Diophantine Equation $ x^n + y^n =z^n (x2) $ I am looking for simple college level algebraic solution to prove that $x$ and $y$ ($x$ < $y$) for the above equation can't be prime numbers. (I know more complex and involved solution using high level of mathematics exists). 
That $y$ can't be a prime number can be shown. But can it be shown for $x$ also ? I know it can be shown for $x$ if we can show, by some simple method, that the following is always false. $$ x^n + y^n =(1+y)^n$$ Can anyone provide me some hints or refer to any online resource. The emphasis is on simple college level algebra. 
I have proved the first part ( $y$ can't be prime ) in the following way:
$$y^n = (z-x)(z^{n-1}+...+x^{n-1})$$
This makes $(z-x)$ have two solutions $(z-x)=1$ or $(z-x)=y^r$ where $r \le n$. It can be easily shown that $z \ne x+1$. Also if $z-x=y^r$ then $x+y > z = x+ y^r$ . which makes $y> y^r$ which is contradictory.
However  as $x<y$ the same is not applicable for $x$. This is where I am stuck.
 A: As a corollary of Zsigmondy's theorem, if $n>2$ and $a,b$ are positive coprime integers, then $a^n-b^n$ has a prime divisor that does not divide $a-b$. Consequently, if  $(x,y,z)$ is a coprime solution to $x^n+y^n=z^n$ ($n>2$) then $z^n-x^n$ and $z^n-y^n$ each have at least two distinct prime divisors. Thus $x$ and $y$ cannot be prime.
A: I understand, Barun, what you are looking for is a proof to a fact of elementary level like, say, “one of x, y, z, must be even” although less easy or maybe difficult but unpretentious. I try here to give you a proof for $n=5$ hoping  you will see the case $p=$ odd prime using this hint.
Recall that it is enough to consider n odd prime and $x, y, z$ coprimes. It is clear for you that if $z-y > 1$ then you finish. One has then being $x=q$ an odd prime (why not $x=2$?)
$q^5 = (y+1)^5 – y^5 = 5y^4+10y^3+10y^2+5y +1$ implies $q^5-1=5y^4+10y^3+10y^2+5y$
$q^5-1=5y^4+10y^3+10y^2+5y=5(y^4+y^3+y^2+y)$ 
 If $q=5$ then $-1≡0 (mod 5)$, absurde, therefore (Fermat's little theorem) $q-1 ≡0 (mod 5)$ and $q= 5m+1$. 
Hence you have
$(5m+1)^5 = 5y^4+10y^3+10y^2+5y +1$;  this gives
$A^5 + 5(A-y)[A^3 + (y+2)A^2 + (y^2+2y+2)A + (y^3+2y^2+2y+1)] = 0$   where $A= 5m$ so
$5^4m^5 + (A-y)[A^3 + (y+2)A^2 + (y^2+2y+2)A + (y^3+2y^2+2y+1)] = 0$ therefore either
$5|(A-y)$  or $5|(y^3+2y^2+2y+1)$  (I leave you the job to see that the first is not possible).
Thus $ y^3+2y^2+2y+1 ≡0 (mod 5)$. This is  possible just for $y=5n-1$ because making $f(x) = y^3+2y^2+2y+1$ we have modulo 5, $(f(0), f(1), f(2), f(3), f(4)) = (1, 1, 1,2,0)$. Hence you have the restriction $y=5n-1$.
I stop here. I wanted help you but the required job seems not easy. I do not remember seeing it on the big miscellany of partial conditions on FLT before Wiles & Company so do not rule out that your problem could be difficult enough . I wanted a proof for you but I could not. 
