.$$$$P(0,y)=f(y)f(-y)^2=f(y)^2$$$$P(y,0)=f(y)f(y)=f(y)^2$$So $f(y)\ne 0\implies f(y)=f(-y).$

Now if$$f(t)=0,t\ne 0\implies P(t,-t)=f(-t+t)=0=(f(-t)+f(t))^2-4t^2\cdot f(t)=f(-t)^2\implies f(-t)=0$$

Hence$$f(y)=f(-y)\implies y \text { is even }$$

Now, we show that $f$ is injective at $0.$ Let $f(t)=0, t\ne 0.$

$$P(t,x)=f(x+t)f(x-t)=f(x)^2$$$$P(t,x)=f(x+t)f(t-x)=f(x)^2-4t^2f(x)$$

But $f(x-t)=f(t-x)\implies 4t^2f(x)=0$

Assuming $f$ is not constant we get $t=0.$ So $f$ is injective at $0.$

$$P(1,1)\implies 0=f(2)\cdot f(0)=4f(1)^2-4f(1)=4(f(1))(f(1)-1)$$

Using injectivity at $0,$ we get $f(1)=1.$

$$P(y,1)=f(y+1)f(y-1)=1+f(y)^2+2f(y)-4y^2$$$$P(1,y)=f(1+y)f(1-y)=1+f(y)^2+2f(y)-4f(y)$$

But $P(1,y)=P(y,1)\implies y^2=f(y).$

Hence the solutions are $\boxed{f(x)=x^2,f=0}.$

[2011 P2]Call a natural number $n$ faithful if there exist natural numbers $a<b<c$ such that $a|b,$ and $b|c$ and $n=a+b+c.$

$(i)$ Show that all but a finite number of natural numbers are faithful.

$(ii)$ Find the sum of all natural numbers which are not faithful.

Proof: Note that for odd $n\ge 7,$ consider $(a,b,c)=(1,2,n-3).$

Moreover, if a number $x$ has an odd divisor $d$ $\ge 7.$ Then consider $(1,2,d-3)\frac{x}{d}$  

In other words, $\exists d \text { such that d is faithful }, d|n \implies \text { such that n is faithful }.$

By checking, we get $1,2,3,4,5,6$ are not faithful. So consider $2^x\cdot 3^y\cdot 5^z.$ Numbers having larger prime factors are faithful as we showed earlier.

Note that $16=1+3+12$ is faithful. So numbers with $v_2\ge 4$ are faithful.

Note that $25,9$ are faithful. So numbers with $v_3,v_5\ge 2$ are faithful.

Hence the numbers which are non-faithful are of the form, $2^x\cdot 3^y\cdot 5^z, x<4,y<2, z<2.$ And hence are finite.

Moreover, we also have $10=1+3+6$ faithful too.

By checking, we get $\boxed{1,2,3,4,5,6,8,12,24}$ non faithful. Summing to $\boxed{65}$

[2012 P2]:Let $p_1<p_2<p_3<p_4$ and $q_1<q_2<q_3<q_4$ be two sets of prime numbers, such that $p_4 - p_1 = 8$ and $q_4 - q_1= 8$. Suppose $p_1 > 5$ and $q_1>5$. Prove that $30$ divides $p_1 - q_1$.

Proof: By $\mod 6$ ( and using the fact that primes are $\pm 1 \mod 6.$)

We get that the primes are of the form $(p_1,p_2,p_3,p_4)=(p_1,p_1+2,p_1+6, p_1+8)$

 and are $(p_1,p_2,p_3,p_4)\equiv (5,1,5,1) \mod 6.$

Considering $\pmod 5$ and using the fact that $(p_1,p_2,p_3,p_4)=(p_1,p_1+2,p_1+6, p_1+8).$

We get $(p_1,p_2,p_3,p_4)\equiv (1,3,7,4) \mod 5.$

Both are unique modulo. Hence we are done.

[2017 P1]: In the given figure, $ABCD$ is a square sheet of paper. It is folded along $EF$ such that $A$ goes to a point $A'$ different from $B$ and $C$, on the side $BC$ and $D$ goes to $D'$. The line $A'D'$ cuts $CD$ in $G$. Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$.

Proof: Enough through show, $$CG+A'C-A'G=FD'+D'G+A'B+EB-( GF+A'E) $$ $$\implies CG+A'C+GF+A'E=FD'+D'G+A'B+EB+A'G$$

Enough to show, $$ \implies AD-FD+AD-BE+AD-A'B=FD+AD+A'B+EB$$

or show, $$ 2s=2(FD+A'B+EB), \text { where s is the side }$$

or enough to show $$s=FD+A'B+EB.$$

or show $$FD+A'B=AE.$$

Drop the altitude from $F$ to $AB$ as $X.$ 

Enough to show that $\delta FXE $ is congruent $\delta ABA'$ or show that $\angle FEX=\angle AA'B$ which is true by angles.  

[2020 p1]: Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.

proof: And my $X,Y$ got exchanged.

Idk how it help but we have $BXY$ collinear.

Claim: $BXY$ collinear.

Proof: Let $\angle ACB=\alpha,\angle ADB= \theta.$ By tangent lemma, we get $\angle CBA=\theta, \angle ABD=\alpha .$ And $\angle CAB=\angle BAD=180-(\theta+\alpha)\implies \angle CBX=\angle CAX=\angle YAD=\angle DBY=90-(\alpha+\theta)/2.$

Claim: $PO_1=PO_2$

Proof: Note that $PO_1$ is perpendicular bisector of $AC$ and $PO_2$ is perpendicular bisector of $AD.$ And by radical axis, we get $O_1O$ is perpendicular bisector of $AB.$

Definig the midpoint of $AC,AD,AB$ as $H,I,E$ we get $HO_1EA$ and $IO_2EA$ cyclic $\implies \angle EO_1H=180-(\theta+\alpha)=\angle EO_2I\implies \angle PO_1O_2=\angle PO_2O_1.$

Claim: $QO_1=QO_2$

Proof: Note that $QO_1$ is the perpendicular bisector of $AX$ and $QO_2$ is perpendicular bisector of $AY.$ 

Definig the midpoint of $AY,AX$ as $F,G$ we get $FO_1EA$ and $GO_2EA$ cyclic $\implies \angle QO_1O_2= \angle EO_1Q=180-(\angle FO_1E)=\angle YAB=\angle XAB=\angle EO_2G=\angle QO_2O_1.$

[2016 P4]:Suppose $2016$ points of the circumference of a circle are coloured red and the remaining points are coloured blue . Given any natural number $n\ge 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.

Proof: Is it a troll? Make the $n$gons with those $2016.$ We get maximum $2016$ $n-$gons with having at least one red vertices but there are infinitely many $n$ gons

[2020 P6]:A stromino is a $3 \times 1$ rectangle. Show that a $5 \times 5$ board divided into twenty-five $1 \times 1$ squares cannot be covered by $16$ strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.)

Proof: Since each cell is covered with  one or two strominos and every strominoe covers $3$ cells implies there are $2$ cells which are covered only one times. 

Consider the colouring:

$$RBGRB$$

$$BGRBG$$

$$GRBGR$$

$$RBGRB$$

$$BGRBG$$

Note that each strominoe, by this colouring will cover three cells of different colours, but there $8 $ Reds, $8$ Greens and $9$ blues. Now since $16$ strominoe covers $16$ of each color implies the two cells not covered twice are Blue cells.

Now consider another colouring,

$$$BRGBR$

$$GBRGB$$

$$RGBRG$$

$$BRGBR$$

$$GBRGB$$

This colouring also implies $2$ blue cells must be covered one times, but there is only one common intersection. So not possible. And we are done!

Thenks to Abdul for correcting the mistake!

[2018 P6] Let $\mathbb N$ denote set of all natural numbers and let $f:\mathbb{N}\to\mathbb{N}$ be a function such that

$\text{(a)} f(mn)=f(m).f(n)$ for all $m,n \in\mathbb{N}$;

$\text{(b)} m+n$ divides $f(m)+f(n)$ for all $m,n\in \mathbb N$.

Prove that, there exists an odd natural number $k$ such that $f(n)= n^k$ for all $n$ in $\mathbb{N}$.

Proof: Let $P(x,y)$ denote the assertion.

Then $P(1,1)\implies f(1)=1.$

Claim: $f(2)=2^k$ for odd $k.$ 

Proof: $$P(1,2)\implies 1+2|f(2)+1\implies f(2)\equiv 2\pmod 3.$$

And $$P(2,2)\implies 2+2|f(2)+f(2)\implies 2|f(2).$$

Now, if an odd prime $p$ divides $f(2)\implies p|f(2)f(\frac{p-1}{2}).$

$$P(p-1,1)\implies p|f(p-1)+1=f(2)f(\frac{p-1}{2})+1\implies p=1$$

So,$f(2)=2^k$ for odd $k$ as $f(2)\equiv 2\pmod 3.$

Note that $f(2^a)=(2^a)^k.$

Claim: $f(n)=n^k$ for odd $k.$ Which implies $f(x)=x^k.$

Proof: Take $t$ to be sufficient large. $$P(2^t,n)\implies 2^t+n|(2^t)^k+f(n)$$ $$2^t+n|(2^t)^k+n^k$$ $$\implies 2^t+n|f(n)-n^k\implies f(n)=n^k$$ 

[2006 P2] Prove that for every positive integer $n$ there exists a unique ordered pair $(a,b)$ of positive integers such that

$$ n = \frac{1}{2}(a + b - 1)(a + b - 2) + a . $$

Proof: I am just writing a sketch cause it's too difficult to write it out..

Let $T_n$ denote the triangular number i.e $\frac{n(n+1)}{2}.$

For now let's fix the sum $a+b$ and varry $a$ and $b$ to get different $n.$ Then we get numbers of the range $$ T_n+1, T_n+2,\dots, T_n+n=T_{n+1}$$

[2018 P3]: Let $\Gamma_1$ and $\Gamma_2$ be two circles with respective centres $O_1$ and $O_2$ intersecting in two distinct points $A$ and $B$ such that $\angle{O_1AO_2}$ is an obtuse angle. Let the circumcircle of $\Delta{O_1AO_2}$ intersect $\Gamma_1$ and $\Gamma_2$ respectively in points $C (\neq A)$ and $D (\neq A)$. Let the line $CB$ intersect $\Gamma_2$ in $E$ ; let the line $DB$ intersect $\Gamma_1$ in $F$. Prove that, the points $C, D, E, F$ are concyclic.

Proof: We claim that $BE, BF$ are the diametres. 

Note that $$\angle 180-\angle \frac{AO_2D}{2}=\angle ABD=\angle ABF, \angle AFB=\angle \frac{AO_1D}{2}\implies \angle FAB=90^{\deg}$$

Similarly for $BE.$

Note that $$CB\cdot BO_2=DB\cdot BO_1\implies 2CB\cdot BO_2=2DB\cdot BO_1\implies CB\cdot BE=DB\cdot BF.$$

[2006 P2] Prove that for every positive integer $n$ there exists a unique ordered pair $(a,b)$ of positive integers such that

$$ n = \frac{1}{2}(a + b - 1)(a + b - 2) + a . $$

Proof: I am just writing a sketch cause it's too difficult to write it out..

Let $T_n$ denote the triangular number i.e $\frac{n(n+1)}{2}.$

For now let's fix the sum $a+b$ and varry $a$ and $b$ to get different $n.$ Then we get numbers of the range $$ T_n+1, T_n+2,\dots, T_n+n=T_{n+1}$$

[2020 P4]:Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$

I actually saw Red Pig's solution quite long time ago and tried it myself. 

Proof: We use induction.  For $n=2,$ we get $ a_1+a_2=4$ and show $$ a_1+2\le a_1a_2$$

which is just $$ (a_1-1)(a_2-2)\le 0 .$$

Let's assume it's true for $n=1,\dots k$ and we will show that $n=k+1$ works. So we have $$a_1+a_2+\dots+a_{k-1}+a_k+a_{k+1}=2k+2\implies a_1++a_2+\dots+a_{k-1}+(a_k+a_{k+1}-2)=2k, a_{k-1}\le (a_k+a_{k+1}-2)$$

And 

$$ a_1+a_1a_2+\dots +a_1a_2\cdots a_{k-1}+2\le a_1a_2\cdots a_{k-1} (a_k+a_{k+1}-2)$$

So, $$ a_1+a_1a_2+\dots +a_1a_2\cdots a_{k-1}+2+ a_1a_2\cdots a_{k-1}a_k\le a_1a_2\cdots a_{k-1} (a_k+a_{k+1}-2)+a_1a_2\cdots a_{k-1}a_k$$

So, if we show that $$a_1a_2\cdots a_{k-1} (a_k+a_{k+1}-2)+a_1a_2\cdots a_{k-1}a_k\le a_1a_2\cdots a_{k-1}a_ka_{k+1}$$ we are done.

Or show $$a_k+a_{k+1}-2 +a_k\le a_ka_{k+1}$$

which is just $$ (a_{k+1}-2)(1-a_k)\le 0$$

[2017 P6] Let $n\ge 1$ be an integer and consider the sum$$x=\sum_{k\ge 0} \dbinom{n}{2k} 2^{n-2k}3^k=\dbinom{n}{0}2^n+\dbinom{n}{2}2^{n-2}\cdot{}3+\dbinom{n}{4}2^{n-k}\cdot{}3^2 + \cdots{}.$$Show that $2x-1,2x,2x+1$ form the sides of a triangle whose area and inradius are also integers.

Proof: Note that by $s=3x$ by herons, we want to show that $3x(x-1)(x+1)(x)$ a square. Or show $x^2-1/3 $ is a square. Or show, $x$ is a solution of  $x^2-1=3y^2.$

But note that $x = \frac{(2+ \sqrt3)^{n}+ (2- \sqrt3)^{n}}{2}$ which is a solution of this pell's equation.. 

[2021 P1]Suppose $r \ge 2$ is an integer, and let $m_1, n_1, m_2, n_2, \dots, m_r, n_r$ be $2r$ integers such that$$\left|m_in_j-m_jn_i\right|=1$$for any two integers $i$ and $j$ satisfying $1 \le i<j \le r$. Determine the maximum possible value of $r$.

Aa this problem was the problem I solved in INMO.  This wasn't the solution I wrote in INMO, more complex but the following works.

Proof: We use parity. Note that by parity. We cant have two coordinates $n_i, n_j$ such that both of them are even. Similarly, for $m_i, m_j.$

Also, we can't have both $m_in_j, m_jn_i$ to be odd or else the difference is even. Moreover, $m_i, n_i$ both cant be even 9i.e the even numbers cant have the same index).

If $r\ge 4,$ then the numbers be $m_1,m_2,\dots, m_r$ and $n_1,n_2,\dots n_r.$ WLOG let $m_1,n_2$ be the even numbers. Then just consider $(m_3,m_4,n_3,n_4).$

So $r\le 3$ and $r=3$ construction is just doable.

Summary of 2021:

2021 was definitely a roller coaster for me. I made a lot of new friends, watched a few kdramas and I am still stanning BTS. I think most memorable is probably my EGMO TSTs which sucked. Thank you God.

Sunaina 💜