A proof of Fermat´s Last Theorem using only Gauss´s Lemma for the roots of monic polynomials with integer coefficients.**
Writing the Fermat equation $$a^n + b^n - c^n = (c-p)^n + (c-q)^n - c^n = 0$$ with $q < p$ being integers and c taken as an independent variable we get for $n=3$ the polynomial equation $$F_3(c,p,q) = c^3 - 3(p + q)c^2 + 3(p^2+q^2)c -(p^3+q^3) = 0.$$ In order to preserve the parity always two of $p, q, c$ must be odd and one even. So for an odd $c$ one of $p$ or $q$ must be odd and the other even (first case), whereas for an even $c$ both $p$ and $q$ must be odd (second case). Therefore in both cases $F_3(c,p,q)$ has an even value and might be zero because zero is an even number.
In order to show that this never can be we compare in the first case the roots of $F_3(c,p,q)=0$ to the roots of $F_3(c,2p,2q)=0$, which for an odd $c$ is odd and therefore cannot be zero. Hence $F_3(c,2p,2q)=0$ according to Gauss´s Lemma must have irrational roots which according to Vieta´s formulas for the roots and coefficients of polynomials are twice the roots of $F_3(c,p,q)=0$, which therefore are also irrational. Thus $F_3(c,p,q)$ in the first case never can be zero for an odd $c$.
In the second case $c$ is even and can be written as $2^t*d$, where $d$ is the odd part of $c$ and $t$ is an integer. Rewriting $F_3(c,p,q)$ as $$F_3(2^td, 2^tp,2^tq) = 2^3t [d^3 – 3 (p + q)d^2 + 3 (p^2+q^2)d -(p^3+q^3)] = 0$$ which never can be zero because of the odd value of the expression in square brackets.
With the substitution of $p$ and $q$ by $2^tp$ and $2^tq$ the roots of $F_3(2^td,2^tp,2^tq)=0$ are $2^t$-fold the roots of $F_3(c,p,q)=0$ and are irrational according to Gauss´s Lemma. Thus the roots of $F_3(c,p,q)=0$ are also irrational which shows that it in the second case also never can be zero for an even $c$.
Therefore the Fermat equation $a^3 + b^3 - c^3 = 0$ for $n=3$ never can be zero and thus cannot have a solution in integers $a,b,c$. This result can easily be extended to all degrees $n$ being odd primes using the features of the binomial coefficients, especially the fact that for all $n$ there is always an even number of odd coefficients (see, please, https://www.researchgate.net/publication/338819565_Proving_Fermats_Last_Theorem_as_a_Mean_Value_Problem_Title_Proving_Fermats_Last_Theorem_as_a_Mean_Value_Problem_Author).
The flaw seems to be in reasoning that if $F_3(x,2p,2q)$ for odd $x=c$ is odd, and so cannot have rational root eventually, then $F_3(x,p,q)$ could not have a rational root for $x=c$ odd. But, $c$ was supposed to be odd in $F_3(x,p,q)$ with $x=c$, while in $F_3(x,2p,2q)$ we expect a root $x=2c$, which is even. You have incorrectly expected that the $F_3(x,2p,2q)$ would have to have its root odd as well.
If you still don't see the flaw, notice that your proof would show that the original equation had no solution, but that is not true. There are known trivial solutions, one is corresponding to $c=1,p=1,q=0$. Here $F_3(x,p,q)=(x-1)^3$ (an odd root), but $F_3(x,2p,2q)=(x-2)^3$ (an even root). Another missed solution is $p=q=c=0$. Try to go through your argument with these specific values and you will see the issue yourself.