Fermat's Last Theorem

Today we started off the class by introducing what a Diophantine Equation is.

Diophantine Equations

Diophantine Equation
is any equation that contains one or more variables that can be solved in the integers.

There are two subclasses of Diophantine equations that we went into depth with today


if you have coefficients a1,a2,….aN $\in \mathbb{Z}$

then the equation of the form..

a1X1+a2X2…+aNXN=b is linear

Applications & Examples

Linear Diophantine Equation

$x + 2y = 3$

Theorem: Let $ax = b$ be a linear Diophantine equation in the one variable x. If $a \nmid b$, then the equation has no solutions; if $a \mid b$, then the equation has exactly one solution, namely, $x = b/a.$

Theorem: Let $ax + by = c$ be a linear Diophantine equation in the two variables x and y and let $d = (a,b)$. If $d \nmid c$, then the equation has no solutions; if $d \mid c$, then the equation has at least one solution.

Example: Is $24x + 60y = 15$ solvable?
$(24,60) = 12$; since 12 does not divide 15, we have that this is not solvable.


This is one of the most enlightening definitions in the book. A Diophantine Equation is nonlinear if it is not linear

Applications & Examples

$x^{2}+y^{2} = z^{2}$ was talked about earlier in pythagorean triples.

$x^{2}+y^{2} = n$ was talked about earlier with the sum of two squares.

$x^{n}+y^{n} = z^{n}$ is the equation we will be looking at today.

Fermat's Last Theorem

We started out by stating what Fermat's Last Theorem actually is

X,Y,Z,n$\epsilon \mathbb{Z}$

$Y^{n}+X^{n}=Z^{n}, n>2 ,$
has No Solutions

Andrew Wiles

A portion of the interview that Andrew Wiles had with Nova about his journey through proving the Theorem was read. The complete version of the really interesting interview is at the link below.

NOVA Online|The Proof|

The First Steps of the Proof

We were not able to go through every single case of what n, the exponent, in the Theorem actually could be. So, we started out by showing how if we are able to prove that n=3 is not a solution then all multiples of 3 are also not solutions.
We are given n and then we break it down into its prime factorization

n=p1p2….pN , p is a prime number

$Y^{n}+X^{n}=Z^{n}, n>2 ,$

We can rewrite this equation as

$Y^{p_1p_2 \cdots p_N}+X^{p_1p_2 \cdots p_N}=Z^{p_1p_2 \cdots p_N} ,$

Then manipulating this equation to look in a way that will help us knock out a whole class of equations

$Y^{(p_1p_3 \cdots p_N)p_2}+X^{(p_1p_3 \cdots p_N)p_2}}=Z^{(p_1p_3 \cdots p_N)p_2}} ,$

Now arbitrarily assign p2 = 3, So, when the case of n=3 is shown to have no solutions this means that every case with a multiple of 3 i.e. 3,6,9….30 and so on, also have no solutions.

Now it's the time for the proof for the case when $n = 3$. The general idea of the proof is the technique called the method of infinite descent. If we start by assuming that there is a solution x, y, y to the Fermat's equation, then we can use this solution to construct another solution that is strictly smaller (in the absolute value), and we can keep applying this method infinitely many times, so we will get infinite strictly decreasing series of solution. However, it is impossible to have infinitely many solutions that are strictly decreasing because we are talking about integer solutions. Hence, the assumption that there is a solution x, y, z to Fermat's equation is false, that is, the Fermat's equation has no equation.

Before we get to the proof of the theorem, we need to first prove some several lemmas.

Lemmas needed for the proof

Lemma 1: Suppose $x^{3}+y^{3} = z^{3}$ has a solution that is not zero, then there is a solution x, y, z that are pairwise relatively prime.

Lemma 2: If $x^3+y^3=z^3$ has a solution, then there exists positive integers $a$ and $b$ such that $(a,b) = 1$, exactly one of a and b is even, and $2a(a^2+3b^2)$ is a cube.

Proof: Suppose $x, y, z$ are solutions of $x^3+y^3=z^3$, and $x, y, z$ are pairwise relatively prime. Then, we know that exactly one of $x, y, z$ is even. Without loss of generality, assume that z is even. This implies that x and y are odd, and so $x+y$ and $x-y$ are even. Hence, we can write
$x+y = 2a$
$x-y = 2b$
Now we can solve for x and y in terms of a and b. We have that
$x = a+b$
$y = a-b$
Since both x and y are odd, this means exactly one of a and b is even. We also have that $(a,b) = 1$.
So the last thing we need to show is that $2a(a^2+3b^2$ is a cube.
Consider $z^3 = x^3+y^3 = (x+y)(x^2-xy+y^2)$. Now if we substitute x with $a+b$ and y with $a-b$, we will get $z^3 = 2a(a^2+3b^2)$. done!!!

Lemma 3: Suppose $(a,b)=1$, exactly one of a and b is even, then $(2a, a^2+3b^2) =$ 1 or 3.

Proof: If $(2a, a^2+3b^2) =$ 1 or 3, then we are done. So suppose that $(2a, a^2+3b^2) \neq$ 1 or 3. Notice that $(2a, a^2+3b^2) \neq 2$ since a and b have opposite parity, so $a^2+3b^2$ is odd. Now suppose that $(2a, a^2+3b^2) > 3$. Then there is an odd prime $p > 3$ that divided d, so $p \mid 2a$ and $p \mid a^2+3b^2$. Hence, we can write
$2a = pm$
$a^2+3b^2 = pn$
Since $2 \nmid p$, we have $2 \mid m$, and so we can write $a = p(\frac{m}{2}) = p \cdot k$.
Also, from $a^2+3b^2 = pn$, we have $3b^2 = pn - a^2 = pn - (pk)^2 = p(n-pk^2)$. Since $p > 3$, we have $p \mid b^2$ which implies $p \mid b$. This is a contradiction because p divides both a and b. Hence, we get that $(2a, a^2+3b^2) =$ 1 or 3.

Lemma 4: if $(a,b) = 1$, then every odd factor of $a^2+3b^2$ also has this form.

Lemma 5: if $(a,b) = 1$, a and b have opposite parity, and $a^2+3b^2$ is a cube, then there exists r and s such that

\begin{equation} a = r^3 - 9rs^2 \end{equation}
\begin{equation} b = 3r^2s-3s^3 \end{equation}

$(r,s) = 1$, and r and s has opposite parity.

Now we can finally move to the last theorem that will prove that there are no non-trivial solutions to $x^3+y^3=z^3$.

Theorem: If x, y, z is a solution to $x^3+y^3=z^3$, and x, y are written in terms of a and b, $(a,b) = 1$, exactly one of a and b is even, and $2a(a^2+3b^2)$ is a cube, then there exists a smaller solution $x_1, y_1, z_1$ to $x^3+y^3=z^3$.

Proof: From lemma 3, we know that $(2a, a^2+3b^2) =$ 1 or 3.
So the first case is when $(2a, a^2+3b^2) = 1$. This implies that $2a$ and $a^2+3b^2$ are cube.
By lemma 5, we can write

\begin{equation} a = r^3 - 9rs^2 \end{equation}
\begin{equation} b = 3r^2s-3s^3 \end{equation}

$(r,s) = 1$, and r and s has opposite parity.
Multiply both sides of $(3)$ by 2 gives $2a = 2r^3 - 18rs^2 = 2r(r-3s)(r+3s)$. The claim here is that $2r$, $r-3s$, and $r+3s$ are pairwise relatively prime.
First, we will show that $(2r, r-3s) = (2r, r+3s) = 1$. We will only show that $(2r, r-3s) = 1$ because the proof of $(2r, r+3s) = 1$ is very similar. Notice that $2 \nmid r-3s$ because r and s have different parity, so $(2r, r-3s) \neq 2$. Also $(2r, r-3s) \neq 3$, or it would mean $3 \mid r$, and so $3 \mid a$ and $3 \mid b$. If $d = (2r, r-3s)$, and $d > 3$, this would mean $d \mid r$, and $d \mid s$, which is also a contradiction. So $(2r, r-3s) = 1$.
Next we need to show that $(r-3s, r+3s) = 1$, but the proof of this is very similar to the argument above, so we will not go into the detail.
Now we know that $2r$, $r-3s$, and $r+3s$ are cube because 2a is a cube, and $2r$, $r-3s$, and $r+3s$ are pairwise relatively prime. This means we can write
$x_1^3 = r-3s$
$y_1^3 = r+3s$
$z_1^3 = 2r$
Clearly, $x_1, y_1, z_1$ is a solution to $x^3+y^3 = z^3$, and they are smaller than the original solution $x, y, z$ because

\begin{equation} 2a = x_1^3y_1^3z_1^3 < 2a(a^2+3b^2) = z^3 \end{equation}

Now we will prove the second case where $(2a, a^2+3b^2 = 3$. Notice that $3 \mid a$, but $3 \nmid b$. This means there exists c such that $a = 3c$. Since $2a(a^2+3b^2)$ is a cube, we have that $8 \mid 2a(a^2+3b^2)$, and so $4 \mid a$. This implies $4 \mid c$. Substituting $a = 3c$ in $z^3 = 2a(a^2+3b^2)$ gives

\begin{equation} z^3 = 6c(9c^2+3b^2) = 18c(3c^2+b^2) \end{equation}

We claim that $(18c, 3c^2+b^2)=1$. This is true because we know that b must be odd, so $2 \nmid 3c^2+b^2$. Also, $3 \nmid 3c^2+b^2$ because $3 \nmid b$. Moreover, $(c, b) = 1$ because $(a,b) = 1$.
Hence, we have that $18c$ and $3c^2+b^2$ are cube, or we can write
$r^3 = 18c$
$s^3 = 3c^2+b^2$
where s is odd and $3 \mid r$.
By lemma 4, we have $s = u^2 + 3v^2$, and by lemma 5, we have
$b = u^3 - 9uv^2$
$c = 3u^2v - 3v^3$
$(u,v) = 1$, and u and v have opposite parity. But because we know that b is odd, so we know that indeed u is odd and v is even. Also, $2v, u+v, u-v$ are pairwise relatively prime.
Since $r^3 = 18c = 54v(u^2-v^2)$, we have
$(\frac{r}{3})^3 = 2v(u+v)(u-v)$, which means all of them are cube. Thus, we can write
$x_1^3 = u-v$
$y_1^3 = 2v$
$z_1^3 = u+v$
Clearly, $x_1, y_1, z_1$ is a solution to $x^3+y^3 = z^3$, and they are smaller than the original solution $x, y, z$ since

\begin{equation} z^3 = 18c(3c^2+b^2) = 54v(u-v)(u+v)(3c^2+b^2) = z_1^3[54v(u-v)(3c^2+b^2)] > z_1^3 \end{equation}



Prove that the Diophantine equation $x^{2}+y^{2} = z^{3}$ has infinitely many solutions.
$x = n^{3} - 3n$
$y = 3n^{2} - 1$

$(n^{3} - 3n)^{2} + (3n^{2} - 1)^{2}$

$(n^{6} - 6n^{4} + 9n^{2}) + (9n^{4} - 6n^{2} + 1)$

$n^{6] + 3n^{4} + 3n^{2} + 1$

$(n^{2} + 1)^{3}$

You should notice that the final answer is a cube so therefore, by plugging any integer in for n, you can come up with a solution to the equation $x^{2}+y^{2} = z^{3}$. Hence, there are infinitely many solutions.

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