关于满足xⁿ + yⁿ = zⁿ（n>2且x,y,z∈N）的三角形存在性的求证咨询

Hey folks,

My friend dropped this problem on me at school, and I’ve been staring at it for hours without making much progress.

My current plan is to try proving that $k>1$ (so I can then say $x>z-y$), but I’m really not confident if this is the right path to take.

Edit: Oops, sorry I missed the most crucial detail earlier! The full problem states that $x,y,z,n \in \mathbb{N}$ and $n>2$, and I’m trying to figure out if a triangle with side lengths $x$, $y$, $z$ can exist under this equation.

Answer #

Hey there! Let’s break this down step by step—you’re actually tangling with a problem tied to one of the most famous theorems in math: Fermat’s Last Theorem.

First, let’s recall the triangle inequality rule: for three lengths to form a valid triangle, the sum of any two sides has to be greater than the third. Since $x,y,z$ are natural numbers, the inequalities $x+z>y$ and $y+z>x$ are automatically true (adding a positive number to one side will always make it bigger than the other). The make-or-break one here is $x+y>z$.

Now, let’s connect this to your equation $x^n + y^n = z^n$ where $n>2$.

• Even if we ignore Fermat’s Last Theorem for a second, let’s compare $z^n$ to $(x+y)^n$. Using the binomial theorem, $(x+y)^n = x^n + nx^{n-1}y + \dots + y^n$. Every term in that expansion after $x^n$ and $y^n$ is positive, so $(x+y)^n$ is definitely larger than $x^n + y^n = z^n$. That means $x+y>z$, which would satisfy the triangle inequality—if such a triple $(x,y,z)$ existed.

• But here’s the kicker: Fermat’s Last Theorem (proven back in the 90s) tells us that there are no natural numbers $x,y,z,n$ with $n>2$ that satisfy $x^n + y^n = z^n$. So those triples don’t exist at all.

So to address your original approach: while trying to prove $k>1$ might lead you to the triangle inequality, the bigger takeaway is that there’s no need to even check—because there are no such $x,y,z$ to form a triangle with in the first place.

In short: No such triangle exists, since there are no natural number solutions to $x^n + y^n = z^n$ when $n>2$.

备注：内容来源于stack exchange，提问作者Ghost