Let's do problems only

Welcome back to this blog! 

My blog completed it's 1-year :D i.e 29 Nov. I am so happy that this blog grew so much and it didn't die! It also crossed 10k views! Thanks a lot!

Enjoy the problems. It's more of a miscellaneous problem set with the level being INMO or less. So here are $20$ INMO level problems.

Problems: 

Problem 1: [IMO 2009/P1] Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ ( k\ge 2)$ be distinct integers in the set $ { 1,2,\ldots,n}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

Problem 2[ USEMO 2021 P4]: Let $ABC$ be a triangle with circumcircle $\omega$, and let $X$ be the reflection of $A$ in $B$. Line $CX$ meets $\omega$ again at $D$. Lines $BD$ and $AC$ meet at $E$, and lines $AD$ and $BC$ meet at $F$. Let $M$ and $N$ denote the midpoints of $AB$ and $ AC$.

Can line $EF$ share a point with the circumcircle of triangle $AMN?$

Problem 3 [ 2006 n5]: Prove that $\frac{x^{7}-1}{x-1}= y^{5}-1$ has no integer solutions.

Problem 4[ELMO 2016 P4]: Big Bird has a polynomial $P$ with integer coefficients such that $n$ divides $P(2^n)$ for every positive integer $n$. Prove that Big Bird's polynomial must be the zero polynomial.

Problem 5[2006 g3]: Let $ ABCDE$ be a convex pentagon such that

$$\angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.$$

The diagonals $BD$ and $CE$ meet at $P$.  Prove that the line $AP$ bisects the side $CD$.

Problem 6[ 2019 g3]: In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_2$, and $\angle PP_2C=\angle BAC$. Similarly, let $Q_2$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_2$, and $\angle CQ_2Q=\angle CBA$.

Problem 7[Japan 1996/2]: Let $m$ and $n$ be odd positive integers with $\gcd(m,n)=1$.

Evaluate $\gcd(5^m+7^m, 5^n+7^n). $

Problem 8 [INMO 1992]: Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect square.

Problem 9 [ Iran MO]:Let $a,b$ be two positive integers and $a>b$.We know that $\gcd(a-b,ab+1)=1$ and $\gcd(a+b,ab-1)=1$. Prove that $(a-b)^2+(ab+1)^2$ is not a perfect square.

Problem 10 [Iran 2005]: Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.

Problem 11 [CGMO 2012]: As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.

Problem 12[GMO 2017 P1]: (1) Find all positive integer $n$ such that for any odd integer $a$, we have $4\mid a^n-1$

(2) Find all positive integer $n$ such that for any odd integer $a$, we have $2^{2017}\mid a^n-1$

Problem 13[ELMO 2009 P1]:Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Problem 14[2013 Iran ]: Find all pairs $(a,b)$ of positive integers for which $\gcd(a,b)=1$, and $\frac{a}{b}=\overline{b.a}$. (For example, if $a=92$ and $b=13$, then $b/a=13.92$ )

Problem 15[ 2012 CGMO P1]: Let $ a_1, a_2,\ldots, a_n$ be non-negative real numbers. Prove that

$\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+$ $\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)} \le 1.$

Problem 16[CGMO 2013 P2]: As shown in the figure below, $ABCD$ is a trapezoid, $AB \parallel CD$. The sides $DA$, $AB$, $BC$ are tangent to $\odot O_1$ and $AB$ touches $\odot O_1$ at $P$. The sides $BC$, $CD$, $DA$ are tangent to $\odot O_2$, and $CD$ touches $\odot O_2$ at $Q$. Prove that the lines $AC$, $BD$, $PQ$ meet at the same point.

Problem 17[ USATST 2021 P1, simplified]: Prove all primes  $s \ge 4$  there doesn't exist positive integers $a$, $b$, $c$, $d$ such that $s = a+b+c+d$ and $s$ divides $abc+abd+acd+bcd$.

Problem 18[ CMO 2012]: Let $x,y$ and $z$ be positive real numbers. Show that $x^2+xy^2+xyz^2\ge 4xyz-4$.

Problem 19[ Canada 1984]: Let $ABC$ be a triangle with $\angle{BAC} = 40^{\circ}$ and $\angle{ABC}=60^{\circ}$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\angle{CBD} = 40^{\circ}$ and $\angle{BCE} = 70^{\circ}$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF$ is perpendicular to the line $BC$.

Problem 20[Iran 2017]: a) Prove that there doesn't exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: gcd(a_i+j,a_j+i)=1$

b) Let $p$ be an odd prime number. Prove that there exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: p \not | gcd(a_i+j,a_j+i)$

Problem 1: [IMO 2009/P1] Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ ( k\ge 2)$ be distinct integers in the set $ { 1,2,\ldots,n}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

Proof: Assume not, then  $a_ka_1 \equiv a_k \pmod n.$

The given conditions can be written as,

$a_i\cdot a_{i+1}\equiv a_i\mod n.$

So $$a_1\cdot a_2\equiv a_1\mod n$$

$$a_2\cdot a_3\equiv a_2\mod n$$

$$\vdots$$

$$a_{k-1}\cdot a_k\equiv a_{k-1}\mod n$$

So we have $$a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1$$

Now, we again reduce, so we get $$ a_1\cdot\dots\cdot a_{k-2}\cdot a_{k-1}\cdot a_k\equiv a_1\cdot\dots\cdot a_{k - 2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$$

So we get $a_1\equiv a_k\mod n.$ Which is not possible, since $ a_1,a_k\in{1,\dots n}$ and are distinct.

Problem 2[ USEMO 2021 P4]: Let $ABC$ be a triangle with circumcircle $\omega$, and let $X$ be the reflection of $A$ in $B$. Line $CX$ meets $\omega$ again at $D$. Lines $BD$ and $AC$ meet at $E$, and lines $AD$ and $BC$ meet at $F$. Let $M$ and $N$ denote the midpoints of $AB$ and $ AC$.

Can line $EF$ share a point with the circumcircle of triangle $AMN?$

Proof: Notice that $BN\parallel CX$. Thus,

$$\measuredangle EBN = \measuredangle EDC = \measuredangle BDC = \measuredangle BAC \implies EA\cdot EN = EB^2$$Moreover, if $\odot(AMN)$ meet $AD$ again at $P$, then $BP\parallel DX$. Thus,

$$\measuredangle FPB = \measuredangle FDC = \measuredangle ADC = \measuredangle ABC \implies FA\cdot FP = FB^2.$$Both equations combined imply that $EF$ is the radical axis of $\odot(AMN)$ and $\odot(B,0)$. As $B$ lies outside this circle, $EF$ can never intersect the $\odot(AMN)$.

Problem 3 [ 2006 n5]: Prove that $\frac{x^{7}-1}{x-1}= y^{5}-1$ has no integer solutions.

Proof: We use the following lemma stated in Evan's handout.

Let $\Phi_n(x)$ denote the $n$th cyclotomic polynomial. If $\Phi_n(x) \equiv 0 \pmod{p}$, then either $p \equiv 1 \pmod{n}$ or $n \equiv 0 \pmod{p}$.

So if prime $p|\frac{x^{7}-1}{x-1}$ then $p=7$ or $p \equiv 1 \pmod{7}$. Hence $y\equiv 0,1 \mod 7\implies y^4+y^3+y^2+y+1 \equiv 5,3 \pmod{7}$

Problem 4[ELMO 2016 P4]: Big Bird has a polynomial $P$ with integer coefficients such that $n$ divides $P(2^n)$ for every positive integer $n$. Prove that Big Bird's polynomial must be the zero polynomial.

Proof: Let $p$ and $q$ be two odd primes.

Note that, by FLT $2^{pq}\equiv \left(2^q\right)^p\equiv 2^q\pmod{p}$

So$$P(2^{pq})\equiv P(2^q)\equiv 0\mod p.$$So for infinite many primes $p,$ $p|P(2^{q}).$ Not possible.

So for infinitely many $x,$ we have $P(x)=0\implies P$ is the zero polynomial.

Problem 5[2006 g3]: Let $ ABCDE$ be a convex pentagon such that

$$\angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.$$

The diagonals $BD$ and $CE$ meet at $P$.  Prove that the line $AP$ bisects the side $CD$.

Proof: Note $A$ is the unique spiral centre taking $BC\rightarrow CD\rightarrow DE.$ So $$\Delta ABD\sim \Delta ACE \implies \angle ABP=\angle ABD=\angle ACE=\angle ACP.$$ So $ABCP$ is cyclic. Similarly, we get that $APCD$ is cyclic. We als have $$\Delta BCD\sim CDE \implies \angle PBC=\angle DBC=\angle ECD=\angle PDC.$$ So $(APDC) $ is tangent to $CD.$ Similarly, we get that $(APED)$ is tangent to $CD.$

Since $AP$ is the radical axis of  $(APDC), (APED)$  and $CD$ common tangent. So $AP\cap CD=M$ is the midpoint of $CD$ by POP ( $MC^2=MP\cdot MA=MD^2$)

Problem 6[ 2019 g3]: In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_2$, and $\angle PP_2C=\angle BAC$. Similarly, let $Q_2$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_2$, and $\angle CQ_2Q=\angle CBA$.

Proof: Let $PQ\cap CA=P_1,~~PQ\cap CB=Q_1,~~B_1B\cap (ABC)=B_2,~~AA_1\cap (ABC)=A_2.$ Then $$\angle CP_2P=\angle CP_1P\implies CP_2P_1P \text{~~is cyclic}.$$

And $$\angle CQ_2Q=\angle CQ_1Q\implies CQ_2Q_1Q \text{~~is cyclic}.$$

By converse of reim, $PQB_2A_2$ is cyclic.

Note that $$\angle CP_2P=\angle CP_1P=\angle CAB=\angle CB_2B\implies CP_2B_2B_1 \text{ is cyclic}.$$ 

So $$\angle B_1B_2P_2=180-\angle P_2CB_1=180-\angle B_1PP_1=\angle B_1PQ\implies P_2B_2PQ \text{ is cyclic}.$$

Similarly we get $PQA_2Q_2$ cyclic.

So we get that $P_2B_2PQA_2Q_2$ is cyclic. Hence $P_2Q_2PQ$ cyclic.

Problem 7[Japan 1996/2]:

Let $m$ and $n$ be odd positive integers with $\gcd(m,n)=1$.

Evaluate $\gcd(5^m+7^m, 5^n+7^n). $

Proof:

Let $d=\gcd(5^m+7^m, 5^n+7^n)$ WLOG m>n.

• $d|5^m+7^m-5^{m-n}\cdot 5^n-5^{m-n}\cdot 7^n\implies d|7^{m-n}-5{m-n}.$
•  Euclid algorithm goes on.. 
• So $d|5^{\gcd(m,n)}+7^{\gcd(m,n)}\implies d|12.$
•  Clearly $12|5^m+7^m\implies \gcd=12.$

Problem 8 [INMO 1992]: Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect square.

Solution: Let $2^m+3^n=x^2 .$ By $$\pmod 3\implies m \text{ is even}.$$ So $$3^n=(x-2^k)(x+2^k)\implies x-2^k=1,x+2^k=3^n, \text{ since} \gcd( x-2^k, x+2^k)|2^k\implies \gcd=( x-2^k, x+2^k)=1.$$

Case 1: When $k$ is odd.

Then we get $$x=1+2^k\implies 3|x. \text{ But} 3| x+2^k.$$ So no solution.

Case 2: When $k$ is even.

Note that $x\equiv 1 \pmod 4.$ Taking $\pmod 4$ on $$x+2^k=3^n\implies n\text{ is even}.$$

Hence, let $n=2n'.$ We get $$1+2^k=3^n-2^k\implies 3^n-1=2^{k+1}\implies (3^{n'}-1)(3^{n'}+1)=2^{k+1}$$ $$\implies n'=1\implies \boxed{n=2, k=4}$$

Problem 9 [ Iran MO]:Let $a,b$ be two positive integers and $a>b$.We know that $\gcd(a-b,ab+1)=1$ and $\gcd(a+b,ab-1)=1$. Prove that $(a-b)^2+(ab+1)^2$ is not a perfect square.

Proof: Note that $$(a-b)^2+(ab+1)^2=a^2+b^2-2ab+{ab}^2+1+2ab=(a^2+1)(b^2+1)= (a+b)^2+(ab-1)^2$$

Note that if $$\gcd(a^2+1,b^2+1)=1\implies a^2+1, b^2+1 \text{ are perfect squares}.$$

Which is not possible.

So let $$d=\gcd(a^2+1,b^2+1)\implies d|a^2-b^2\implies d|(a+b)(a-b).$$Also note that $$ 1=\gcd(a-b,ab+1)=gcd(a^2+b^2+2).$$

Now, let $$r=\gcd(a-b,a+b), r|d \implies r|2a,2b.$$ Now,$2\nmid r$ else $2|\gcd(a-b,a^2+b^2+2).$

So $$r|a,r|b, r|a^2+1, r|b^2+1\implies r=1.$$

So $d|a-b \text { or } d|a+b.$

Case 1: $d|a-b\implies d|(a^2+1))(b^2+1)-(a-b)^2$

$$ \implies d|(ab+1)^2\implies d=1$$

Case 2: $d|a+b\implies d|(a^2+1))(b^2+1)-(a+b)^2$

$$ \implies d|(ab-1)^2\implies d=1$$

Problem 10 [Iran 2005]: Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.

Proof: Note that $$n\le p-1, p\le n^3-1\implies p|n^2+n+1 .$$ Let $nk=p-1$ $$ \implies kn+1|n^2+n+1\implies kn+1\le n(n+1)+1\implies k\le n+1$$

Also $$ nk+1|n^2k+nk+k \implies nk+1|nk+k-n$$ $$\implies nk+1\le nk+k-n \implies n+1\le k$$

$$ \implies n+1=k\implies p=n^2+n+1\implies 4p-3=4n^2+4n+1=(2n+1)^2$$

Problem 11 [CGMO 2012]: As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.

Proof: Note that by Fact 5, $A-I-O.$ And then note that $\Delta DIO=\Delta EIO$

Problem 12[GMO 2017 P1]: (1) Find all positive integer $n$ such that for any odd integer $a$, we have $4\mid a^n-1$

(2) Find all positive integer $n$ such that for any odd integer $a$, we have $2^{2017}\mid a^n-1$

Proof: 

(1) Any $n$ even works

(2) We use LTE as $2|a-1.$ So $$ v_2(a^n-1)=v_2(a-1)+v_2(a+1)+v_2(n)-1.$$ We know that $$v_2(a-1)+v_2(a+1)\ge 3.$$ So $\min(v_2(n))\le 2015.$

Clearly, taking $a=5,$ we get by LTE $\min(v_2(n))= 2015.$

So all numbers such that $v_2(n)\ge 2015$ suffice.

Problem 13[ELMO 2009 P1]:Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Proof: Let $$a^2-bc=x^2\implies (a-x)(a+x)=bc.$$ 

By Four-Numbers theorem, there exist positive integers $p,q,r,s$ with $$a-x=pq, a+x=sr, b=ps \text{ and }c=rq.$$ Hence

$$2a+b+c=(a-x)+(a+x)+b+c=pq+sr+ps+rq=(p+s)(q+r)$$

which is composite.

Problem 14[2013 Iran ]: Find all pairs $(a,b)$ of positive integers for which $\gcd(a,b)=1$, and $\frac{a}{b}=\overline{b.a}$. (For example, if $a=92$ and $b=13$, then $b/a=13.92$ )

Proof: Note that$$b+1>\frac{a}{b}>b\implies b^2+b>a>b^2. .$$And since $\frac{a}{b}$ is not recurring, we get $b=2^x\cdot 5^y.$

Now, we have$$\frac{a}{b}=\frac{10b+a}{10}\implies 10a=10b^2+ab\implies a(10-b)=10b^2\implies 0<b<10$$Since $b=2^x\cdot 5^y,$ hence possible values of$$b=2,4,5,8$$and range of $a$ are$$ b=2, a\in ( 2,6)$$$$ b=4, a\in (16,20)$$$$b=5, a\in (25,30), $$$$b=8, a\in (64,72)$$Bashing the cases, we get $\boxed{b=2, a=5}$ the only satisfying case.

Problem 15[ 2012 CGMO P1]: Let $ a_1, a_2,\ldots, a_n$ be non-negative real numbers. Prove that

$\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+$ $\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)} \le 1.$

Proof: I was in a gsolve, where atul just solved it! We eliminate $a_n$ and work out.. 

The identity which is helpful is 

$$1-\frac{1}{1+a_1}=\frac{a_1}{1+a_1}$$

And we get $$1-\Big(\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\dots

+\frac{ a_1 a_2... a_{n-1}}{(1+ a_1)(1+ a_2)...(1+ a_n)}\Big)=\frac{a_1a_2...a_n}{(1+a_1)(1+a_2)...(1+a_n)}\ge0 $$

Equality holds when $$\frac{a_1a_2...a_n}{(1+a_1)(1+a_2)...(1+a_n)}=0\implies a_1,a_2,\dots a_n=0 $$

Problem 16[CGMO 2013 P2]: As shown in the figure below, $ABCD$ is a trapezoid, $AB \parallel CD$. The sides $DA$, $AB$, $BC$ are tangent to $\odot O_1$ and $AB$ touches $\odot O_1$ at $P$. The sides $BC$, $CD$, $DA$ are tangent to $\odot O_2$, and $CD$ touches $\odot O_2$ at $Q$. Prove that the lines $AC$, $BD$, $PQ$ meet at the same point.

Proof: Let $AD\cap BC=X, AC\cap BD=Y.$ Note that $AYB$ is similar $CYD.$ So enough to show that $$\frac{AP}{PB}=\frac{CQ}{QD}.$$

Now note that $\odot O_1$  is the excircle of $XAB$ and $\odot O_2$ is incircle of $XDC.$ As $PB=AL$ where $L$ is the $X$ incircle touch point in $XAB$ we get $$\frac{AP}{PB}=\frac{BN}{AN}=\frac{CQ}{QD}.$$

Problem 17[ USATST 2021 P1, simplified]: Prove all primes  $s \ge 4$  there doesn't exist positive integers $a$, $b$, $c$, $d$ such that $s = a+b+c+d$ and $s$ divides $abc+abd+acd+bcd$.

Proof: Suppose not.

$$s|abc+abd+acd+bcd \implies s|abc-(a+b+c)(ab+ac+bc)=a^2b+ab^2+ac^2+a^2c+b^2c+bc^2+2abc$$

So $$ s|(a+b)(b+c)(c+a). \text{ But } s> a+b, s>b+c, s>c+a$$ Not possible as $s$ is prime.

Problem 18[ CMO 2012]: Let $x,y$ and $z$ be positive real numbers. Show that $x^2+xy^2+xyz^2\ge 4xyz-4$.

They killed it. She killed it. I just did.

Proof: Note that $$4+ x^2+xy^2+xyz^2= 4+x^2 + \frac{xy^2}{2} + \frac{xy^2}{2} + \frac{xyz^2}{4} + \frac{xyz^2}{4} + \frac{xyz^2}{4} + \frac{xyz^2}{4} .$$

Applying AM-GM, we get, 

LHS $\ge 4xyz.$

Problem 19[ Canada 1984]: Let $ABC$ be a triangle with $\angle{BAC} = 40^{\circ}$ and $\angle{ABC}=60^{\circ}$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\angle{CBD} = 40^{\circ}$ and $\angle{BCE} = 70^{\circ}$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF$ is perpendicular to the line $BC$.

Proof: Let $G$ be a point on $BC$ such that $AG\perp BC\implies \angle BAG=30,~~\angle GAC=30.$

Now note that$$\frac{sin\angle BAG}{sin\angle GAC}\cdot \frac{sin\angle DCE}{sin\angle ECB}\cdot \frac{\sin \angle CBD}{sin\angle DBA}

= \frac{sin\:30}{sin\:10}\cdot \frac{sin\:10}{sin\:70}\cdot \frac{\sin 40}{sin\:20}$$$$=\frac{1}{2}\cdot \frac{2\cdot \:\cos 20\:\sin 20}{sin\:70\cdot sin\:20}=1$$

Hence by converse of trig ceva, we get $A-F-G$ collinear. We are done.

Problem 20[Iran 2017]: a) Prove that there doesn't exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: gcd(a_i+j,a_j+i)=1$

b) Let $p$ be an odd prime number. Prove that there exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: p \not | gcd(a_i+j,a_j+i)$

Proof: Part a) Suppose it does exist. 

We consider parity of $a_1$. And it just works. 

Solved with Gunjan and Swastika.

Case 1: If $a_1$ is even. Then since $\gcd(a_1+2,a_2+1)=1\implies a_2 \text { is even}.$ Similarly, we get $a_4$ is even. Hence $\gcd(a_2+4,a_4+2)\ne 1.$

Case 2: If $a_1$ is odd. Then since $\gcd(a_1+3,a_3+1)=1\implies a_3 \text { is even}.$ Since $\gcd(a_3+2,a_2+3)=1\implies a_2 \text { is even}.$Similarly, we get $a_4$ is even. Hence $\gcd(a_2+4,a_4+2)\ne 1.$

Part b) We consider the index $\mod p.$ 

For $i\not\equiv 0\pmod p, a_i= i\cdot p.$ For $p|i, a_i=1.$ Clearly this works.

Yeah! That's it for this blog post! It took a lot of time.. I hope you enjoyed it.

Sunaina💜