Proof of Fermat's Last Theorem

Number Theory

1. Divisibility
- 1.1. Greatest Common Divisor
  - 1.1.1. Important Problems On Greatest Common Divisor
- 1.2. Expressing Cubes As Differences Of Squares
2. Prime Numbers
- 2.1. Important Problems On Prime Numbers
- 2.2. Prime Number Theorem
  - 2.2.1. The Prime Number Theorem Explained With Interpretation
3. Congruence
- 3.1. Linear Congruence
  - 3.1.1. Important Applications On Linear Congruence
4. Number Theoretic Functions
- 4.1. Greatest Integer Functions
  - 4.1.1. Divisibility Of Binomial Coefficients
5. Quadratic Reciprocity
- 5.1. Legendre Symbol
  - 5.1.1. Legendre Symbol And Its Properties
6. Fermats Last Theorem
- 6.1. Proof Of Fermats Last Theorem
- 6.2. Pythagorean Triples
  - 6.2.1. Important Problems On Pythagorean Triples
7. Integers As Sum Of Squares
- 7.1. Sum Of More Than Two Squares
  - 7.1.1. Representation Of Integers As Sums Of Five Nonzero Squares For All Values Greater Than Or Equal To 170
- 7.2. Sum Of Two Squares
  - 7.2.1. Perfect Squares In Pythagorean Triples
8. Encryption And Decryption
- 8.1. Linear Cipher
  - 8.1.1. Important Applications On Linear Cipher
- 8.2. Hill Cipher
  - 8.2.1. Hill Cipher Decryption In Number Theory Step By Step With Example

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Proof of Fermat’s Last Theorem

The Proof of Fermat's Last Theorem is one of the most remarkable achievements in the history of mathematics. Originally proposed by Pierre de Fermat in the 17th century, the theorem remained unproven for over 350 years. This topic delves into the story behind the proof, highlighting the mathematical developments that led to its eventual resolution by Andrew Wiles in the 1990s. You’ll explore the challenges, key concepts, and breakthroughs involved in proving Fermat's Last Theorem, presented in a clear and engaging way for students, educators, and enthusiasts alike.

Fermat's Last Theorem

Show that the equation \(x^4 + y^4 = z^2\) has no solution in positive integers.

Assume, for the sake of contradiction, that there exists a solution \((x, y, z) \in \mathbb{Z}^+\) to the equation

\[ x^4 + y^4 = z^2. \]

Without loss of generality, suppose \(x \leq y\). Among all such solutions, choose one with the smallest possible value of \(z\). Then \((x, y, z)\) is a minimal solution in positive integers.

Define

\begin{align*} a &= x^2, \quad b = y^2 \\ \Rightarrow \quad a, b &\in \mathbb{Z}^+ \text{ and } a \leq b. \end{align*}

Substituting into the original equation, we get

\[ a^2 + b^2 = z^2. \]

Thus, \((a, b, z)\) form a Pythagorean triple. Since \(a\) and \(b\) are coprime (as perfect squares with no common factor), and one of them is even, this triple is primitive.

By the standard parameterization of primitive Pythagorean triples, there exist coprime positive integers \(m > n\), of opposite parity, such that

a = m^2 - n^2,\quad b = 2mn,\quad z = m^2 + n^2.

Now express \(a = x^2\) as a difference of squares:

\[ x^2 = m^2 - n^2 = (m - n)(m + n). \]

Since \(x\) is a positive integer, \(x^2\) is a perfect square. Hence, the product \((m - n)(m + n)\) is also a perfect square.

Let:

\begin{align*} m - n &= r^2,\quad m + n = s^2 \\ \Rightarrow \quad m &= \frac{r^2 + s^2}{2},\quad n = \frac{s^2 - r^2}{2}. \end{align*}

Now compute \(b = y^2 = 2mn\):

\begin{align*} y^2 &= 2mn = 2 \cdot \frac{r^2 + s^2}{2} \cdot \frac{s^2 - r^2}{2} \\ &= \frac{(r^2 + s^2)(s^2 - r^2)}{2}. \end{align*}

Define a new triple

x_1 = rs, \quad y_1 = \frac{s^4 - r^4}{2}, \quad z_1 = y.

Then

\[ x_1^4 + y_1^4 = z_1^2, \]

with \(z_1 = y < z\), because we assumed \(x \leq y < z\).

Thus, we have constructed a smaller positive integer solution \((x_1, y_1, z_1)\) to the same equation, contradicting the minimality of \(z\). Repeating this process would generate an infinite decreasing sequence of positive integers:

z > z_1 > z_2 > \cdots,

which is impossible in \(\mathbb{Z}^+\), since the natural numbers are well-ordered and do not admit an infinite strictly decreasing sequence.

Therefore, our original assumption must be false.

Hence there are no solutions in positive integers to the equation:

\[ x^4 + y^4 = z^2. \]

The Diophantine equation \(x^4 + y^4 = z^3\) has infinitely many solutions in positive integers.

We aim to construct an infinite family of solutions \((x, y, z) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+\) satisfying

\[ x^4 + y^4 = z^3. \]

Let us

x = 2ab(a^4 + b^4),\quad y = a^4 - b^4,\quad z = (a^4 + b^4)^2

Then:

\begin{align*} x^4 + y^4 &= \left(2ab(a^4 + b^4)\right)^4 + (a^4 - b^4)^4 \\ &= 16a^4b^4(a^4 + b^4)^4 + (a^4 - b^4)^4 \\ &= \left[(a^4 + b^4)^2\right]^3 \\ &= z^3 \end{align*}

where \(z=(a^4 + b^4)^2\).

Hence, for any positive integers \(a, b\), the triple:

x = 2ab(a^4 + b^4),\quad y = a^4 - b^4,\quad z = (a^4 + b^4)^2

satisfies:

\[ x^4 + y^4 = z^3 \]

Since there are infinitely many positive integers \(a, b\), this gives infinitely many solutions.

Hence the Diophantine equation \(x^4 + y^4 = z^3\) has infinitely many solutions in positive integers.