In an elementary "proof" (most likely, a wrong one) of Fermat's Last Theorem, I spotted an inconsistency I want you to confirm. I know, I am probably wasting my time doing this, but still, spotting a fallacy is satisfying.
The author assumes $n$ to be an odd prime, then considers the $n$ th degree Pythagorean triplet to be relatively coprime, with $x<y$ and $h = z-x$. From there, he goes on to prove $1<h<y$ and farther, by expanding binomially $x^n+y^n= (x+h)^n$, the expression for $y^n$ becomes : $y^n= \sum \limits_{p=1}^{n} {n \choose p}x^{n-p}h^p $. From there, he concludes $h \mid y^n$, and thus, $\gcd(y,h)=k>1$. Upto this point, everything is pretty clear.
From here, it gets a bit suspicious:
The proof says: We write $h=\alpha k^r$ and $y^n= \beta k^{m+n} $ for some positive $r$ and nonnegative $m$ and $ k \nmid \alpha, \ k\nmid \beta$. He says, then, CLEARLY : $0 \leq r-1 \leq m$.
Is the inequality at all true? Since, we know that there cannot be a solution, how do we generate a counterexample to this particular argument?
It is indeed pretty easy to construct a counterexample (please do check valid or not) to this seemingly false argument:
Consider $x=4, y = 2135 = 5 \times 7 \times 61, z= 1229 $ (a prime); $h=z-x = 1225 = (5 \times 7)^2.$ $\gcd(y,h)= 35 $; $h = 35^2$, $y^3 = (35)^3 \times 61^3 $. Now, $h \mid y^3$, $n=3$ and $m=0, \ r = 2$ and consequently $0 \leq r-1 \leq m \implies 0 \leq 1 \leq 0$
In the counterexample, we have followed every previous assumptions, i.e $n =$ an odd prime, $1 < h < y$; $ x, y, z$ are coprime (not mentioned pairwise or not) and $x<y$ and the rest of the conditions regarding $ \alpha$, $\beta$.
Please do make sure my counterexample holds.