Mathematics : Important Theorems

Here is a short, accurate response you can use for an interview.

Part 1: What is Fermat’s Last Theorem?

“It’s a theorem from number theory, first stated by Pierre de Fermat in 1637.

It states that no three positive integers $a$, $b$, and $c$ can satisfy the equation $a^n + b^n = c^n$ for any integer value of $n$ greater than 2.

For $n=2$ (the Pythagorean theorem), there are infinite solutions, like $3^2 + 4^2 = 5^2$. Fermat’s theorem states that for $n=3$, $n=4$, and so on, there are zero solutions.”

Part 2: What is the Proof Outline?

“Fermat’s claim of having a ‘marvelous proof’ is unproven; it’s believed he was mistaken. The actual proof was a monumental achievement by Andrew Wiles in 1995, and it’s over 100 pages long.

The strategy is a proof by contradiction that connects two completely different areas of mathematics: elliptic curves and modular forms.

The short outline is:

Assume Fermat is Wrong: You start by assuming a solution $(a, b, c)$ for $a^n + b^n = c^n$ does exist.
Create an Elliptic Curve: In the 1980s, a mathematician named Gerhard Frey showed that if a solution existed, you could use it to create a very specific and strange elliptic curve (now called a Frey curve).
Identify the Problem: Frey and Ken Ribet proved that this hypothetical curve would be so ‘weird’ that it could not be modular. This means it could not be related to a modular form.
Wiles’s Breakthrough: Andrew Wiles spent seven years proving a part of the Modularity Theorem (which was then a conjecture). Wiles proved that all curves of this type must be modular.
The Contradiction: This created a logical impossibility.
- If a solution to Fermat’s equation exists, it creates a curve that cannot be modular (by Ribet).
- But Wiles proved that this exact type of curve must be modular.

Since a curve cannot both be modular and not be modular, the only logical failure is the initial assumption. Therefore, the assumed solution $(a, b, c)$ cannot exist, which proves Fermat’s Last Theorem.”

Elliptic Curve

An elliptic curve $E$ over $\mathbb{Q}$ is a smooth cubic curve defined by two main conditions:

The Equation: It is the set of solutions $(x, y)$ to a Weierstrass equation, which can be simplified to:$$y^2 = x^3 + Ax + B$$where $A$ and $B$ are rational numbers.
The Smoothness Condition: The curve must have no cusps or self-intersections. This is guaranteed by the discriminant being non-zero:$$\Delta = -16(4A^3 + 27B^2) \neq 0$$
The Group Structure: Its set of points forms an abelian group under a geometric “point addition” rule, with the point at infinity serving as the identity element.

Conclusion: The Four Color Theorem

The Four Color Theorem is a simple-sounding but deeply complex proof.

The Guarantee: It guarantees that four colors are always enough to color any map on a flat surface (or a globe).
The Rule: The only rule is that two regions sharing a border line (not just a single point) must have different colors.
The Logic: It works because the “what if” scenario that would require a fifth color—a map where five countries all mutually touch each other—is mathematically impossible to draw on a flat surface.
The Proof: The theorem was famously proven in 1976 by Appel and Haken, who used a computer to check over 1,900 “problem” configurations that no human could verify by hand. This proved that no map, no matter how complex, could ever be constructed to break the four-color rule.

The Collatz Conjecture is one of the most famous unsolved problems in mathematics. It’s extremely simple to describe, but no one has been able to prove it.

It’s also known as the $3n + 1$ problem or the hailstone problem.

🧮 The Rules

The conjecture proposes a simple sequence for any positive integer:

Start with any positive integer, $n$.
If the number is even, divide it by 2 ($n / 2$).
If the number is odd, multiply it by 3 and add 1 ($3n + 1$).
Take the new number and repeat the process.

The conjecture is: No matter what positive integer you start with, you will always eventually reach 1.

📈 An Example (Starting with 7)

Let’s see how this “hailstone” sequence works:

7 is odd $\rightarrow$ $3 \times 7 + 1 = $ 22
22 is even $\rightarrow$ $22 / 2 = $ 11
11 is odd $\rightarrow$ $3 \times 11 + 1 = $ 34
34 is even $\rightarrow$ $34 / 2 = $ 17
17 is odd $\rightarrow$ $3 \times 17 + 1 = $ 52
52 is even $\rightarrow$ $52 / 2 = $ 26
26 is even $\rightarrow$ $26 / 2 = $ 13
13 is odd $\rightarrow$ $3 \times 13 + 1 = $ 40
40 is even $\rightarrow$ $40 / 2 = $ 20
20 is even $\rightarrow$ $20 / 2 = $ 10
10 is even $\rightarrow$ $10 / 2 = $ 5
5 is odd $\rightarrow$ $3 \times 5 + 1 = $ 16
16 is even $\rightarrow$ $16 / 2 = $ 8
8 is even $\rightarrow$ $8 / 2 = $ 4
4 is even $\rightarrow$ $4 / 2 = $ 2
2 is even $\rightarrow$ $2 / 2 = $ 1

…and if you reach 1, it gets stuck in a loop: $1 \rightarrow 4 \rightarrow 2 \rightarrow 1…$

🤔 Why Is It a “Big” Problem?

The conjecture is so famous because it looks deceptively simple, but it’s impossibly hard to prove.

It’s Unpredictable: As you saw with the number 7, the sequence can shoot up to high numbers (like 52) before suddenly “crashing” down to 1. This “up and down” motion is why it’s called the “hailstone problem.”
No One Can Prove It: No mathematician has been able to prove that it’s true for all numbers. A proof would need to show that no number can:
- Go to infinity.
- Get stuck in a different loop that doesn’t include 1.
No One Can Disprove It: To disprove it, you only need to find one starting number (a “counterexample”) that never reaches 1.
- Computers have tested all numbers up to $2^{68}$ (a number with 21 digits), and every single one of them eventually goes to 1. But this isn’t a proof—it just shows it’s true for a lot of numbers.

The famous mathematician Paul Erdős said about the Collatz Conjecture: “Mathematics may not be ready for such problems.” It highlights a deep and mysterious gap in our understanding of simple arithmetic.

Here’s a final conclusion on Russell’s Paradox.

Conclusion: Russell’s Paradox

The Paradox: The paradox arises from asking a simple question about a seemingly valid collection:”The set $R$ of all sets that do not contain themselves.”When we ask, “Does $R$ contain itself?” we get a perfect logical contradiction: It must contain itself if and only if it does not.
What It Proved: The paradox proved that the old, intuitive definition of a set (“any collection of objects”) was fundamentally broken and led to contradictions.
The Solution: The crisis was solved by creating modern, axiomatic set theory. These new rules (axioms) are stricter and detail exactly how a “well-defined” set can be built.
The Final Answer: Under these modern rules, the collection $R$ is not a set. The rules make it illegal to even form this paradoxical collection, thus saving mathematics from the contradiction.

Here is a final conclusion on the Poincaré Conjecture.

Conclusion: The Poincaré Conjecture

The Problem: It was a central, century-old question in topology (the study of shapes). It asks: How can we mathematically prove that a 3D shape is just a sphere in disguise?
The “Rubber Band Test”: The conjecture states that if any 3-dimensional shape is “closed” (finite with no edges) and “simply connected” (meaning any loop you draw on it can be shrunk down to a single point), then it must be a 3-sphere.
The Solution: The conjecture was proven true by the Russian mathematician Grigori Perelman in 2002-2003.
The Importance: Perelman’s proof was a monumental achievement. He didn’t just solve the conjecture; he solved the much harder Geometrization Conjecture, which classifies all possible 3-dimensional shapes. He famously declined both the $1 million Millennium Prize and the Fields Medal for this work.

. Zero Volume, Zero Surface Area

Example: A single point.
Explanation: This is a theoretical object with no dimensions (no length, width, or height). As such, it has no surface and encloses no volume.

2. Zero Volume, Infinite Surface Area

Example: A Menger Sponge (a 3D fractal).
Explanation: This is a paradoxical but real mathematical object. You start with a cube, then drill holes through it. You then drill smaller holes in the remaining cubes, and so on, infinitely.
- At each step, you remove volume. In the infinite limit, the total volume of the object becomes zero.
- However, each time you drill a hole, you create new surface area. In the infinite limit, the total surface area becomes infinite.

3. Infinite Volume, Zero Surface Area

Example: Impossible.
Explanation: This is a logical contradiction. An object must have a surface to “contain” its volume. If the surface area is zero, the object can only be a point (or a collection of points), which has zero volume, not infinite volume.

4. Infinite Volume, Infinite Surface Area

Example: An infinitely long cylinder (or an infinite 3D prism).
Explanation: This is the most intuitive case. If you take a simple shape like a cylinder and extend its length forever, both its volume (the space it fills) and its surface area (the material needed to cover it) will also extend to infinity.

FINITE volume but an INFINITE surface area.

This object is called Gabriel’s Horn (or Torricelli’s Trumpet).¹

🎺 Gabriel’s Horn

This is a 3D shape created by a simple process:

Take the graph of the function $y = 1/x$ for all values of $x \ge 1$.
Rotate this curve around the x-axis.

This creates a long, thin “horn” that stretches to infinity, getting infinitely narrower.

The Paradox

Here is the paradox, which is proven with basic calculus:

1. It has an INFINITE Surface AreaTo find the surface area, you calculate the integral (a way of summing up infinite pieces):$$A = \int_{1}^{\infty} 2\pi y \sqrt{1 + (y’)^2} \,dx$$This integral diverges, meaning the surface area is infinite.2
2. It has a FINITE VolumeTo find the volume, you use the disk method:$$V = \int_{1}^{\infty} \pi y^2 \,dx = \int_{1}^{\infty} \pi (1/x)^2 \,dx$$This integral converges to a specific, finite value: $\pi$ (pi).

The “Paint” Paradox

This leads to a mind-bending conclusion:

You can fill the entire horn with a finite amount of paint ($\pi$ cubic units), but you would never have enough paint to cover its inside surface.

Euler’s Identity is often called “the most beautiful equation in mathematics.”

It is a simple, profound statement that connects the five most fundamental constants in all of mathematics:

$$e^{i\pi} + 1 = 0$$

👑 The “Five Kings” of Numbers

What makes this identity so special is that it links these five constants, which all come from completely different areas of math:

$e$ (Euler’s Number $\approx 2.718…$)
- The base of the natural logarithm. It’s the “king” of calculus and describes all types of exponential growth (like compound interest or population growth).
$i$ (The Imaginary Unit = $\sqrt{-1}$)
- The “king” of complex numbers and algebra. It is the foundation for an entire branch of mathematics that involves “imaginary” or two-dimensional numbers.
$\pi$ (Pi $\approx 3.141…$)
- The “king” of geometry. It is the ratio of a circle’s circumference to its diameter and is essential for all things related to circles, spheres, and trigonometry.
$1$ (One)
- The “king” of arithmetic. It is the identity for multiplication (any number times 1 is itself) and the foundation of all other integers.
$0$ (Zero)
- The “king” of algebra. It is the identity for addition (any number plus 0 is itself) and represents “nothingness.”

Euler’s Identity shows that these five numbers, which seem to have no relationship, are actually linked together in one perfectly simple and elegant equation.

Euler’s Formula (for Polyhedra)¹⁵

This is a fundamental theorem in topology and graph theory (the study of shapes and networks).¹⁶

It states that for any simple, connected polyhedron (a 3D shape with flat faces and no holes, like a cube or a pyramid), the number of Vertices (V), Edges (E), and Faces (F) are related by this simple formula:

$V – E + F = 2$

Here is a detailed expansion of Euler’s Theorem and its direct relationship to Fermat’s Little Theorem. They are both fundamental to number theory and cryptography.

1. 🥇 Fermat’s Little Theorem

This is the simpler, more specific theorem. It only works with prime numbers.

It states that if $p$ is a prime number, then for any integer $a$ that is not divisible by $p$:

$$a^{p-1} \equiv 1 \pmod{p}$$

$\equiv 1 \pmod{p}$ is modular arithmetic. It means that when you calculate $a^{p-1}$ and divide it by $p$, the remainder is 1.

Simple Example:

Let’s use $a = 2$ and $p = 7$ (a prime number).
The theorem says $2^{7-1} \equiv 1 \pmod{7}$, which is $2^6 \equiv 1 \pmod{7}$.
Let’s check: $2^6 = 64$.
When you divide 64 by 7, you get 9 with a remainder of 1 ($64 = 7 \times 9 + 1$).
It works.

What it’s for: This theorem is a powerful tool for reducing very large powers of a number. For example, if you need to find the remainder of $2^{70}$ when divided by 7, you can use the theorem:

$2^{70} = (2^6)^{11} \times 2^4$
Since $2^6 \equiv 1 \pmod{7}$, this becomes $(1)^{11} \times 2^4 \equiv 16 \pmod{7}$.
The remainder of 16 divided by 7 is 2.

2. 👑 Euler’s Theorem (The Generalization)

Fermat’s Little Theorem is great, but it only works for prime moduli (like 7). What if you want to find a remainder when dividing by a non-prime number, like 10?

That’s what Euler’s Theorem is for. It generalizes Fermat’s theorem to work for any positive integer $n$.

To do this, it introduces a crucial new tool:

The Key: Euler’s Totient Function ($\phi(n)$)

What it is: The totient function $\phi(n)$ (pronounced “fee of n”) counts how many positive integers less than or equal to $n$ are coprime to $n$.
“Coprime” means their greatest common divisor (GCD) is 1.

Example 1: $\phi(7)$ (a prime number)

The numbers less than or equal to 7 are {1, 2, 3, 4, 5, 6, 7}.
Which ones are coprime to 7? {1, 2, 3, 4, 5, 6}. (All of them except 7 itself).
So, $\phi(7) = 6$.
Notice: For any prime number $p$, $\phi(p) = p-1$. This is the link to Fermat’s theorem!

Example 2: $\phi(10)$ (a non-prime number)

The numbers less than or equal to 10 are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Which ones are coprime to 10? (i.e., not divisible by 2 or 5).
{1, 3, 7, 9}
So, $\phi(10) = 4$.

Euler’s Theorem Stated

Now for the theorem itself. It states that for any two positive integers $a$ and $n$, as long as they are coprime ($\text{GCD}(a, n) = 1$):

$$a^{\phi(n)} \equiv 1 \pmod{n}$$