Modulo $5$, the relation becomes $1 \equiv 2^5 - 3^5 = -211 \pmod{5}$, which is patently false.

My solution is pretty messy. Please tell me if I made a mistake, I'm pretty new to this! Expand $(x+2)^5$ to get $x^5+10x^4+40x^3+80x^2+80x+32$. We replace the LHS by $(x+2)^5-10x^4-40x^3-80x^2-80x-31+y^5=(x+2)^5+(y-3)^5$ so $-10x^4-40x^3-80x^2-80x-31+y^5=(y-3)^5$ Modulo 3, this becomes: $-x^4-x^3+x^2+x = 1\pmod{3}$ If we assume x is a multiple of 3, then a contradiction arises, so we know x is relatively prime to 3. So we can use FLT to get: $-x^4-x+1+x=1\pmod{3}$ which reduces to: $x=0\pmod{3}$ which is a contradiction, so there are no solutions.

By Fermat's Little theorem $x+y+1\equiv (x+2)+(y-3)\pmod{5}$, so $1\equiv -1\pmod{5}$, impossible.

gethd wrote: By Fermat's Little theorem $x+y+1\equiv (x+2)+(y-3)\pmod{5}$, so $1\equiv -1\pmod{5}$, impossible. Isn't a Fermat's Little theorem is applied for only in case x or y is relatively prime to p (in our case p=5)? I mean you didn't consider the case when x and y are divisible by 5.

The relation $a^5\equiv a\pmod{5}$ is true for all $a$. And am I the only one that noticed "are are" in the question?

WakeUp wrote: Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

it is easy to prove that $x^5\equiv x mod 3$ so we must find all x,y such that $x+y+1\equiv x+y-1 mod 3$ which is contradiction.