Simple Method of Solution of $X^p+Y^p=(X+1)^p$ I am looking for simple prove of non existence of solution of $$X^p+Y^p=(X+1)^p$$
for $p>2$
I know it is partial case of Fermat Last Theorem. But I am looking for simple method without going into the complex part of group theory and algebra.
I have simply found from the Fermat Little Theorem that $$Y\equiv1\pmod p$$
But this the maximum I was able to get. Is there any idea how to solve this using simple methods?
 A: EDIT This proof is incorrect.  The calculations are right, I think, but then I jump to a completely unjustified (and untrue) conclusion at the end.  My inclination is to delete this answer, but I'm leaving it at the request of the OP.  
I think I know how to do this.  I'll just give you my idea, and let you fill in the details.
As you said, $Y \equiv 1 \pmod p,$ and as I remarked in my comments, we know $X \equiv 0 \pmod p.$ Write $X=px, Y=py+1,$ and expand the equation by the binomial theorem.$$\begin{align}p^px^p + \sum_{k=0}^p{\binom{p}{k}p^ky^k}&=\sum_{k=0}^p{\binom{p}{k}p^kx^k}\\ \sum_{k=1}^p{\binom{p}{k}p^ky^k}&=\sum_{k=1}^{p-1}{\binom{p}{k}p^kx^k}\end{align}$$ 
Here we have subtracted $p^px^p$ from the sum on the right, subtracted the $k = 0$ term from both sums.(Many thanks to user159517 who pointed out the need for this step.)
Now I claim that $p^3$ divides the coefficient of each of these terms, except the $k=1$ terms.  When $k=1$ the coefficient is $p^2.$  When $k\ge 3$ the coefficient is obviously divisible by $p^3.$  When $k=2,$ the coefficient is $$\binom{p}{2}p^2 = \frac{p(p-1)p^2}{2} \equiv 0 \pmod {p^3},$$
since $p$ is an odd prime.
Now dividing through by $p^2$ and reducing both sides mod $p$ gives $x \equiv y \pmod p$ since all term with $k>1$ vanish.  Now we can say $Y \equiv 1 \pmod {p^2}\text{ and }X \equiv 0 \pmod {p^2}.$ (No we can't!  This is FALSE even in the case $x=y$.) It should be possible to prove by induction that $Y \equiv 1 \pmod {p^n}\text{ and }X \equiv 0 \pmod {p^n}$ for every $n,$  thus proving $X=0, Y=1.$
EDIT Example for $p=3,$ requested by OP.
In this case, we have $X=3x, Y=3y+1,$ so our equation becomes $$\begin{align}(3x)^3+(3y+1)^3 &= (3x+1)^3\\27x^3+27y^3+27y^2+9y+1 &=27x^3+27x^2+9x+1\\27y^3+27y^2+9y&=27x^2+9x\\3y^3+3y^2+y&=3x^2+x\\y &\equiv x \pmod 3\end{align}$$ 
A: This is an open question. See Fermat’s Last Theorem for Amateurs by Paulo Ribenboim. Specifically following results of Catalan are shown: 

For $p$ odd prime number, $0<x<y$ integers such that $x^p+y^p=(y+1)^p$, the following is true:
  
  
*
  
*$py(y+1)$ divides $x^p-1$
  
*$p\nmid x$,  $p\mid x-1$
  
*If $q$ is a prime dividing $y+1-x$ then $q$ divides $x-1$
  
*$\gcd(x+y,y+1-x)=1$
  
*$\gcd(2x-1,2y+1)=1$
  
*$x$ is the only integer such that $(py^{p-1})^{1/p} < x < (p(y+1)^{p-1})^{1/p}$
  

