Based on https://www.youtube.com/watch?v=EymVXkPWxyc
Fermat's Last Theorem states that no three positive integers $a$ , $b$ , and $c$
satisfy $a^n + b^n = c^n$ for any integer value of $n$ greater than 2.
https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
We have
a is an integer > 0
b is an integer > 0
c is an integer > 0
n is an integer > 2
a^n + b^n = c^n
If a, b, c, n satisfies the above conditions, it is notated here as (a b c FLT n)
Without loss of generality, let $a \le b$
Since addition is commutative
$a \le b$ or $b \le a$ . so (a b c FLT n) or (b a c FLT n) satisfies this property.
If a = b, then c^n = a^n + a^n = 2 * a^n
So c = 2^(1/n) * a
But 2^(1/n) is irrational.
a and b are positive, so their sum is greater than both.
(This ignores the powers, which don't affect anything. Dealing with them is left as an exercise to the reader.)
$$ \text{We have: } a, b, c \in \mathbf{N}, a < b < c, 2 < c $$
$$\begin{align}
a^n + b^n &= c^n \\\
\\\
a^n &= c^n - b^n \\\
&= (c-b) \sum_{i=1}^{n}c^{n-i}b^{i-1} \\\
&> (c-b) \sum_{i=1}^{n}a^{n-i}a^{i-1} \\\
&= (c-b) \sum_{i=1}^{n}a^{n-1} \\\
&= (c-b) na^{n-1} \\\
\\\
a &> (c-b) n \\\
&> n
\end{align}$$
Am I crazy, or can we do
$$\begin{align}
a^n + b^n &= c^n \\\
(a/c)^n + (b/c)^n &= 1^n \\\
\\\
(a/c) &> n
\end{align}$$
Edit: this is wrong
In the sum, instead of replacing c and b with a, replacing with b gets
$$
\begin{align}
a^n &> (c-b) nb^{n-1} \\
a &> (c-b)^{\frac{1}{n}} n^{\frac{1}{n}} b^{\frac{n-1}{n}}
\end{align}
$$
As $n$ approaches infinity, the right hand side approaches $b$ , but $a < b$ . So $a$ is getting squeezed.
However, there's still plenty of room: the larger a and b are, the more room there is.
Slightly stricter bound for a We have 0 < b < c, 2 < n. Note that the 2 < n is important for ">" in the first half, because i=2 and c^(n-i) = b^(n-i) when n-i=0
$$
\begin{align}
c^n-b^n &= (c-b) \sum_{i=1}^{n}c^{n-i}b^{i-1} \\
&= (c-b) (c^{n-1} + \sum_{i=2}^{n}c^{n-i}b^{i-1}) \\
&> (c-b) (c^{n-1} + \sum_{i=2}^{n}b^{n-i}b^{i-1}) \\
&= (c-b) (c^{n-1} + \sum_{i=2}^{n}b^{n-1}) \\
&= (c-b) (c^{n-1} + (n-1)b^{n-1}) \\
\\
a^n &> (c-b) (c^{n-1} + (n-1)b^{n-1}) \\
&= (c-b) (\frac{c^{n-1}}{b^{n-1}} + n-1) b^{n-1} \\
a &> (c-b)^{\frac{1}{n}} (\frac{c^{n-1}}{b^{n-1}} + n-1)^{\frac{1}{n}} b^{\frac{n-1}{n}}
\end{align}
$$
$1 < \frac{c}{b}^n < 2$ The lower bound of 1 is trivial.
$$\begin{align}
a^n &= c^n - b^n \\\
\frac{a^n}{b^n} &= \frac{c^n}{b^n} - 1 \\\
\\\
a &< b \\\
a^n &< b^n \\\
\frac{a^n}{b^n} &< 1 \\\
\frac{c^n}{b^n} - 1 &< 1 \\\
\frac{c^n}{b^n} &< 2 \\\
\end{align}$$
In other words: $$c < b\sqrt[n]{2}$$
Weaker bound: $(c-b)n < b$