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.
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.