This is my first naïve attempt at stating and proving a Lemma. Is it valid? Thanks!
Definitions.
Twin Pythagorean Triple. A Pythagorean Triple (a,b,c) for which two values are consecutive integers. Some examples are $(3, 4, 5), (5, 12, 13), (7, 24, 25),…$
Fermat's Last Theorem. For any integer n > 2, the equation $a^n+b^n=c^n$ has no positive integer solutions.
Mihăilescu's Theorem. The integers $2^3$ and $3^2$ are two powers of natural numbers whose values ($8$ and $9$, respectively) are consecutive. The theorem states that this is the only case of two consecutive powers.
Rules.
Assuming FLT has not yet been proved!
Lemma.
For $n$ even, Fermat's Last Theorem has no Twin Pythagorean Triples.
Proof.
Let $n = 2k$ $(k \in \mathbb N, k>1)$
$a^n+b^n=c^n$
can be expressed as
$(a^k)^2+(b^k)^2=(c^k)^2$
Notice the Pythagorean Triple
$(a^k,b^k,c^k)$
Assume Twin Triple, then
$(b^k-a^k = 1)$ or $(c^k-b^k = 1)$
resulting in pairs of two consecutive powers
$(a^k,b^k)$ or $(b^k,c^k)$
However, by Mihăilescu's Theorem, the only case of two consecutive powers is
$(2^3,3^2)$
Consequently
$(b^k-a^k > 1)$ or $(c^k-b^k > 1)$
Therefore, no Twin Triple exists for $k>1$. $\square$
Best Answer
$FLT$ having already been proved it is clear that this is not possible. Then by "rule of the game" we assume that $FLT$ has not yet been proved. $$(x^n)^2+(y^n)^2=(z^n)^2$$ By definition we want that $y^n-x^n=1$ or $z^n-y^n=1$. This is trivially impossible because $$y^n-x^n=(y-x)P(x,y)\\z^n-y^n=(z-y)Q(z,y)$$ Do we need more explanation? I have not understood the problem?