Unique approach at certain polynomials of n degree

Lets consider following equations for integers
Z;A;B where Z>A and Z>B and also Z;A;B of gcd=1 :
what substitutes X^n + Y^n = Z^n
for X = Z-A ; Y = Z-B;
For n=2 :
Z^2 -2(A+B)Z +A^2 +B^2 = 0
Essentially such equation for integers means,
that A^2 +B^2 polynomial should be divided by Z.
Using well known Pythagorean triplets and some
substitution of them we can show it clear...
For n=3 :
Z^3 - 3(A+B)Z^2 +3(A^2 +B^2)Z -A^3 - B^3 = 0
Using properly developed parameters we can write:
For the case Y/3 once Y = T+A; X = T+B;
Z = t*s ; s = t^2 -3^u ab; A = 3^(3u-1) a^3;
B = b^3; and from Z = A+B+T; T = 3^u abt
A+B = t(s -3^u ab)
then slightly to show Z|(A^3 + B^3):
A^3 + B^3 = (A+B)[(A+B)^2 -3AB ]
where (A+B)/t but not s of gcd=1 to t;a;b;3
then 3^2u a^2 b^2 t^2 - 3^3u a^3 b^3 =
3^2u a^2 b^2(t^2 - 3^u ab ) /s
and it looks like such parameters could be
some true solution once using them:
Z = t*s|(A^3 + B^3)
(similar for the case Z/3 : Z = 3^u t*s
once X/3 means B = 3^(3u-1) and is inverse
of Y/3 )
HOWEVER FOR n=5 AND FOR BIGGER PRIMES A^n +B^n
POLYNOMIAL WOULD LIKE TO SHOW SOME
SIMPLE IMPOSSIBILITY FOR DIVISION BY Z :
once Z = t*s or Z = n^u t*s
and appropriate A = a^n or n^(nu-1) a^n
and let B = b^n ...
Could it be taken as some valid proof ?
Once from n=5 and bigger primes it should be
recognized one more parameter: let it be p
then once a;b;t;s;p;n of gcd =1
so from A^n +B^n division by Z
looks as a or b should be anyhow divided by n
for 1-st case of FLT what is not permitted,
then for 2-nd case of FLT as p should be divided
by n what is not permitted ...
Also once p of gcd=1 to a and b so it could
be only as p/n or p=1.

I'll show more details very soon.
Best Regards
Ro-Bin
.