Skip to main content

Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :)

Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end!


Problems discussed in the blog post

Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic.

Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

Problem 3: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$.

Problem 4: Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Problem 5: In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.



STEMS Mathematics 2023 Category B P3

Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic.

Proof: Define $G$ as midpoint of $LD$, $J$ as midpoint of $FD$, $K$ as midpoint of $ED$. Note that $G-J-K$ collinear.

Claim: $BJKC$ cyclic

Proof: Note that$$\angle JKD=\angle FED$$Also $BJ \perp FD, CK\perp ED$. So$$\angle JKC=90+\angle FED=90+\angle FDB=90+90-\angle JBD\implies BJKC \text{ cyclic }.$$

Note that $(L,D;B,C)=-1$. So $GB\cdot GD=GD^2$.


Claim: $PXJK$ cyclic

Proof: Invert $PBC$ about the incirlce, we get $B\rightarrow J, C\rightarrow K$.


Claim: $G-X-P$ collinear

Proof: Radical axis on $(DFE),(BJKC),(XPJK)$


So $GD^2=GB\cdot GC=GJ\cdot GK=GX\cdot GP$.

So $GD$ is tangent $(XPD)$. So $\angle XPD=\angle XDL$.


To show $XHDL$ cyclic, it is enough to show $LH$ is tangent to $(XPH)$.


Note that $LP=LH$ as $EF$ is perpendicular bisector of $PH$.


Now, we have the reduced problem.

Problem: Given $ABC$ a triangle, define $E=AA\cap BC$. Define $H$ as the orthocenter and $D'=AH\cap (ABC)$. Define $X=(AHE)\cap (ABC)$. Prove that $EH$ is tangent to $(HXD')$.

Proof:$$\angle EHX=\angle EAX=\angle ACX=\angle AD'X=\angle HD'X.$$So $EH$ is tangent to $(HXD')$.

We are done~~

Brazilian Math Olympiad 2007, Problem 5

Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

Proof: Define $QO\cap AB=X,PO\cap BC=Y$. Note that by ceva-menalaus, we have $(P,X;A,B)=-1=(B,C;F,E)$. Since $\angle POX=90\implies \angle AOX=\angle XOB$. By angle bisector harmonic lemma. Similarly, we have $\angle BOF=\angle COF$. Done!


USA TSTST 2016 Problem 2, by Evan Chen

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$.

Proof: Note that $Q$ is the humpty point. So $AG,HQ,BC$ concur at say $X$. Note that by brocards $N-H-G$ collinear and $HG\perp AG$. So $O-M-P||N-H-G$. Also note that $Q\in (GNX)$.

Claim: $PGMA$ cyclic

Proof: Since $M$ is centre of $(GMH)$, we get$$\angle GMA=2\angle GHA=2\angle PGA=2(90-\angle GPM)=180-\angle GPA.$$

Claim: $MP$ diameter of $AGMP$

Proof:$$\angle MAG=\angle MGA=90-\angle GMP= 90-\angle PAG.$$

Define $Y=(APGM)\cap (BMC)$. By radicals on $(MBC),(APGM),(ABCG)$ we get $BC,AG,MY$ concurring at $X$.

Define $E,F$ as feet of the altitude on $AC,AB$. Radicals on $(AH),(FEBC),(ABC)\implies E-F-X$ Note that $MYEF$ cyclic as$$XY\cdot XM=XB\cdot XC=XE\cdot XF.$$

Claim: $P-Y-N$ collinear

Proof: As $MN$ is the diameter of $N_9$ we get$$90=\angle MEN=\angle MYN.$$Since $\angle MYP=90\implies P-Y-N.$

So $Y\in (XGQN)$. Hence $Y$ is the desired concurrency.

Iranian TST 2020, second exam day 2, problem 4

Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proof: Define $X=AD\cap BC$. 

Claim: $N$ is the centre of $\omega$. 

Proof: Define $N'$ such that $N'I\perp AI$ and $N'\in AC$. Note that $$\angle N'CI=\angle BCI=\angle N'IC\implies N'\text{ is the center of }\omega\implies N'=N.$$

Define $P=BC\cap \omega$. Define $P'$ as the reflection of $P$ wrt $N$. Then note that $$\angle QP'C=90\implies P'\in \omega\implies QP'||BC.$$

Claim: $P'\in AC$

Proof: $$\angle AP'Q=\angle QPD=\angle 90-\angle DPX=\angle AXB\implies P'\in AD.$$

So by homothety, we get $X-N-M$ collinear. 


2022 IZHO P3

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

Proof: We begin with the claim. 

Claim: $AMNK$ cyclic

Proof: Let $\angle MAB=\theta\implies \angle ABM=180-2\theta.$ Let $$\angle MBC=\alpha \implies \angle ABC=180-2\theta+\alpha$$ $$\implies \angle MCD=\angle BCD-\angle MCB=2\theta-\alpha-\alpha=2\theta-2\alpha.$$ $$\angle MDC=90-\theta+\alpha\implies \angle ADM=\angle ADC-\angle MDC$$ $$=180-2\theta+\alpha-(90-\theta+\alpha)=90-\theta\implies \angle DNK=\theta\text{ as } DM\perp NK.$$

So $$\angle MNK=\angle DNK-\angle DNC=\angle DNK-(180-\angle NCD-\angle NDC)$$ $$=\angle DNK-(180-2\theta+2\alpha-180+2\theta-\alpha)=\theta-\alpha.$$

Moreover, $$\angle MAN=\angle BCD-\angle MAB=2\theta-\alpha-\theta=\theta-\alpha.$$

So $$\angle MAN=\angle MNK\implies AMNK\text{ is cyclic }.$$

Claim: $MK||AB||BC$

Proof: $$\angle MKD=\angle MND=\angle DNC=\alpha.$$ And $$\angle KDC=\angle ADC-\angle NDK=180-2\theta+\alpha-(180-2\theta)=\alpha.$$

So $$\angle MKD=\angle KDC=\alpha\implies MK||AB||BC.$$

Define $O$ as the circumcenter of $(AMD)$. 

Claim: $O-L-M-K$ collinear

Proof: $$\angle NMO=\angle AMO-\angle AMN=\frac{180-\angle AOM}{2}-\angle AMN$$ $$=90-\angle ADM-\angle AMN$$ $$=\angle DNK-\angle ANK=\angle DNK-(\angle DNM-\angle MNK)=2\angle MNK.$$

Note that $AB=BM=MC=MD$. Note that $OC$ is the perpendicular bisector of $MOD$.

Claim: $CKNO$ cyclic

Proof: $$\angle KOC=\angle MAD=\angle MAN=\angle MKN=\angle MNK.$$

Infact, $MO=MC=CD$ and $NOCK$ is isosceles trapezoid. 

Define $(AMD)\cap (CNKO)=P,Q$. We use radical axis on $(CKNOP),(AKDO),(AMDP)$. We get $PQ,OK,AD$ concurring at $L$.

We use radicals on $(ANMK),(AMD),(NKC)$ we get $NK,PQ,AM$ concurring at say $X$. 

Claim: $M$ is incenter of $AKD$

Proof: $$\angle AKM=\angle DNM=\angle MKD$$ and $MN=MK$. 

But note that $AMDP$ is appollinius circle i.e define $M'$ as antipode of $M$ in $(AMD)$, we have $(M',M;L,K)=-1$. So $\angle MPK=\angle MPQ$.

To finish, note that $$\angle CPM=\angle CPK+\angle MPK$$ $$=\angle CNK+\angle MPK$$ $$=\angle MKN+\angle MPK$$ $$=\angle MAD+\angle MPK$$ $$=\angle MPD+\angle MPK$$ $$=\angle MPD+\angle MPQ=\angle DPL.$$

Some random problems I did

Problem 1: Let $ABCD$ a convex quadrilateral. Suppose that the circumference with center $B$ and radius $BC$ is tangent to $AD$ in $F$ and the circumference with center $A$ and radius $AD$ is tangent to $BC$ in $E$. Prove that $DE$ and $CF$ are perpendicular.

Proof: Note that $BF\perp AF, AE\perp BE\implies AFEB \text{ is cyclic}$.

Note that$$\angle FCE=\angle CFB=90-\frac{\angle FBE}{2}=90-\frac{\angle DAE}{2}=\angle FDE\implies DCEF\text{ is cyclic}.$$To finish,$$\angle DEC=180-(\angle DEA+\angle AFD)=180-(\angle ADE+90)=90-\angle ADE=90-\angle FCE\implies DE\perp FC.$$


Problem 2: Consider an acutangle triangle $ABC$ with circumcenter $O$. A circumference that passes through $B$ and $O$ intersect sides $BC$ and $AB$ in points $P$ and $Q$. Prove that the orthocenter of triangle $OPQ$ is on $AC$.

Proof: Define $R$ as the $(AQO)\cap AC.$ We claim that $R$ is the orthocenter of $OPQ$.

For $RO\perp PQ$, note that $\angle OQP=90-A$ and $A=\angle QAR=180-\angle QOR$. So $RO\perp PQ$. For $RP\perp QO$, note that $\angle ORP=\angle OCP=90-A$ and$$\angle QPR=\angle QPO+\angle RPQ=\angle OBA+\angle OCA=90-C+90-B\implies RP\perp QO.$$

Random configurations pics




Diagram creds to mogmog8

We use a lot of stuffs in USATST 2013 P2. Firtly, we use prism lemma, which states like, if two lines $l_1$ and $l_2$ meet at say $X$ and $A_1,B_1,C_1$ are points on $l_1$ and $A_2,B_2,C_2$ are points on $l_2$. We have $A_1A_2,B_1B_2,C_1C_2$ concur iff $(A_1,B_1;C_1,X)=(A_2,B_2;C_2,X)$. Concurrency and prism lemma goes well with each other.


We also use pole and polars/brokards.  And we use radicals too, Pretty good problem.

I even tried desargues, which led to thinking of cross ratios. 



Nine point circle configuration is pretty cute but I tend to forget it. I think most important is knowing the circle exists and the radius is $R/2$. 



 


Comments

Post a Comment

Popular posts from this blog

Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral. Solution:  Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral....
3 comments
Read more

Just spam combo problems cause why not

This post is mainly for Rohan Bhaiya. He gave me/EGMO contestants a lot and lots of problems. Here are solutions to a very few of them.  To Rohan Bhaiya: I just wrote the sketch/proofs here cause why not :P. I did a few more extra problems so yeah.  I sort of sorted the problems into different sub-areas, but it's just better to try all of them! I did try some more combo problems outside this but I tried them in my tablet and worked there itself. So latexing was tough. Algorithms  "Just find the algorithm" they said and they died.  References:  Algorithms Pset by Abhay Bestrapalli Algorithms by Cody Johnson Problem1: Suppose the positive integer $n$ is odd. First Al writes the numbers $1, 2,\dots, 2n$ on the blackboard. Then he picks any two numbers $a, b$ erases them, and writes, instead, $|a - b|$. Prove that an odd number will remain at the end.  Proof: Well, we go $\mod 2$. Note that $$|a-b|\equiv a+b\mod 2\implies \text{ the final number is }1+2+\dots ...
2 comments
Read more

IMO 2023 P2

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang...
1 comment
Read more

My experiences at EGMO, IMOTC and PROMYS experience

Yes, I know. This post should have been posted like 2 months ago. Okay okay, sorry. But yeah, I was just waiting for everything to be over and I was lazy. ( sorry ) You know, the transitioning period from high school to college is very weird. I will join CMI( Chennai Mathematical  Institue) for bsc maths and cs degree. And I am very scared. Like very very scared. No, not about making new friends and all. I don't care about that part because I know a decent amount of CMI people already.  What I am scared of is whether I will be able to handle the coursework and get good grades T_T Anyways, here's my EGMO PDC, EGMO, IMOTC and PROMYS experience. Yes, a lot of stuff. My EGMO experience is a lot and I wrote a lot of details, IMOTC and PROMYS is just a few paras. Oh to those, who don't know me or are reading for the first time. I am Sunaina Pati. I was IND2 at EGMO 2023 which was held in Slovenia. I was also invited to the IMOTC or International Mathematical Olympiad Training Cam...
11 comments
Read more

Problems with meeting people!

Yeah, I did some problems and here are a few of them! I hope you guys try them! Putnam, 2018 B3 Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n - 1$, and $n-2$ divides $2^n - 2$. Proof We have $$n|2^n\implies n=2^a\implies 2^a-1|2^n-1\implies a|n\implies a=2^b$$ $$\implies 2^{2^b}-2|2^{2^a}-2\implies 2^b-1|2^a-1\implies b|a\implies b=2^c.$$ Then simply bounding. USAMO 1987 Determine all solutions in non-zero integers $a$ and $b$ of the equation $$(a^2+b)(a+b^2) = (a-b)^3.$$ Proof We get $$ 2b^2+(a^2-3a)b+(a+3a^2)=0\implies b = \frac{3a-a^2\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}$$ $$\implies a^4-6a^3-15a^2-8a=a(a-8)(a+1)^2\text{ a perfect square}$$ $$\implies a(a-8)=k^2\implies a^2-8a-k^2=0\implies \implies a=\frac{8\pm\sqrt{64+4k^2}}{2}=4\pm\sqrt{16+k^2}. $$ $$ 16+k^2=m^2\implies (m-k)(m+k)=16.$$ Now just bash. USAMO 1988 Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1...
3 comments
Read more

Solving Random ISLs And Sharygin Solutions! And INMO happened!!

Some of the ISLs I did before INMO :P  [2005 G3]:  Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$ Solution: Note that $$\Delta LDK \sim \Delta XBK$$ and $$\Delta ADY\sim \Delta XCY.$$ So we have $$\frac{BK}{DY}=\frac{XK}{LY}$$ and $$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$$ Hence $$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY} $$ $$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$$ $$\implies \Delta CBK\sim \Delta LDK$$ And we are done. We get that $$\angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2$$ Motivation: I took a hint on this. I had other angles but I did...
4 comments
Read more

Let's complex bash Part 1

I have to learn complex bash. And almost everyone knows that I am notes taking girl so thought why not make a post on complex bash ( so that I don't get emotionally demotivated lol).😇 There wasn't any need for learning complex bash, but it was in my dream checklist i.e " To learn a bash." And since I am not loaded with exams, I think it's high time to learn Bash and new topics.  Also if anyone from the "anti-bash" community is reading, sorry in advance and R.I.P.  Notes:- 1. Complex numbers are of the form $z=a+ib,$ where $a$ and $b$ are real numbers and $i^2=-1.$ 2. In polar form, $z=r(\cos \theta+~~i\sin\theta)=~~re^{i\theta},$ where $r=~~|z|=~~\sqrt{a^2+b^2},$ which is called the magnitude. 3. Here we used euler's formula i.e $\cos \theta+~~i\sin\theta=~~e^{i\theta}.$ 4. The $\theta $ is called the argument of $z,$ denored $\arg z.$ ( $\theta$ can be considered in $\mod 360$ and it is  measured anti-clockwise). 5. The complex conjugate of $z$ is ...
4 comments
Read more

How to prepare for RMO?

"Let's wait for this exam to get over".. *Proceeds to wait for 2 whole fricking years!  I always wanted to write a book recommendation list, because I have been asked so many times! But then I was always like "Let's wait for this exam to get over" and so on. Why? You see it's pretty embarrassing to write a "How to prepare for RMO/INMO" post and then proceed to "fail" i.e not qualifying.  Okay okay, you might be thinking, "Sunaina you qualified like in 10th grade itself, you will obviously qualify in 11th and 12th grade." No. It's not that easy. Plus you are talking to a very underconfident girl. I have always underestimated myself. And I think that's the worst thing one can do itself. Am I confident about myself now? Definitely not but I am learning not to self-depreciate myself little by little. Okay, I shall write more about it in the next post describing my experience in 3 different camps and 1 program.  So, I got...
7 comments
Read more

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$.  Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
7 comments
Read more