MHB Why does it suffice to show it for n=4 and n=p?

[evinda]

Hi! (Wave)

The Fermat equation $X^n+Y^n=Z^n, n \geq 3, X \cdot Y \cdot Z \neq 0$ has no integer solutions.

Remark: It suffices to prove the hypothesis for $n=4$ and $n=p \in \mathbb{P} (p \neq 2)$.

Could you explain me why, in order to show that the Fermat equation has no integer solutions, it suffices to show it for $n=4$ and for $n=p$ ? (Thinking)

 

[Nono713]

Hi! (Wave)

The Fermat equation $X^n+Y^n=Z^n, n \geq 3, X \cdot Y \cdot Z \neq 0$ has no integer solutions.

Remark: It suffices to prove the hypothesis for $n=4$ and $n=p \in \mathbb{P} (p \neq 2)$.

Could you explain me why, in order to show that the Fermat equation has no integer solutions, it suffices to show it for $n=4$ and for $n=p$ ? (Thinking)

Hint: if $x^{pq} + y^{pq} = z^{pq}$, then $(x^p)^q + (y^p)^q = (z^p)^q$. Use the fundamental theorem of arithmetic, suitably modified to take into account the $n = 4$ case, and conclude.

 

[mathbalarka]

Just a note to why Bacterius's reasoning doesn't work for $n = 4$ :

$$X^4 + Y^4 = Z^4 \Longrightarrow (X^2)^2 + (Y^2)^2 = (Z^2)^2$$

But that doesn't conclude anything, as there can be infinitely many solutions in general to $x^2 + y^2 = z^2$.

 

[evinda]

Hint: if $x^{pq} + y^{pq} = z^{pq}$, then $(x^p)^q + (y^p)^q = (z^p)^q$. Use the fundamental theorem of arithmetic, suitably modified to take into account the $n = 4$ case, and conclude.

So, because of the fact that all numbers $n \geq 3, n \neq 4$ can be written as a product of prime numbers, at least one of which is different from $2$, we can show that $x^n+y^n=z^n$ has no integer solutions, when $n$ is a prime? (Thinking)

How can I use the fundamental theorem of arithmetic, to show that $X^4+Y^4=Z^4$ has no integer solutions? (Thinking)

 

[mathbalarka]

So, because of the fact that all numbers $n \geq 3, n \neq 4$ can be written as a product of prime numbers, at least one of which is different from $2$, we can show that $x^n+y^n=z^n$ has no integer solutions, when $n$ is a prime? (Thinking)

Minor correction : it's because of all numbers $> 4$ can be written as a product of primes with at least one prime $> 2$.

How can I use the fundamental theorem of arithmetic, to show that $X^4+Y^4=Z^4$ has no integer solutions?

As a matter of fact you cannot, as I have indicated in my last comment. You need to use some of the "magic tricks" like infinite descent to handle this one. If you are interested, I could post the idea behind.

 

[evinda]

Minor correction : it's because of all numbers $> 4$ can be written as a product of primes with at least one prime $> 2$.

(Nod)

As a matter of fact you cannot, as I have indicated in my last comment. You need to use some of the "magic tricks" like infinite descent to handle this one. If you are interested, I could post the idea behind.

Yes, it would be nice.. (Smile)