Zsigmondy's theorem.

First, suppose, $p\mid x$. Let, $p^a \mid \mid x$, and, $p^b \mid \mid y$, and suppose wlog that $a \leq b$. We have, $$ p^{ap}x'^p+p^{bp}y'^p = p^z \implies x'^p + p^{bp-ap}y'^p = p^{z-ap}. $$Now if $z=ap$, we don't have much choice, and simply have a contradiction (LHS $\ge 2$, while, RHS $=1$). If, $z>ap$, then, if $bp-ap>0$, we would get a contradiction, through modulo $p$ (that, $p\mid x'$). Hence, in this case, we must have, $b=a$, and thus, we can wlog assume that $p \nmid x,y$. Now, write, $x^p+y^p=p^z \implies (x+y)(x^{p-1}-x^{p-2}y+\cdots+y^{p-1})=p^z$, assuming, $p$ is an odd prime. We need to study the largest exponent of $p$, dividing $x+y$. $\bullet$ If, $x+y \equiv 0 \pmod{p^2}$, then, $x\equiv - y\pmod{p^2}$. Hence, $$ x^{p-1}-x^{p-2}y + x^{p-3}y^2 - \cdots + y^{p-1} \equiv px^{p-1} \pmod{p^2}. $$Since, $p\nmid x$, we have that, $x^{p-1}=ap+1$ for some $a$ (FLT), thus, $px^{p-1} = p^2 a + p$ for some $a$, hence in particular, it can only be divisible by $p$, not by $p^2$. Therefore, $$ \left(x^{p-1}-x^{p-2}y + x^{p-3}y^2 - \cdots + y^{p-1}\right)^2 = \left(\frac{x^p+y^p}{x+y}\right)^2 \mid x+y \implies (x^p+y^p)^2 \leq (x+y)^3. $$This can be handled via some basic inequalities. Otherwise, suppose that, $x+y = p$. Now, letting $y=p-x$, we have, thanks to binomial expansion that, $$ x^p+y^p = x^p + (p-x)^p = \underbrace{\sum_{i=3}^p \binom{p}{i}p^i (-x)^{p-i}-\frac{p(p-1)}{2}p^2 x^{p-2}}_{\equiv 0 \pmod{p^3}} +p^2 x^{p-1}=up^3 +p^2x^{p-1} = p^2\underbrace{(up+x^{p-1})}_{\equiv 1 \pmod{p}}. $$Hence, not leaving out much cases, unless, $z=2$. Remark Bulgaria 1994.

lte theorem

xdiegolazarox wrote: Zsigmondy's theorem. Would you like to tell me the difference between Zsigmondy theorem and LTE theorem

Here is my solution Let's assume p is greater than $2$ Now it's obvious that p is odd $L.H.S$=$(x+y)(x^{p-1}-x^{p-2}y+\cdots+y^{p-1})=p^z$ Now applying zsigmondy theorem which states That $a^{n}+b^{n}$ has at least one primitive prime factor than $a^{k}+b^{k} $where $n>k$,where n is odd We get that lhs must have a prime other than one dividing $x+y$ which is obsurd unless $x^p+y^p=x+y$ which is only possible when $x=y=1$ so we get a solution $(x, y, p, z) =(1,1,2,1)$ So clearly p is not an odd prime So let's consider the case when p=2 $x^2+y^2=2^z$ we can see that same power of 2 must divide x and y. Let $2^k|x,y$ So $2^{2k}a^2+2^{2k}b^2=2^z$ where α and β are odd. We get $a^2+b^2\equiv2(mod4)$ But $a^2+b^2$ must be a power of 2 Hence $a^2+b^2=2$ and $a=b=1$ So we can infer that $x=y=2^k$ for some positive integer k. EDIT:forgot to mention the exception to zsigmondy theorem which is $2^3+1^3=3^2$ so this is also a solution $(x, y, p, z) =(2,1,3,2)$