Cheenta Statistics Competition for IIT JAM MS ISI MStat CUET Examinations

Demo Post to Try out sTUFF

How to Attend IIT JAM Statistics Exam - [A Data Analysis]

This year Cheenta Statistics Department has done a survey on the scores in each of the sections along with the total score in IIT JAM MS. Here is the secret for you!

We have normalized the score to understand in terms of percentage.

There are three questions, we ask

The general performance for the IIT JAM MS 2021 exam.
How did the students answer questions in the exam? (General)
What did the toppers do differently? What mistakes did the low scorers do? (Genera;)

General Overview for IIT JAM MS

Total Score in IITJAM 2021 MS

Normal QQPlot for the data (Umm… it kinda follows normal!)

Summary

Total Score in IITJAM 2021 MS ~ Normal(35.8, 12.7²)

Section A

Normal QQPlot

Summary

% in IITJAM 2021 MS Section A ~ Normal (38.6 15.7²)

Section B

Normal QQ Plot (it doesn’t look like normal!)

Histogram (It looks like it follows exponential distribution)

Summary

% in IITJAM 2021 MS Section B~ Exponential(Mean = 30.18)

Section C

Normal QQPlot

Summary

% in IITJAM 2021 MS Section C~ Normal (35.1, 18.1²)

Question 2:

How did the students in general answer the questions?

What is the percentage of marks the people scored in section A?

Summary

Mean = 53.37
Standard Deviation = 11.11

What is the percentage of marks the people scored in section B?

Summary

Mean = 16.67
Standard Deviation = 12.14

What is the percentage of marks the people scored in section C?

Summary

Mean = 30.11
Standard Deviation = 12.68

Question 3:

What mistakes did the low scorers do? What did the toppers do differently?

Low Scorers

Section A: 46
Section B: 17
Section C: 38
Total Score: 18

Their marks decreased with Section B score.

Their marks increased with Section C score.

Mid Scorers

Section A: 54
Section B: 17
Section C: 29
Total Score: 37

Their marks remains unperturbed by what they did.

Top Scorers

Section A: 57
Section B: 17
Section C: 26
Total Score: 54

Their marks decreased with Section A score.

Their marks increased with Section B score.

Question 4:

What is the conclusion? Which sections do I need to give more importance to? How do I manage my time?

Score in Section A (MCQ) to 50 % of the correct answers
Score in Section C (NAT) to 30 % of the correct answers
Score in Section B (MSQ) to 20% of the correct answers
MSQ will make a difference in the top ranks.

Exercise

Why did the score in MSQ follow an exponential distribution?

Information for You:

There is a distribution called Weibull Distribution, which fits all the cases equally well, and its limiting case is Exponential Distribution.

Do you think marks can be well modelled by Weibull Distribution?

Thanks,

Cheenta Statistics Department

How to Measure the Length of your Earphone from a Pic?| Cheenta Probability Series

This is our 5th post in the Cheenta Probability Series. This article teaches how to mathematically find the length of an earphone wire by its picture.

Let's explore some truths.

A Line is made up of Points.
A Curve is made up of Lines.
Earphones are as Messy.

This article is all about connecting these three truths to a topic in Probability Theory $ \cap $ Geometry, which is not so famous among peers, yet its' usefulness has led one to make an app in Google Play Store, which has only one download (by me). The sarcasm is towards the people, who didn't download. (obviously).

Today is the day for Cauchy Crofton's Formula, which can measure the length of any curve using random straight lines by counting points. It is all encoded in the formula.

$ L(c)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta $

This is the begining of a topic called Integral Geometry.

This formula intuitively speaks the truth

The Length of a Curve is the Expected Number of Times a "Random" Line Intersects it.

Describing a Line (like seriously?).

I mean a "Random" Line.

Thus the Space of all Lines in $ \mathbf{R}^2 $ can be represented in "kinda" polar coordinates S = {${(\theta, p) \mid 0 \leq \theta \leq 2 \pi, p \geq 0},$} where $p, \theta$ are as indicated.

Now, we are now uniformly selecting this $ (\theta, p) $ over S. This is how we select a "Random" Line.

Understanding Crofton's Gibberish

If C is the curve, whose length we are interested to measure. Then, $ L(C)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta $.

$ L(C) $ = The Length of the curve

$ n(\theta, p) $ = The Number of times "Random Line" corresponding to $ (\theta, p) $ intersects C.

Crofton meant the following on a pretty, differentiable, and cute (regular) curve C that:

$ L(C)= \frac{1}{2} \times E_{(p, \theta)} [n(p, \theta)] $

The First Truth

Here, we will find the length of a Line using "Random Lines". We will see, how counting points on the line, can help us find the length of a line.

Here, we choose C = Line.

A Line intersects another Line exactly once. $ n(p, \theta) = 1 $. Hence,

$ L(C)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta = \frac{1}{2} \times \iint dp d\theta = \frac{1}{2} \times \int_{0}^{2 \pi}\left(\int_{0}^{|\cos \theta|(1 / 2)} d p\right) d \theta = \frac{1}{2} \times \int_{0}^{2 \pi} l / 2|\cos \theta| d \theta= l $

Thus, Crofton's Formula holds true for a Line.

The Second Truth

Approximate Curves by Lines.

Refer to Approximation of Curves by Line Segments by Henry Stone.

Just Limit!

Now, take apply the result on these lines. By linearity of expectation and the uniform convergence of the Lines to the Curve, we get the same for the curve.

Suppose that the curve c is parameterized on the interval [a,b]. Define a partition $ P $ to be a collection of points that subdivides the interval, say { $ a = a_0, a_1, a_2, ..., a_n = b$ }.

Define $ L_{r}(c)=\sum_{i=1}^{n} |c\left({a}_{i-1}\right)-c\left({a}_{i}\right)| $

Then $ L(c)= sup_{r} L{r}(c) $

Now, each of the $ |c\left({a}_{i-1}\right)-c\left({a}_{i}\right)|$ are approximated by the Crofton's Formula. Now, taking the limit gives ust eh required result. QED.

Measuring the Length of an Earphone

Create a Mesh of Lines (Coordinate System) - A Fixed Sample representing the Population of the Random Lines. Then count the number of intersections using those mesh of lines only and add it up.

The mathematics of the above argument is as follows.

$ L(c)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta \cong \frac{1}{2} \times n r \times \frac{\pi}{3} $

An example is show below. This is done in the app named Cauchy-Crofton App, which unfortunately is not working now and I have reported.

Future Articles and Work

There are lots of corollaries coming up like Isoperimetric Inequality, Barbier's Result, etc. I will discuss these soon. Till then,

Stay Tuned!

Stay Blessed!

Outstanding Statistics Program with Applications

The Mathematics of How Virus can Grow

Beautiful Mathematics Seminar

We have a beautiful mathematics seminar every week on Thursday at 8 p.m. IST. Today's topic is "The Mathematics of How Virus Can Grow".

Prerequisites

Basic Probability
Probabilistic Versions of Addition Principle and Multiplication Principle
The Basic Idea of Limit

Theory

We will discuss some basic questions. We will develop the theory while seeking answers to these questions only. This kind of pedagogy has always been our approach to mathematics. So, let's start to explore coronavirus by asking some questions.

Question 1.1 Suppose you are trying to find some vaccine of coronavirus. You know that every single second, a single cell divides into two. Some coronavirus cells were kept in a jar. You go out of the room and came back after 1 minute to find out the whole jar is full of the virus. When was the jar half full of virus?

Question 1.2 What would have been the answer, if the virus divides at a faster rate, say a single cell divides into 5 cells in a second?

Question 2.1 Consider some coronavirus cells in a lab jar. A single virus either survives all along or it gives rise to another virus when it dies. Will a colony of the virus ever become extinct?

Question 2.2 Consider a single coronavirus cell. Now, in each second it dies and gives rise to another child virus with probability p < 1. Will the virus colony become extinct?

Probability of Extinction Let $X_n$ is the number of virus cells after $n^{th}$ second. The extinction probability at $n^{th}$ second, $u_{n} = P({X_{n}=0})$ for $n \geq 1.$ What is $u_{\infty}$, the probability of ultimate extinction?

Question 2.3 Consider a bunch of coronavirus cells. Now, in each second a single cell dies and gives rise to another child virus with probability p < 1. Will the virus colony become extinct?

Research Question 3 Consider a bunch of coronavirus cells. Now, in each second a single cell dies or gives rise to two child virus cells with probability p < 1. Will the virus colony become extinct? Does it depend on the values of p? Explore.

Reference

Refer to the following document.

The Mathematics of growth Cornovirus of a Different Flavour

Note: The models described above may or may not truly describe the growth of the corona virus. Not sufficient research has been there on how coronavirus grows.

The Exaggerated Triangle Inequality

Triangle Inequality is an exaggerated version of the Basic Idea of the Euclidean Plane, something we follow every day in our lives -

The shortest distance between two points is the straight-line distance between the two points.

It tells you, if you move from A to B via another point P, then the distance of travel will be larger than traveling along AB.

But, we can play with and nurture this idea to solve interesting puzzles. Let's do some Triangle inequality Problems and Solutions. A single diagram is sufficient for the arguments.

Excited right? Let's explore!

Problem 1:

A, B, C, D are four points. Find a point P such that PA + PB + PC + PD is minimum.

Solution:

ABCD is rotated and translated in such a way ABCD -> CEFG, and A, C, and F lie on a straight line. H -> I.

Now, observe, that HA + HB + HC + HD = AH + HC + CI + IF, which is minimized if H lies on AC and I lies on CF. I lie on CF is the same as H lies on BD. So, it is minimized if H lies on AC and BD both, which is the intersection point of the diagonal.

Problem 2:

A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point N on A such that NL + NM is minimum.

Solution:

LO + OM is minimized if L, O, M lie on a straight line. Therefore, N is the required point.

Problem 3:

A is a line. L and M are two points in the same plane on the same side of the line. Find a point N on A such that NL + NM is minimum.

Solution:

M' is the reflected point of M along line A (JK). Hence, NM' = NM. Hence, LO + OM = LO + OM'. Hence, it is minimized if L, O, M' lie on a straight line. Therefore, N is the required point.

Problem 4:

A is a line. L and M are two points in the same plane on the same side of the line. Find a point P on A such that |PL - PM| is minimum.

Solution:

LP = PM. Just draw the perpendicular bisector of LM, and see its intersection with line A(JK).

Problem 5:

A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point P on A such that |PL - PM| is minimum.

Solution:

Problem 6:

A is a line. L and M are two points in the same plane on the same side of the line. Find a point O on A such that |OL - OM| is maximum.

Solution:

Observe that |LP-PM| $ \leq $ LM. Hence, it's maximum value is AB. This occurs for the position O, when O, L, and M lie on a straight line. O is the intersection of the line LM and JK (Line A).

Problem 7:

A is a line. L and M are two points in the same plane on the opposite side of the line. Find a point O on A such that |OL - OM| is maximum.

Solution:

Observe that |LP-PM|$\leq$ LM. Hence, its maximum value is AB. This occurs for the position O when O, L, and M lie in a straight line. O is the intersection of the line LM and JK (Line A).

Problem 8:

An ant is caught on one corner of a cuboid with sides l, b, and h. It wants to reach the diagonally opposite corner, However, the ant can perform a walk only along with the faces of the cuboid. What is the least amount of distance that the ant needs to walk to reach the other corner?

Solution:

More, similar problems to come. Stay tuned!

Geometric Median |Understand the concept

Geometric Median is an important concept in the intersection of Geometry, Data Analysis and Algorithms. This article explores the concept.

Given a set of numbers on the real line, say $ a_1, a_2, a_3, ..., a_n$, the median of these numbers is the middle number when you arrange these numbers in ascending/descending order. To understand more about the median for a set of numbers, go through this.

Let's ask a different question altogether.

Given a set of numbers on the real line, say $ a_1, a_2, a_3, ..., a_n$, what is the number $ x$ such that $ |x-a_1| + |x-a_2| + ... |x-a_n|$ is minimized?

In other words, what is the point on the real number line, such that the sum of the distances of that point from the given set of points is minimum? In some sense, this point gives an idea about the central behavior of the set of points.

In fact, this turns out to be the median of the set of numbers { $ a_1, a_2, a_3, ..., a_n$}. To understand the proof, go through this.

But, I suggest you try this yourself, in inductive steps. Start with two points, any point in between them is the median. Now, add a point in between those two points, Observe that the new median is the new point added. Continue similarly in this inductive process.

My contentment is to ask questions and seek. Therefore, a lot of questions inpoured.

What happens if we take those n points in 2D space?
What happens if we take those n points in 3D space?
What happens if we take those n points in n-D space?
What happens if we take a different distance idea instead of the normal distance?

Well, the intelligent way is to answer the easiest of all and then generalize the idea - the first question.

Given a set of numbers in the 2D space, say $ a_1, a_2, a_3, ..., a_n$, what is the point $ x$ such that $ |x-a_1| + |x-a_2| + ... |x-a_n|$ is minimized?

Let's take the easiest distance concept of all - The Euclidean Distance $d(x,y) = |x-y|= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} $.

This concept is known as Geometric Median in the literature.

Now, the problem seems difficult even in the case of three points.

Let's explore it for a quadrilateral ( $ n = 4 $ ).

Step 1: Does there exist a better point like that?

Yes, there is! Let's explore.

Observe, I is the point such that AG + FG = BI + EI. You can argue this by continuity. But, here I have explicitly constructed it by using the idea of an ellipse, where the intersection point of BE and the ellipse with focal points A and F and passing through G gives the required point.

Observe that (AI + IF) + (BI + IE) = (AG + GF) + (BE) < (AG + GF) + (BG + GE). We are just using the triangle inequality.

Hence, G is a better point than I.

Step 2: Does there exist a better point than I?

Albeit, there is!

Just observe that the intersection point of the diagonals J is a better point

(AI + IF) + (BI + IE) = (AI + IF) + (BE) = (AI + IF) + (BJ + JE) > (AG + GF) + (BG + GE). We are just using the triangle inequality.

Thus, this algorithm results in the converges to the intersection of the diagonals as a special point. This is easy.

But what about, $ n = 3$. It may be easy. No, it's not.

The case for triangles is not at all easy. In the case of triangles, the point is called Fermat's Point.

I will not repeat the solution again. The following pic gives a cute solution. Using the same idea, that the distance between two points is minimum along a straight line.

Now, the rest of the questions remain too difficult to solve. Even there hardly exists a good algorithm to solve it.

Now, this left to your imagination. Stay tuned!

Examples & Counterexamples - A Way to Build Your Own Mathematics

What is an example? What is a counterexample? You are at least sure that they do counter each other very often.

George Polya and John Conway have already invested a lot of their time to help and share their views on how to build Problem Solving Skills. Do check out his book "How to Solve It?".

If you love solving problems and observing patterns in everything, then you will also find the pattern of how your mind starts the exploration, while you feed it with a problem.

The mind has a very sharp intellect. He (She) starts to explore and search the database (memory) whether there has been a similar scenario or not and cut it down into pieces like a knife just to solve your problem.

As you learn Mathematics and transcend into Advanced Mathematics, you will start to hear people saying the word "Mathematical Maturity" too often. You will get perplexed with the inferiority complex crawling into your mind, thinking that they must be really geniuses. No, they are not. You can do that too. We, are here to help you.

What is an Example?

You ask a question. If not you must. Or if you give a definition. Whether you have a solution or not. These examples and counterexamples are the step stones to help you take steps one at a time towards the solution or understanding the definition.

Let's take an example to understand this better.

Suppose, you give the definition of a prime number.

The next step will always be showing some examples, which numbers are primes and which are not. This will help you and the students to understand why we give such a definition. Also, this will help you to explore why such a definition is valid.

Examples are powerful. You can understand it, even more, when you see that you need examples to understand why examples are important.

In our Teachers For Tomorrow Training Program, we will give such examples more. Stay Tuned.

What is a Counterexample?

Counterexamples are interesting. You ask a question and it may be false. The question has a solution in negative. Counterexamples are even more interesting. They are tricky to find.

Suppose, you are understanding Pythagorean Triplets i.e. Positive Integer Solutions to $ a^2 + b^2 = c^2 $. Now you take examples to understand the solutions better.

You get (3,4,5) as a solution. You try to find the next where 5 is increased a bit. You get (6,8,10).

If you are a child at heart. You may think, "Wow, are they multiples of each other always?" This is the first step. This is called the art of Generalization. Let that be a topic for a different day.

Now, you try to explore more. Then you get the numbers (5,12,13). You are sad. The question, you asked excitedly has no answer in positive. This solution has an answer giving a negative example to your question. This is what Counterexample is. But, this is an opportunity to ask another question.

Then, can I find a general formula among these numbers? And you give conjectures ( questions ) and give examples and counterexamples and make your own universe of Mathematics.

Your Questions, Your Examples, Your Counterexamples, Your Proofs - Thus you discover your World of Mathematics.

There, you find yourselves blissed out and drunk in Mathematics for hours and days. Sometimes, you may forget to eat and sleep. Then, you know it is working. :p

Don't forget to keep a journal where it is just you and your Mathematics. 🙂

Stay Tuned.

Interesting Reads

Number Theory, Ireland MO 2018, Problem 9

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$ .
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

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Ireland MO 2018, Problem 9 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]8/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskaryacharya Prathisthan [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"], wIt is so important to know and use the modulo technqiue at the right time. We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

Cubic(3rd) Power Residue of 9 is {0,1,-1}.
6th Power Residue of 9 is {0,1}
Quadratic(2nd Power) Residue of 9 is {0,1,4,7}

We will use these ideas here. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$ . Note that, $a_{k+1}=a^6+2018$ . Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: $ a_k = 0$ mod 9 Then, the sequence modulo 9 will be $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2: $ a_k = 1,-1$ mod 9 Then, the sequence modulo 9 will be $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. QED [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Watch video

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Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Math Olympiad Program" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on"]

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Number Theory, France IMO TST 2012, Problem 3

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that:
$a^p+b^p=p^c.$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.0" open="on"]France IMO TST 2012, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="4.0"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Observe that we will try to fundamental solutions to $ a^p + b^p = p^c$. A fundamental solution (a,b,c,p) gives infinitely many solutions $ (a.p^k, b.p^k, c+k, p)$. A fundamental solution is, therefore (a,b,c,p) if gcd(a,b) = 1. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]We will focus on the fundamental solutions. We will deal with two cases: Case 1: p = 2 The equation reduces to $ a^2 + b^2 = p^2 $. As gcd(a,b) = 1, it implies a and b are odd. Now any odd square = 1 mod 4. So, $ a^2 + b^2 = 2 mod 4 $. Hence, the only fundamental solution is (1,1,1) = (a,b,c) We have that the following solutions are: $(1, 1, 1)$ and $\left(2^k, 2^k, 2^{2k+1}\right)$ .

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Case 2: p is an odd prime This now requires the idea of Lifting the Exponents. Please read here if you don't know it. It is an advanced technique to deal with Diophantine Equations. Let's check that the conditions of the LTE are satisfying here. p is an odd prime. gcd(a,b) = 1. p doesn't divide a or b as we are looking for fundamental solutions. $ a^p = a mod p; b^p = b mod p $. Hence, $ a^p + b^p = a + b mod p $. So, p | a+b, and p don't divide a or b. Hence, we can apply LTE. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Using the LTE idea, we get $1+v_p(a+b)=v_p\left(a^p+b^p\right) = c$ Then since $a+b | a^p+b^p$ , we have that $a+b = p^{c-1}$ , so $a^p+b^p = pa+pb$ Now, you see this can't happen for large p, as the LHS is exploding too fast like exponential as p increases and RHS is linear in p. So, we will apply inequality to prove this and find a bound for p for which it works and search in that bound. Note that $a^p \geq pa, b^p \geq pb$ or $a^{p-1} \geq p, b^{p-1} \geq p$ if $a, b \geq 2$ . Therefore, we must have that $a=1, b=p^{c-1}-1$ (or vice versa). But clearly $\left(p^{c-1}-1\right)^p > p^c$ , so no solution. QED [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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