I like this problem!

If $p=2$ then any $(x,y)$ with $x+y$ a power of two is okay. Hence assume $p > 2$. Then if $\nu_p(x) \ge \nu_p(y) > 0$ we get an immediate contradiction, thus we may assume $p \nmid x,y$. So by Fermat's Little Theorem, $x$ and $y$ are $-1 \pmod p$. It is easy to check that when $p > 2$ we cannot have $x=y$, since otherwise $x(x^{p-2}+1)$ is a power of $p$, which is clearly impossible when $p=2$. Moreover, if $p > 2$ then $x^{p-1}+y \neq y^{p-1}+x$, since otherwise $(x-y)(x^{p-2}+\dots+y^{p-2}) = x-y$, which is impossible unless $x=y$. Thus, suppose $y^{p-1}+x < x^{p-1}+y$, which is equivalent to $y < x$. Then in particular $y^{p-1}+x$ divides $x^{p-1}+y$, so \[ y^{p-1} + x \mid (-y^{p-1})^p + y \implies y^{p-1}+x \mid y^{p(p-2)}+1. \] By Lifting the Exponent, we thus deduce that \[ \nu_p(y^{p-1}+x) \le 1 + \nu_p(y+1). \] Actually, since LHS is a power of $p$, this informs us that \[ y^{p-1} + x \mid p \cdot (y+1) \implies y^{p-1} + x \le p \cdot (y+1). \] Since $x > y$, this forces \[ y^{p-1} + y \le p \cdot (y+1). \] Also, $y \ge p-1$ since $y \equiv -1 \pmod p$. This can only occur if $y = 2$ and $p = 3$. So, it remains to find all $x > 2$ such that $x^2+2$ and $4+x$ are powers of $3$. Letting $4+x=3^b$, we find that \[ \left( 3^b-4 \right)^2 + 2 = 3^{2b} - 8 \cdot 3^b + 18 \] is supposed to be a power of $3$, which can only occur for $b \le 1$. Checking, this gives the last solution $(x,y,p) = (5,2,3)$.

have another solution: Obviously if $p=2, x+y=2^s$ is a solution if the problem so assume that $p\ge 3$ and let $x^{p-1}+y=p^{\gamma}(1), x+y^{p-1}=p^{\beta}$(2) Consider two cases: 1) $p\mid x$ then it's obvious that $p\mid y$ let $v_p (x)=r, v_p (y)=t$ then from (1),(2) we get $(p-1)r=t,(p-1)t=r$ which is contradiction for $p\ge 3$ so there isn't any solution in this case. 2) $p\nmid x,y$ note that if $x=y$ then $x^{p-1}+x=p^{\gamma} \longrightarrow p\mid x$ which is a contradiction so without loss of generality assume that $x>y\longrightarrow \gamma>\beta$ note that from (1): $y=p^{\gamma}-x^{p-1}$ substituting this in (2) we get $x+(p^{\gamma}-x^{p-1})^{p-1}=p^{\beta}$(3) now taking this equality modulo $p^{\beta+1}$ we have $x+(p^{\gamma}-x^{p-1})^{p-1}\equiv x+x^{(p-1)^2}\equiv p^{\beta}\pmod{p^{\beta+1}}\longrightarrow v_p (x^{(p-1)^2-1}+1)=\beta$ using lifting exponent lemma we get $\beta=v_p (x^{(p-1)^2-1}+1)=v_p (x+1)+v_p ((p-1)^2-1)=v_p (x+1)+1\longrightarrow v_p (x+1)=\beta-1\longrightarrow x\equiv -1\pmod{p^{\beta-1}}, x=p^{\beta-1}l+1$(4). Now notice that using (4): $p^{\gamma}-x^{p-1}\equiv -1\pmod{p^{\beta-1}}\longrightarrow p^{\gamma}-x^{p-1}=p^{\beta-1}r-1$ now substituting this in (3) we get $x+(p^{\beta-1}r-1)^{p-1}=p^{\beta}$ but since $p-1\ge 2, r\ge 1, x=p^{\beta-1}l-1$: $p^{\beta}=x+(p^{\beta-1}r-1)^{p-1}\ge (p^{\beta-1}-1)^{p-1}+p^{\beta-1}-1=(p^{\beta-1}-1)(1+(p^{\beta-1}-1)^{p-2})$ from this inquality we immediately get $p=3,\beta=2$ which gives us the solution $(3,5,2)$. Q.E.D

If $p=2$ then any $x+y$ being a power of two will do.Let $p \ge 3$. Let $x^{p-1}+y=p^k$ and $y^{p-1}+x=p^l$.If $x=y$ then we have $x(x^{p-2}+1)=p^k$ which is not possible because $gcd(x,x^{p-2}+1)=1$.So wlog $x>y$.Then $k>l$. If $p|x,y$ then let $p^{\alpha}||x,p^{\beta}||y$.Then we must have $\alpha(p-1)=\beta$ and $\beta(p-1)=\alpha$ simultaneously which is not possible.So now we work on the case when $x,y$ are not divisible by $p$. Let $p^j||x-y$.From the two relations it is easy to see that $x^p-y^p=p^kx-yp^l$.Thus $v_p(p^kx-p^ly)=j+1 \implies l=j+1$.Thus $y^{p-1}+x=p^{j+1} \implies y^{p-1}+y+x-y=p^{j+1} \implies p^j||y^{p-1}+y \implies p^j||y^{p-2}+1=(y+1)(y^{p-3}-y^{p-4}+\cdots+1)$.From $y \equiv -1{\pmod{p}}$ we get that the second factor leaves remainder $p-2$ upon division by $p$.So $p^j||y+1 \implies y \ge p^jl-1$.Thus $p^{j+1}=y^{p-1}+x>y^{p-1} \ge (p^j-1)^{p-1}$.This is not possible for $p>3$.Also for $p=3$ it is possible only for $j=1$.Checking into this possibility we have to have $y^2+x=9$.The only solution in $(x,y)$ satisfying this such that $x^2+y$ is also a power of three is $(2,5)$. So the solutions are $(2,x,y)$ where $x+y$ is a power of two,$(3,2,5),(3,5,2)$.

If $p = 2$ then $x + y$ a power of two works. Assume $p \geq 3$ Let $x^{p - 1} + y = p^{\alpha}$ and $y^{p - 1} + x = p^{\beta}$. If one of $x, y$ is divisible by $p$ then the other is also divisible by $p$, in this case we have $\upsilon_p(y) = \upsilon_p(x^{p - 1}) = (p - 1)\upsilon_p(x)$ as otherwise $x^{p - 1} + y > p^{\max\{(p - 1)\upsilon_p(x), \upsilon_p(y)\}}$ but this number does not divide $x^{p - 1} + y$. Similarly we have $\upsilon_p(x) = (p - 1)\upsilon_p(y)$, but it is impossible for these two relations to hold as $p > 2$. Hence $x, y$ are not divisible by $p$ and so by FLT they are congruent to $-1 \mod{p}$. Additionally we check that $\alpha \neq \beta$ as otherwise we have $\displaystyle\frac{x^{p - 1} - y^{p - 1}}{x - y} = 1$ which is impossible as $p - 1 > 1$. We also have $x \neq y$ as $x(x^{p - 2} + 1)$ is not a power of $p$ since $p \nmid x$. Now assume $x > y$ so $\alpha > \beta$. We have $x^p + xy = p^{\alpha}x$ and $y^p + xy = p^{\beta}y$. Substracting gives us $x^p - y^p = p^{\alpha}x - p^{\beta}y$ so that $\upsilon_p(x - y) + 1 = \upsilon_p(x^p - y^p) = \beta$ by LTE. Put $x - y = p^{\beta - 1}k$. Substituting gives us $y^{p - 1} + p^{\beta - 1}k + y = p^{\beta} \implies y(y^{p - 2} + 1) = p^{\beta - 1}(p - k)$. As $p \nmid y$ this implies $p^{\beta - 1} \mid (y^{p - 2} + 1)$. Thus $$p^{\beta -1}(p - 1) \geq p^{\beta - 1}(p - k) = y(y^{p - 2} + 1) \geq (p - 1)(p^{\beta - 1})$$Where the last equality follows from $p \mid y + 1 \implies y \geq p - 1$. Thus $y = p - 1$, $k = 1$ and $y^{p - 2} + 1 = p^{\beta - 1}$ as all equalities must hold. Thus by LTE $\beta - 1 = \upsilon_p(y^{p - 2} + 1) = \upsilon_p(y + 1) = 1$ as $p - 2$ is odd and $y + 1 = p$. Hence $\beta = 2$. Thus $(p - 1)^{p - 2} + 1 = p$ giving $p = 3$. Finally $y = 2$ and $x = 5$ from $x = p^{\beta - 1}k + y$, which clearly works.

If $p = 2$, any pair $(x, y)$ with $x + y$ a power of $2$ works. Henceforth, assume that $p \ne 2$ and write $x^{p - 1} + y = p^{\alpha}$ and $x + y^{p - 1} = p^{\beta}$ with WLOG $1 \le \alpha \le \beta.$ Clearly, \[0 \equiv y^{p - 2} \cdot p^{\alpha} - p^{\beta} \equiv x\left[(xy)^{p - 2} - 1\right] \pmod{p^{\alpha}}. \quad (\star)\]If $p \mid xy$, then evidently $p \nmid (xy)^{p - 2} - 1.$ Hence, $(\star)$ forces $x \equiv 0 \pmod{p^{\alpha}}$, which is obviously false for size reasons. Therefore $p \nmid xy$, so from $(\star)$ we obtain $(xy)^{p - 2} \equiv 1 \pmod{p^{\alpha}}.$ Moreover, Fermat's Little Theorem applied to $x^{p - 1} + y = p^{\alpha}$ and $x + y^{p - 1} = p^{\beta}$ gives $x \equiv y \equiv -1 \pmod{p}.$ Hence, $p \mid xy - 1$, so LTE yields \[\alpha \le \nu_p\left[(xy)^{p - 2} - 1\right] = \nu_p(xy - 1) \implies xy \equiv 1 \pmod{p^{\alpha}}.\]Consequently, \[x^{p} + xy = x \cdot p^{\alpha} \implies x^{p} + 1 \equiv 0 \pmod{p^{\alpha}}.\]Lifting the Exponent, \[\alpha \le \nu_p(x^p + 1) = 1 + \nu_p(x + 1) \implies p^{\alpha} \le p^{1 + \nu_p(x + 1)} \le p(x + 1).\]Substituting $x^{p - 1} + y = p^{\alpha}$, it follows that \[x(x^{p - 2} - p) \le p - y. \quad (\dagger)\]On account of $y \equiv -1 \pmod{p}$, we see that $p - y \le 1.$ Therefore, $x^{p - 2} - p \le 1$, and since $x \equiv -1 \pmod{p}$, this forces $p - 2 = 1$ and $x = p - 1.$ Thus, $p = 3, x = 2$ and so $-2 \le 3 - y$ by $(\dagger).$ Using $y \equiv -1 \pmod{p}$, it remains to check $y \in \{2, 5\}$, which gives the solution $(p, x, y) = (3, 2, 5)$ and its conjugate $(p, x, y) = (3, 5, 2).$

If $p=2$, then $x+y$ is power of $2$, then any triplets in form of $(2,x,2^k-x), x<2^k, k\in \mathbb{N}$, are the only solutions. So we assume $p\ge 3$. Claim 1. $x,y\equiv -1 \pmod p$. Proof. If $p\mid x$, then $x^{p-1}+y$ is a power of $p$ greater than $p^{p-1}$, so $p^{p-1}\mid x^{p-1}+y \Rightarrow p^{p-1}\mid y$, and so $x+y^{p-1}$ is a power of $p$ greater than $p^{(p-1)^2}$, so $p^{(p-1)^2}\mid x+y^{p-1} \Rightarrow p^{(p-1)^2}|x$, and we can continue this process to get $p^{(p-1)^3}\mid y \Rightarrow p^{(p-1)^4}\mid x \Rightarrow p^{(p-1)^5}\mid y \Rightarrow \cdots $, a contradiction since $p-1 \ge 2$ we have $v_p(x)\ge (p-1)^{2n}$ for all $n$ which tends to infinity for large $n$, which cannot hold. So $p$ does not divides $x$, so as $y$. By Fermat Little Theorem, $x^{p-1}\equiv 1 \pmod p \Rightarrow y\equiv -1\pmod p$. Similarly, $x\equiv -1 \pmod p$. Claim 2. $p^2$ does not divides $x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1$. Proof. By Lifting the Exponent Lemma, since from Claim 1, $p|x+1$, and $p$ does not divides $x$ and $1$, we have $v_p(x^{(p-1)^2-1}+1)= v_p(x+1)+v_p((p-1)^2-1)= v_p(x+1)+1$. Now note that $x^{(p-1)^2-1}+1=(x+1)(x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1)\Rightarrow v_p(x^{(p-1)^2-1}+1)= v_p(x+1)+v_p(x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1) \Rightarrow v_p(x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1)=1$. Claim 3. $x^{p-1}+y\mid x^{(p-1)^2-1}+1$. Proof. Suppose $x\le y$, then since $x^{p-1}+y$ and $x+y^{p-1} $ are both distinct powers of $p$, then $x^{p-1}+y\mid x+y^{p-1}$, and since $p$ is odd, $x^{p-1}+y\mid x^{p(p-1)}+y^p$. Together we have $x^{p-1}+y\mid x^{p(p-1)}+y^{p}-y(x+y^{p-1}) \Rightarrow x^{p-1}+y\mid x^{p(p-1)}-xy $. Since $p$ does not divides $x$, $(p,x)=1$, so $ x^{p-1}+y\mid x^{p(p-1)}-xy \Rightarrow x^{p-1}+y\mid x^{p(p-1)-1}-y \Rightarrow x^{p-1}+y\mid x^{p(p-1)-1}-y+(x^{p-1}+y) \Rightarrow x^{p-1}+y\mid x^{p(p-1)-1}+x^{p-1} \Rightarrow x^{p-1}+y\mid x^{(p-1)^2-1}+1$. Claim 4. $x^{p-1}\le p(x+1)$ Proof. From Claim 3, $x^{p-1}+y\mid x^{(p-1)^2-1}+1=(x+1)(x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1)$. Since from Claim 2, $p\mid\mid x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1 $, then we have $x^{p-1}+y\mid (x+1)(x^{(p-1)^2-2}-x^{(p-1)^2-3}+\cdots +1) \Rightarrow x^{p-1}+y\mid p(x+1) \Rightarrow x^{p-1}< p(x+1) $. Claim 5. The only solutions for odd primes $p$ are $(3,2,5)$ and $(3,5,2)$. Proof. From Claim 4, $x^{p-1}< p(x+1) \Rightarrow x^{p-1}-px-1< 0$. The derivative of the function $f(x)=x^{p-1}-px+1$ is $(p-1)x^{p-2}-p$, which is greater than $0$ for all $x\ge 2$ and for all $p$. Since $f(2)\ge 0$ for $p\ge 5$, $f(x)< 0$ for $x\ge p-1\ge 2$ has no solutions. So the only case is $p=3$, which we get $x^2< 3(x+1) \Rightarrow x\le 3$, but since $x\equiv -1 \pmod 3$, then $x=2$. Then $4+y, 2+y^2$ is power of $3$, so $$y+4\mid 2+y^2=(y+4)(y-4)+18 \Rightarrow y+4\mid 18$$Since $y+4$ is power of $3$ greater than $4$ and divisor of $18$, we must have $y+4=9 \Rightarrow y=5$. So $(3,2,5)$ is a solution for $x\le y$, and by symmetry, $(3,5,2)$ is also a solution. So the only solutions for odd primes $p$ are $(3,2,5)$ and $(3,5,2)$. In conclusion, the only solutions are $(2,x,2^k-x), x<2^k, k\in \mathbb{N}$, $(3,2,5)$ and $(3,5,2)$, and these are easily checked to be solutions.

LTE and use some inequalitites Which can be proven by İnduction.

Let's first take the obvious case of $p=2$. We can just take $x=k, y=2^n-k$ where $n$ and $k$ are positive integers. Suppose that $x^{p-1}+y=p^k \leq y^{p-1}+x=p^r$. Then $k \leq r$. Notice that $x^{p-1}+y=p^k$, so we can say $x^{p-1} \equiv -y \mod p^k$. Similarly we get $y^{p-1} \equiv -x \mod p^k$. We can deduce that $x^{(p-1)^2} \equiv -x \mod p$ and since $p$ does not divide $x,y$ for $p>2$, we can deduce that $x^{2p(p-2)} \equiv 1 \mod p^k$. By FLT and the gcd lemma, we can see that $x^{gcd((p-1)p^{k-1},2p(p-2)} \equiv x^{2p} \mod p^k$. For size reasons, we immediately rule out the possibilities of $p^k|x-1,x+1$, so $p^k|x^p-1$ or $p^k|x^p+1$. Notice that FLT on the original equations gives $x,y \equiv -1 \mod p$, so it must be the second case. LTE now gives $k \leq v_p(x+1)+1$, implying $x \geq p^{k-1}-1$. This allows us to see that $x^{p-1}+y \geq x^{p-1}+1 \geq (p^{k-1}-1)^{p-1}+1>p^k$, giving contradiction for all odd primes as long as $k>3$. First we handle $x^{p-1}+y=p$. We get that $x<2$, implying $x+1=2$, contradiction. Now consider $x^{p-1}+y=p^2$. We can see that $x$ must be no more than $3$ for $p \geq 3$, implying that $x+1=2,3,4$, only the middle of which is divisible by an odd prime. Similarly $x^{p-1}+y=p^3$, implying that $x \leq 5$, implying that $x+1=2,3,4,5,6$, only $3$ and $5$ of which are odd. However $5$ fails because $4^{p-1}>p^3$ for $p>3$ and $(4,11)$ fails. Thus $p=3$. So $x^2+y$ and $y^2+x$ are both powers of $3$. $x^2+y$ must then equal $9,27$. From here, we can easily get the only other solution to be $x=2, y=5$. Thus all solutions are in the forms $(2,x,y)$ where $x+y$ is a power of two,$(3,2,5),(3,5,2)$.

Solution $\boxed{Claim-1}$ $\boxed{p\mid x \implies no solution}$ Proof::if $p\mid x$ then $p\mid y$ but we can see that $v_p({x^{p-1}})=v_p({y})$ and $v_p({y^{p-1}})=v_p({x})$ which is absurd.$\square$ $\boxed{Claim-2}$ $\boxed{p\nmid x\implies p=2,3 }$ Proof :we assume W.L.O.G $y>x$ and $p\geq 3$ Denote $x^{p-1}+y=p^k$,$y^{p-1}+x=p^l$ $x^{p-1}+y\equiv 1+y\pmod{p}$ $y^{p-1}+x\equiv 1+x\pmod{p}$ Now as $y>x$ $x^{p-1}+y \mid y^{p-1}+x$ $$\implies x^{p-1}+y \mid x^{p-2}y^{p-1}+x^{p-1}$$$$\implies x^{p-1}+y \mid x^{p-2}y^{p-1}+x^{p-1}-(x^{p-1}+y)$$$$\implies x^{p-1}+y \mid (xy)^{p-2}-1$$By F.L.T,$(xy)^{p^{k-1}.(p-1)}\equiv 1\pmod {p^k}$ $(p^{k-1}.(p-1),(p-2))=1$ for$p\geq 3$ $\implies (xy)\equiv 1\pmod {p^k}$ $$\implies x^p+xy\equiv 0\pmod {p^k}$$$$\implies x^p\equiv -1\pmod {p^k}$$Lifting the exponent ,

THUS our claims are proved and Now we are left with case $p=2$ we have infinite solutions with $x+y=2^k$$\blacksquare$ $Q.E.D$

Bashy solution: The solutions are $p=2$ and any $x,y$ with $x+y$ is a power of $2$, and the solution $(p=5,\{x,y\}=\{2,5\})$. It is easy to check that these work. Firstly, if $p=2$, then we must have $x+y$ is a power of $2$, and we have that solution. From now on, we assume $p\ge 3$. Claim 1: If $p\nmid x,y$. Proof of Claim 1: Suppose $p^k\mid x,y$ with $k\ge 1$, and furthermore suppose this is the largest $k$ with the previous properties. Let $x=p^ka$ and $y=p^kb$. Then, \[p^r=x^{p-1}+y=(p^ka)^{p-1}+p^kb.\]Firstly, note that $r>k(p-1)$. Note that $k(p-1)\ge 2k>k$, so taking mod $p^{k(p-1)}$, we have that \[p^{k(p-1)}\mid p^k b,\]so in particular, $p\mid b$. We can very similarly show that $p\mid a$, so we have $p^{k+1}\mid x,y$, which is the desired contradiction. $\blacksquare$ Note by Fermat's little theorem, we must have that $y+1,x+1\equiv 0\pmod{p}$. Let $x=p^ka-1$ and $y=p^kb-1$ where $k\ge 1$, and is again the greatest $k$ with given properties. Claim 2: Unless $p=3$ and $k=1$ and $(a=1\text{ or }b=1)$, we have that $x^{p-1}+y,y^{p-1}+x> p^{k+1}$. Proof of Claim 2: Suppose $x^{p-1}+y\le p^{k+1}$. Therefore, $x^{p-1}<p^{k+1}$, so \[(p^k a-1)^{p-1}<p^{k+1}.\]If $a\ge 2$, then \[(p^ka-1)^{p-1}\ge p^{k(p-1)}\ge p^{2k}\ge p^{k+1},\]which is a contradiction. Therefore, $a=1$, so \[(p^k-1)^{p-1}<p^{k+1}.\]If $k\ge 3$, then \[(p^k-1)^{p-1}\ge p^{2(k-1)}\ge p^{k+1},\]which is a contradiction, so $k\le 2$. If $k=2$, then we have \[(p^2-1)^{p-1}\ge (p^2-1)^2,\]and it is easy to see that for $p\ge 3$, we have $(p^2-1)^2\ge p^3$, so there is a contradiction. Thus, $k=1$, so \[(p-1)^{p-1}<p^2.\]It is easy to see that the only solution for $p\ge 3$ is $p=3$, so we must have $p=3$. Therefore, in this case $p=3,k=1,a=1$. In the case $y^{p-1}+x<p^{k+1}$, we have $p=3,k=1,b=1$, so the claim is proved. $\blacksquare$ We will revisit the case $p=3,k=1,(a=1\text{ or }b=1)$ later, so for now assume that $x^{p-1}+y,y^{p-1}+x\ge p^{k+1}$. We see that for $r\ge k+2$, we have that \[p^r=(p^ka-1)^{p-1}+(p^kb-1)=p^k(a+b)+ap^{k+1}\left[-1+a\binom{p-1}{2}p^{k-1}\right]+p^{k+2}(\text{other integer terms}).\]Therefore, taking mod $p^{k+1}$, we see that $p\mid a+b$, and taking mod $p^{k+2}$, we have that \[a\left[-1+a\binom{p-1}{2}p^{k-1}\right]\equiv -\frac{a+b}{p}\pmod{p}.\]Doing the same for $y^{p-1}+x$, we see that \[a\left[-1+a\binom{p-1}{2}p^{k-1}\right]\equiv b\left[-1+b\binom{p-1}{2}p^{k-1}\right]\pmod{p}.\]If $k=1$, then $a(a-1)\equiv b(b-1)\pmod{p}$, so $a^2-b^2\equiv b-a\pmod{p}$, so $a\equiv b\pmod{p}$. Combined with $p\mid a+b$, this means that $p\mid a,b$. If $k\ge 2$, then this simply says that $-a\equiv -b\pmod{p}$, so again $p\mid a,b$. However, this violates the maximality of $k$, which is a contradiction. Therefore, we must have that $p=3,k=1$, and WLOG $a=1$. Therefore, $x=2$ and $y=3b-1$. Thus, $2^2+(3b-1)=3(b+1)$ is a power of $3$, so $b+1$ is a power of $3$. Let $b=3^h-1$. We also have that $(3b-1)^2+2$ is a power of $3$, or that $9b^2-6b+3=3(3b^2-2b+1)$ is, so $3b^2-2b+1$ is a power of $3$. Therefore, \[3(3^h-1)^2-2(3^h-1)+1=3^{2h+1}-2\cdot 3^{h+1}+3-2\cdot 3^h+2+1\]is a power of $3$. If $h\ge 2$, then taking mod $9$ is a contradiction, so we must have $h=1$, so $b=2$. Therefore, $x=2,y=5,p=3$, which is the only other claimed solution. $\blacksquare$

ANSWER: The answers are $(p,x,y)=(2,x,2^z-x) \text{ or } (3,2,5) \text{ or } (3,5,2)$. It's easy to see that this works. SOLUTION: The case $p=2$ easily gives the first set of answers. So from now on assume that $p$ is an odd prime. Let $$x^{p-1}+y=p^m \text{ and } x+y^{p-1}=p^n \text{ where } m,n \in \mathbb{N}$$It's easy to see that $x=y$ gives no solutions, cause then $x(x^{p-2}+1)$ is a power of $p$, which is not possible for $p>2$ (as $x(x^{p-2}+1)$ must be an even integer). So WLOG assume that $x>y$, which also gives $m>n$. Our solution uses the following important claim- CLAIM| The integers $x$ and $y$ are of the form $x=ap^{n-1}-1$ and $y=bp^{n-1}-1$ for some positive integers $a$ and $b$ such that $p \nmid a,b$. Proof of Claim Our proof to this claim is divided into three parts as given below. We first show that $p \nmid x,y$. Suppose, to the contrary, that $p \mid x$. Then $p \mid y$ must also hold true. WLOG take $\nu_p(x) \geq \nu_p(y)$. Then this means that $\nu_p(x^{p-1})>\nu_p(y)$, and so we get that $\nu_p(x^{p-1}+y)=\nu_p(y)$, or equivalently that $\nu_p(y)=m$. However, $p^m=x^{p-1}+y>y \Rightarrow m>\nu_p(y)$, and so we arrive at a contradiction. Thus, we must have that $p \nmid x,y$. $\text{ }$ Now we will show that $\nu_p(x+1)=\nu_p(y+1)>0$. Note that, as $p \nmid x,y$, so by FLT, we have $0 \equiv x^{p-1}+y \equiv 1+y \pmod{p}$, and so $p \mid y+1$. In a similar fashion, we get that $p \mid x+1$. Again (for brevity) take $\nu_p(x+1)=r$ and $\nu_p(y+1)=s$. Then we wish to prove that $r=s$. So let us assume to the contrary that $r \neq s$. Now, by Lifting The Exponent Lemma, we have $$\nu_p(x^{p-1}+x)=\nu_p(x)+\nu_p(x^{p-2}+1)=0+\nu_p(x+1)+\nu_p(p-2)=r$$Similarly, we get that $\nu_p(y^{p-1}+y)=s$. Then, using our assumption (i.e. $r \neq s$), we have $$\nu_p(p^m+p^n)=\nu_p((x^{p-1}+x)+(y^{p-1}+y)) \Rightarrow \min(r,s)=n$$ If $r=n$, then $x+1=kp^n \Rightarrow x+1=k(x+y^{p-1})$, which is not possible unless $y=k=1$. But then $$p^m=x^{p-1}+y=(p^n-1)^{p-1}+1 \Rightarrow 0 \equiv (p^n-1)^{p-1}+1 \equiv 2 \pmod{p} \rightarrow \text{CONTRADICTION}$$ If $s=n$, then $y+1=kp^n \Rightarrow y+1=k(x+y^{p-1})$, which is not possible because $x>y$. Thus, we arrive at a contradiction in both the cases, and so we must have $r=s$. $\text{ }$ Take $\nu_p(x+1)=\nu_p(y+1)=e$, i.e. $x+1=ap^e$ and $y+1=bp^e$, where $\nu_p(a)=\nu_p(b)=0$. We wish to show that $e=n-1$. Note that $p^e \leq x+1<x+y^{p-1}=p^n$, and so we must have $e<n$. As $n<m$, so we have $e<n<m$. Then we get that $$p^m=x^{p-1}+y=(ap^e-1)^{p-1}+(bp^e-1)=(1-a(p-1)p^e+ \dots +(ap^e)^{p-1})+(bp^e-1)=p^e(b-a(p-1)+wp^e)$$where $w$ is a positive integer. As $m>e>0$, this gives that $$a+b-ap+wp^e=p^{m-e} \Rightarrow p \mid a+b$$Now, notice that $p \nmid a-b$, since otherwise $p \mid (a-b)+(a+b) \Rightarrow p \mid 2a$, which is not possible as $\nu_p(2)=\nu_p(a)=0$. So this means that $\nu_p(x-y)=\nu_p(p^e(a-b))=e$. Then, by LTE, we have that $$\nu_p(x^p-y^p)=\nu_p(x-y)+\nu_p(p)=e+1$$But, using our original equations, we get $$x^p-y^p=x(x^{p-1}+y)-y(x+y^{p-1})=xp^m-yp^n=p^n(xp^{m-n}-y) \Rightarrow p^n \mid x^p-y^p$$$$\Rightarrow n \mid \nu_p(x^p-y^p) \Rightarrow n \mid e+1 \Rightarrow n \leq e+1 \Rightarrow \text{As } e<n \text{, so we must have } e=n-1 \text{.}$$ Summarizing our three steps, we get that $x=ap^{n-1}-1$ and $y=bp^{n-1}-1$. Return to the problem at hand. Note that our Lemma gives that $x,y \geq p^{n-1}-1$. Also, $n \geq 2$, since otherwise $p \nmid x+1,y+1$. Now, with a bit of brute force calculations (I'll probably write them down later), one can easily show that this inequality is not true for $p \geq 5$ and $n \geq 3$ (For the diligent reader, try proving this by contradiction). So we get that $p=3$ and $n=2$. This gives $x=3a-1$ and $y=3b-1$. Thus, we have $$x+y^2=3^2 \Rightarrow 9b^2-6b+3a=9 \Rightarrow b<2 \Rightarrow b=1,a=2 \Rightarrow x=5,y=2 \text{ } \blacksquare$$

Really nice, but surprisingly easy for a N5. hajimbrak wrote: Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p -1} + y$ and $x + y^ {p -1}$ are both powers of $p$. Proposed by Belgium We claim that the only solutions are $(p,x,y)=(3,2,5),(3,5,2),(2,n,2^n-1),$ $(2,2^n-1,n).$ These clearly work. Now we prove these are the only ones. Assume that $$x^{p-1}+y=p^k, \qquad y^{p-1}+x=p^\ell$$where $x<y$ and $k<\ell.$ Treat with the case $p=2$ seperately. Claim: We have $\nu_p(y-x) \ge k-1.$ Proof: Firstly, by Fermat $x \equiv -1 \equiv y \pmod{p}$ and so $p \nmid x,y$ and $p \mid y-x.$ We have $$x^{p-1}+y \mid y^{p}+xy-x(x^{p-1}+y)=y^p-x^p$$So by LTE $\nu_p(x-y)+1 \ge k.$ Suppose $k>2.$ Claim: We have $x \equiv -1 \pmod{p^2}.$ Proof: Since $k>2,$ hence $p^2 \mid x-y$ and hence $$0 \equiv p^k=x^{p-1}+y \equiv x^{p-1}+x \pmod{p^2}$$Thus $x^p \equiv -x^2 \pmod{p^2}.$ (notice that $\gcd(x,p)=1$ so this is all valid). But also, by Euler we have $x^{p(p-1)} \equiv 1 \pmod{p^2}$ and thus \begin{align*} -x \equiv y^{p-1} \equiv \left( -x^{p-1} \right)^{p-1} \equiv x^{(p-1)^2} \pmod{p^2} \\ \implies x^{p^2-2p} \equiv -1 \overset{\text{Euler}}{\equiv} -x^{p^2-p} \implies x^{p} \equiv -1 \pmod{p^2} \end{align*}So, $-1 \equiv x^p \equiv -x^2$ and so $x^2 \equiv 1 \pmod{p^2}.$ Write $x=pm-1$ for some $m>0.$ Then $1 \equiv (pm-1)^2 \equiv 2pm+1$ and so $m \equiv 0 \pmod{p}.$ So $x \equiv -1 \pmod{p^2},$ as desired. $\square$ Now, we get $$x^{p-1}+y \mid p(x-y)+p(x^{p-1}+y)=px(x^{p-2}+1)$$Now, clearly $\gcd(x,y)=1$ else we must have $\gcd(x,y)=p^z$ but then $p^k=x^{p-1}+y=p^z(\alpha),$ where $\gcd(\alpha,p)=1$ and $\alpha>1,$ which is not possible. Hence, we get $x^{p-1}+y \mid p(x^{p-2}+1)$ and so $x^{p-2}(x-p)+y \le p.$ If $x>p \ge 3,$ then this gives $x^{p-2}<p.$ This is impossible since $p>x^{p-2} \ge x>p.$ Thus, $x \le p$ and so $x=p-1.$ Thus the second claim gives $p-1 \equiv -1 \pmod{p^2}$ which is impossible. Thus, $k=2$ which gives $p=x^{p-1}+y > (p-1)^{p-1}$ and so $p=3.$ Then $x=3-1=2$ and so $y=5,$ and we get the solution, and we are done. $\blacksquare$

If $p=2$, we can choose any pair $x,y$ such that $x+y=2^k$ for some $k$ and we are done. So assume that $p$ is odd. Also, obviously $x$ cannot be equal to $y$ and $x,y$ can't be multiples of $p$ since $p-1>1$ here. From similar logic, $x,y$ are coprime. Also, FLT gives that both $x,y$ are $p-1$ mod $p$. Thus, $x,y \geq p-1$. So, let $x>y$ WLOG. Let $x^{p-1}+y=p^a, y^{p-1}+x=p^b$ with $b<a$. Note that $b \geq 2$ since $x,y \geq p-1$ and $p>2$. Obviously, we get $y^{p-1}+x|x^{p-1}+y$. From here we can trivially get $y^{p-1}+x|(xy)^{p-2}-1$ by adding a suitable multiple of $y^{p-1}+x$ and from the fact that $x,y$ are coprime. Since $y^{p-1}+x$ is a power of $p$ and $p-2$ is coprime to $p-1$ we get $y^{p-1}+x|xy-1$. Let $k(y^{p-1}+x)=xy-1$ for some positive integer $k$. Solving for $x$ gives $x=\frac{ky^{p-1}+1}{y-k}$. Substitute this value to get that $\frac{y^p+1}{y-k}=p^b$. By using the fact that $x$ is a positive integer, and also using LTE lemma gives two things: 1.$p^{b-1}|y+1$ 2.$y^{p-1}<\frac{y^p+1}{y}<p^b$. Since $y$ is a positive integer, from $1$ we get that $y \geq p^{b-1}-1$. This gives that $(p^{b-1}-1)^{p-1}<p^b$ and since $b \geq 2$ as we saw earlier so there are no solutions for $p \geq 5$. For $p=3$ this inequality only holds true for $b=2$ which produces the solution $(3,2,5)$ or $(3,5,2)$. So the solutions are given by: $(2,x,2^k-x)$(where $k,x$ are positive integers such that $x<2^k$), $(3,2,5),(3,5,2)$. Proved.

If $p=2$, then any $x+y$ being a power of 2 works. We claim the only other solutions are $(p,x,y)=(3,2,5),(3,5,2)$. It is easy to see that these work. Assume $p$ is an odd prime. WLOG $x\le y$. Let $x^{p-1}+y=p^k$, so that $y=p^k - x^{p-1}$. Then $x+y^{p-1}=x+(p^k-x^{p-1})^{p-1}$ is a power of $p$. In particular, it is a multiple of $p$, so $x+(p^k-x^{p-1})^{p-1} \equiv x+x^{(p-1)^2} = x\big(1+x^{p(p-2)} \big) \equiv 0\pmod{p}$. This means that either $x\equiv 0 \pmod{p}$ or $1+x^{p(p-2)} \equiv 0 \pmod{p}$. In the latter case, $-1\equiv x^{p(p-2)} \equiv x^{p-2} \equiv 1/x \pmod{p}$, so $x \equiv -1\pmod{p}$. Therefore, either $x\equiv 0 \pmod{p}$ or $x\equiv -1\pmod{p}$. Case 1: $x\equiv 0 \pmod{p}$. We know $x+(p^k-x^{p-1})^{p-1}$ is a power of $p$. Let $x=p^a \ell$ for $\gcd(\ell,p)=1$. Then \begin{align*} x+(p^k-x^{p-1})^{p-1} &= p^a\ell + (p^k-p^{a(p-1)}\ell^{p-1})^{p-1} \\ &= p^a \left[ \ell + p^{(p-1)\cdot \min(k,a(p-1))}(--), \right] \end{align*}where $--$ is not divisible by $p$. Since $\ell$ is also not divisible by $p$, and since the exponent of $p$ in front of $--$ is at least 1, the second factor is not divisible by $p$, which is a contradiction. So there are no solutions in this case. Case 2: $x\equiv -1 \pmod{p}$. We know $x+(p^k-x^{p-1})^{p-1}$ is a power of $p$. Expanding with the binomial theorem gives us $x+x^{(p-1)^2} \equiv 0 \pmod{p^k}$. Since $p\nmid x$, we get $x^{p(p-2)} \equiv -1 \pmod{p^k}$, so $\nu_p(x^{p(p-2)} + 1) \ge k$. By LTE, we get $\nu_p \big(x^{p(p-2)} - (-1)^{p(p-2)} \big) = 1+\nu_p(x+1)$, so $\nu_p(x+1) \ge k-1$. This means $p^{k-1} \mid x+1$. In particular, $x \ge p^{k-1}-1$. This is a very strong inequality, let's see if we can push it to finish. We know $x^{p-1}+y=p^k$, so \[ p^k \ge (p^{k-1}-1)^{p-1} + y > (p^{k-1}-1)^{p-1}.\]If $k=1$, the inequality is satisfied. If $k=2$, we get $p^2 > (p-1)^{p-1}$. This only holds if $p=3$; for $p\ge 5$ it clearly does not hold. If $k\ge 3$, we claim there are no solutions. Indeed, clearly $(p^{k-1}-1)^{p-1}-p^k$ is increasing for $p\ge 5$ past $k\ge 1$, so if we can show the inequality does not hold for $p=3$ we are done. For $p=3$, we get $3^k > (3^{k-1}-1)^2$, which clearly never holds for $k\ge 3$. Therefore, the only solution for odd primes $p$ is $(p,x,y)=(3,2,5),(3,5,2)$.

v_Enhance wrote: If $p=2$ then any $(x,y)$ with $x+y$ a power of two is okay. Hence assume $p > 2$. Then if $\nu_p(x) \ge \nu_p(y) > 0$ we get an immediate contradiction, thus we may assume $p \nmid x,y$. So by Fermat's Little Theorem, $x$ and $y$ are $-1 \pmod p$. It is easy to check that when $p > 2$ we cannot have $x=y$, since otherwise $x(x^{p-2}+1)$ is a power of $p$, which is clearly impossible when $p=2$. Moreover, if $p > 2$ then $x^{p-1}+y \neq y^{p-1}+x$, since otherwise $(x-y)(x^{p-2}+\dots+y^{p-2}) = x-y$, which is impossible unless $x=y$. Thus, suppose $y^{p-1}+x < x^{p-1}+y$, which is equivalent to $y < x$. Then in particular $y^{p-1}+x$ divides $x^{p-1}+y$, so \[ y^{p-1} + x \mid (-y^{p-1})^p + y \implies y^{p-1}+x \mid y^{p(p-2)}+1. \]By Lifting the Exponent, we thus deduce that \[ \nu_p(y^{p-1}+x) \le 1 + \nu_p(y+1). \]Actually, since LHS is a power of $p$, this informs us that \[ y^{p-1} + x \mid p \cdot (y+1) \implies y^{p-1} + x \le p \cdot (y+1). \]Since $x > y$, this forces \[ y^{p-1} + y \le p \cdot (y+1). \]Also, $y \ge p-1$ since $y \equiv -1 \pmod p$. This can only occur if $y = 2$ and $p = 3$. So, it remains to find all $x > 2$ such that $x^2+2$ and $4+x$ are powers of $3$. Letting $4+x=3^b$, we find that \[ \left( 3^b-4 \right)^2 + 2 = 3^{2b} - 8 \cdot 3^b + 18 \]is supposed to be a power of $3$, which can only occur for $b \le 1$. Checking, this gives the last solution $(x,y,p) = (5,2,3)$. Who can explain why is this true: (1) $y^{p-1} + x \mid p \cdot (y+1)$ (2) $y \ge p-1$ since $y \equiv -1 \pmod p$. This can only occur if $y = 2$ and $p = 3$. ? @Evan: lol, I got that $v_p (y^{p-1}+x)=p^k $ for some $k $. You said $LHS $ but $LHS=v_p (y^{p-1}+x)$. However is my bad haha. Don't need a strong justification that "size reasons"?

(1) Since $y^{p-1}+x$ is power of $p$, so follows by $\nu_p(y^{p-1}+x) \le \nu_p( p\cdot(y+1) )$. (2) The two inequalities $y^{p-1}+y \le p \cdot (y+1)$ and $y \ge p-1$ have no simultaneous integer solutions except for $(y,p) = (2,3)$. Just size reasons.

$p = 2$ has sol $x + y = 2^k$. For odd $p$: WLOG $x > y$, $x^{p-1}+y = p^a, x+y^{p-1} = p^b$ then $x+y^{p-1} \mid x^{p-1}+y$ so $p^b \mid y^{(p-1)^2}+y$ so $p^b \mid y^{p(p-2)}+1$ but by FLT $p \mid y + 1$ so by LTE $p^{\max \{1, b-1\}} \mid y + 1$ so since $p^b > y^{p-1}$, $(p^{\max \{1, b-1\}} - 1)^{p-1} < p^b$. This fails for all $p, b$ pairs except $3, 2$ where there is a solution $p = 3, x = 5, y = 2$.

Hard problem for us. There was so much to try and see that it took us a long time, but we did it. Solved with proxima1681. Solution: We will show that the only solution triples $(p,x,y)$ are $(3,5,2)$, $(3,2,5)$ and $(2,x,2^n - x)$ for arbitrary natural number $n$ such that every quantity is positive. These solutions clearly work. We will handle the case $p = 2$ separately. This gives one of the solution claimed which clearly works. From here on, assume that $p$ is an odd prime. Set \begin{align} x^{p-1} + y = p^a \label{1} \\ y^{p-1} + x = p^b \label{2} \end{align}for some positive integers $a$ and $b$. Since the problem statement is symmetric about $x$ and $y$, without loss of generality assume that $x \ge y$ which in turn gives $a \ge b \ge 1$. One can easily checkthat $a = b \iff x = y$. Claim: $x \ne y$ with $\gcd(x,p) = \gcd(y,p) = \gcd(x,y) = 1$ and $p \mid x-y$. Proof: If $x,y$ are non-distinct, then observe that $(1)$ and $(2)$ translate to $x(x^{p-2} + 1) = p^a$. Only possible solution here is achieved when $p = 2$ which is just contradiction. Everything else is just impossible since $\gcd(x,x^{p-2}-1) = 1$. We will now move on to the harder part. Assume on the contrary that $p \mid x$. One will get that $p \mid y$ as well from $(1)$. We now consider two separate cases on $\nu_p(x)$ and $\nu_p(y)$. If $\nu_p(x) \ge \nu_p(y)$, then $\nu_p(y) = a$ from $(1)$. Now observe that $\nu_p(LHS) > \nu_p(p^b)$ in $(2)$ which is just impossible. For the second case, we need to consider $\nu_p(x) < \nu_p(y)$. From $(2)$ we clearly have that $\nu_p(x) = b$. This would give a size contradiction again in $(2)$ and this shows the gcd part. Reducing $(1)$ and $(2)$ modulo $p$ with the help of ``Fermat's Little Theorem'' we get that $x \equiv y \equiv -1 \pmod{p}$ giving $p \mid x-y$ which finally completes the proof to the claim. $\square$ Claim: $\nu_p(y+1) = \nu_p(x-y) = b-1$. Proof: $x \times (1) - y \times (2)$ gives us \begin{align} x^p - y^p = p^b(xp^{a-b}-y). \tag{3} \end{align}From ``Lifting the Exponent Lemma'' in $(3)$, we get that \[\nu_p(x^p-y^p) = \nu_p(x-y) + 1 = b \iff \nu_p(x-y) = b - 1.\]We have now established one of the desired equalities. We now proceed to show the next one. Observe that \begin{align*} y^{p-1} + x &\mid x^{p-1} + y \\ x^{p-1} + y &\mid x^{p-1} + y - x^{p-2}(x^{p-1} + y) \\ x^{p-1} + y &\mid y^{p-1}x^{p-2} + y. \end{align*}Now just substitute $x \equiv -y^{p-1} \pmod{y^{p-1}+x}$ and cancel off $y$. This would boil down everything to $y^{p^2 - 2p} \equiv -1 \pmod{y^{p-1}+x}$. Since we have a power of a prime dividing a quantity, we must have that \[\nu_p(y^{p^2 - 2p} + 1) \ge b\]Invoking ``Lifting the Exponent Lemma'' to get \[\nu_p(y^{p^2 - 2p} + 1) = \nu_p(y+1) + 1 \ge b \iff \nu_p(y+1) \ge b-1\]For the sake of contradiction, assume that $\nu_p(y+1) \ge b$. Reducing $(2)$ modulo $p^b$ we would get $x \equiv -1 \pmod{p^b}$ which would give $p^b \mid x-y$ which is absurd since $\nu_p(x-y) = b - 1$. $\square$ The critical observation is to notice that $p^{b-1} \mid y + 1 \implies p^b \mid p(y+1)$. Putting stuff from $(2)$ we get $y^{p-1} + x \mid p(y+1)$. Observe that $y = 1$ would give $p = 2$ which is illegal which gives $y >1$. Note that \[y^{p-1} < y^{p-1} + x \le p(y+1) < p(2y) \implies y^{p-2} < 2p.\]Since $y \ge 2$, we have $2^{p-2} \le p$ from the above. This forces $p=3$ and $y=2$ from $y^{p-2} \le 2p$. Put $p = 3$, $y = 2$ in $p^{b-1} \mid y+1$ to get $b = 2$ and finally put everything back in $(1)$ to get $x = 5$ finishing the problem. $\blacksquare$

Let $x^{p-1}+y=p^a$ and let $y^{p-1}+x=p^b$ where $a>b$, $x>y$, and $p$ is an odd prime (when $p=2$, all solutions are trivial; we will deal with the case $a=b$ later). Note that we have $x\equiv y\equiv 0\pmod p$ or $x\equiv y\equiv -1\pmod p$. In the former, note that \[(x^{p-1}-y^{p-1})-(x-y)=p^b(p^{a-b}-1)\]so if $p>2$ we find that $\nu_p(x-y)=b$ (by factoring out $x-y$). However this is impossible; therefore assume that $x\equiv y\equiv -1\pmod p$. We then have \[x^p-y^p=xp^a-yp^b\implies \nu_p(x-y)=b-1\]which implies that \[\nu_p(y^{p-1}+y)=b-1\implies py^{p-2}+p\ge y^{p-1}+x\]so that either $x=p-1$ or $y=p-1$. This implies that $y=p-1$ since we have assumed $x>y$. Now we have \[\nu_p(y^{p-1}+y)=\nu_p(y^{p-2}+1)=\nu_p(y+1)=1\]which means that $b=2$. At this point we easily find $p=3$ which gives the solutions $(2,5,3)$ and $(5,2,3)$. When $p=2$ we find that all $(x,y,2)$ where $x+y$ is a power of $2$ are fine. When $a=b$ we find $x=y$ so that $x=p^k$ for some $k$ (where $p$ is an odd prime) which gives that \[x^{p-1}+x=p^k(p^{k(p-2)}+1)\]is a power of $p$. Then $k=0$ which implies $p=2$ which is a contradiction. Therefore we have already characterized all solutions and we are done. $\blacksquare$

used hints to solve this one . plix gib dis nub some IQ

handwritten solution images attached

For $p=2$, the solutions are $(2,x,2^a-x)$. Assume $p>2$. Let $p^a=x^{p-1}+y,p^b=y^{p-1}+x, a>b$. If $a=b$, then $x=y$ as $f(t)=t^{p-1}-t$ is strictly increasing over the positive integers. This would imply $p|x,p|x^{p-2}+1$, which is impossible, so assume $a\neq b$. By FLT, we have $x\equiv y\equiv -1 \pmod{p}$. Now we also have $xp^a=x^p+xy$ and $yp^b=y^p+xy$ or \[xp^a-yp^b=x^p-y^p\]which implies that $\nu_p(x-y)=b-1$. In addition, consider \[ p^b=\gcd(x^{p-1}+y,y^{p-1}+x) = \gcd(x^{p-1}+y,(xy)^{p-2}-1)\]By LTE, this implies $\nu_p(xy-1)=p^b$. Thus, \[\nu_p(xy-1-x+y)=\nu_p((x+1)(y-1))=b-1\]so $\nu_p(x+1)=b-1$, and similarly $\nu_p(y+1)=b-1$ (subtract the two instead of adding them). From here, size finishes: \[ p^b=y^{p-1}+x>(p^{b-1}-1)^{p-1}+p^{b-1} \geq (p^{b-2})^{p-1}+p^{b-1}\]which forces $b=1$ or $2$. If $b=1$, then $p=y^{p-1}+x$. If $y=1$, then $p^a=x^{p-1}+1$ which is impossible by Catalan's (or whatever you fancy), and if $y>1$, then $y^{p-1}\geq 2^{p-1}>p$. Now consider $b=2$. If $y=1$, we have $p^a=1+(p^2-1)^{p-1}$ which is impossible also by Catalan's. So if $y>1$, then $y^{p-1} \geq 2^{p-1}>p$ except for when $p=3,5$. Testing $3$ produces the solutions $(3,2,5),(3,5,2)$. Testing $5$ produces no solutions, so we are done.

If $p=2$ then clearly any $(x,y)$ whose sum is a power of two works. Suppose $p>2$. If $p\mid x$ then let $\nu_p(x)$. Clearly, $\nu_p\left(x^{p-1}\right)=\nu_p(y)$ and thus the same thing can be applied to $y$, which gives a contradiction. If $x^{p-1}+y=y^{p-1}+x$ then $x^{p-1}-x=y^{p-1}-y$. However, this function is increasing on $(1,\infty)$ so $x=y$, which gives $x^{p-1}+x$ is a power of $p$. However, then $x$ is a power of $p$, so $x=1$. Absurd. The smaller one is divisible by the bigger. Suppose $x<y$, so $x^{p-1}+y\mid y^{p-1}+x$. We have $y\equiv -x^{p-1}\pmod {x^{p-1}+y}$ so \[0\equiv \left(-x^{p-1}\right)^{p-1}+x\equiv x^{p^2-2p+1}+x\pmod x^{p-1}+y\]We have $\gcd(x^{p-1}+y,x)=1$ because the former is a power of $p$. Note that $y^{p-1}\equiv 1\pmod p$ so $x\equiv -1\pmod p$. Indeed, we can write \[\nu_p(x^{p-1}+y)\le \nu_p(x^{p^2-2p}+1^{p^2-2p})=\nu_p(x+1)+1\]Since $x^{p-1}+y$ is actually a power of $p$, $x^{p-1}+x\le x^{p-1}+y\le p(x+1)$. Note that the difference, $x^{p-1}-(p-1)x-p$, has a derivative that is nonnegative on $x\ge 1$. Clearly, for $p\ge 5$ we have $2^{p-1}+2\le 3p$. Thus, we must have $p=3$, so $x^2+x\le 3(x+1)\implies x\le 3$ so $x=2$. We have $4+y\mid 2 + y^2$. Thus, $4+y\mid 18$. $y$ is equal to one of the following: $2$, $5$, $14$. Clearly, only only $5$ works. We are done.

o wow didn't expect to solve this so soon after last post (how so oar) $p=2$ is trivial and is left as an exercise to the reader. The only other solution is $(x,y,p) = (2,5,3)$ up to switching $x$ and $y$. Henceforth assume that $p$ is an odd prime. WLOG assume $x^{p - 1} + y \mid y^{p - 1} + x$. Then it is easy to obtain the relations \[ x^{p - 1} + y \mid y^{p - 1} + x - (y^{p - 1} + x^{p - 1}y^{p-2}) \]and \[ x^{p - 1} + y \mid y^{p-1}x^{p-2} + x^{p-1} - x^{p-1} - y \] which imply that \[ x^{p-1} + y \mid \gcd{(x,y)}((xy)^{p-1} - 1). \] Let $x=da$, $y=db$ with $\gcd{(a,b)}=1$. Then we obtain \[ d^{p-1}a^{p-1} + b \mid (d^2ab)^{p-1} - 1. \]Now observe that $p$ must divide the left hand side, since we are dealing with positive integers and we divided out a power of $p$. If $d > 1$ then $p \mid d$, and it cannot divide right hand side, contradiction. Hence, we obtain \[ x^{p-1} + y \mid (xy)^{p-1} - 1.\]With a bit of fiddling around, one obtains \[ x^{p-1} + y \mid y^p + 1, \]and with Fermat's Little Theorem it is clear that $y \equiv -1 \pmod{p}$. Moreover by dividing out the other way, one obtains \[ x^{p-1} + y \mid x^{p(p-2)} + 1. \](multiply lefthandside by $(xy)^{p-1} \cdot y^{-1}$ and go from there) Fermat's Little Theorem again says that $x^{-1} \equiv -1 \pmod{p}$, hence $x \equiv -1 \pmod{p}$. Now observe that we need \[ \nu_p(x^{p(p-2)} + 1) \ge \nu_p(x^{p-1} + y). \]By LTE since $p \nmid x$, we obtain \[ \nu_p(x^{p(p-2)} + 1) = 1 + \nu_p(x + 1) \ge \nu_p(x^{p-1} + y).\]Let $x + 1 = kp^r$. Then since $x^{p-1} + y$ is a perfect power of $p$ we require that \[ (kp^r - 1)^{p-1} < p^{r + 1} \]for the $\nu_p$ inequality to hold at all ($y$ is a positive integer). However this becomes trivially false for $k>1$, and even considering $k=1$ limits the options greatly; $p>3$ clearly does not work. For $p=3$, $r=1$ provides the only solution of $x=2$, $y=5$.

$p = 2$ is trivial since we just take $(n,2^k-n)$ for positive integers $n$ and $k$. The initial idea is to get a divisibility relation in the form of $c \mid f(n)$ i.e the RHS is just in one variable for $p > 2$. Now assume $p > 2$. WLOG $x \geq y$, which is the same thing as saying $x^{p-1}+y \geq x+y^{p-1}$. It follows that $y^{p-1}+x \mid x^{p-1}+y$ which is the same thing as saying $$x^{p-1}+y \equiv 0 \pmod {y^{p-1}+x}$$However, in lieu of the idea we can also rewrite this as $$(-y^{p-1})^{p-1}+y \equiv 0 \pmod{y^{p-1}+x}$$and simplifying yields $$y^{p-1}+x \mid y(y^{p(p-2)}+1)$$ Now note that clearly $p \nmid x,y$. FLT gives us $x^{p-1} + y \equiv 1+y \equiv 0 \pmod p$ which means $y \equiv -1 \pmod p$. Similarly $x \equiv -1 \pmod p$. This means that LTE gives us $$v_p(y(y^{p(p-2)}+1)) = v_p(y^{p(p-2)}+1) = v_p(y+1)+1$$It follows that $v_p(y^{p-1}+x) \leq v_p(y+1)+1$. Now the key step is that since $y^{p-1}+x$ is actually a power of $p$, we can say that $$y^{p-1}+x \mid p(y+1)$$However, this means that $y^{p-1} + x \leq p(y+1)$ but $y \equiv -1 \pmod p \geq p-1$ so we must have $p=3$ and $y=p-1 = 2$. Plugging this back into the equation and working out some details that I will omit yields that $x=5$ so the only triple is $(3,5,2)$ and we're done. remark - usually we simply add and subtract stuff from RHS to simplify a divisibility relation but in this case we had to use $v_p$ at the end. I guess I should've known $v_p$ would come up at some point given the "power of prime" in the problem statement haha

Notice p=2 we take (x,y)=(k,2^n-k), so henceforth assume p>2; it's pretty easy to prove x,y are relatively prime to p. (For example, $\nu_p(x) \ge \nu_p(y)\implies\nu_p(y)\ge p-1$ from $x^{p-1}+y=p^{x_0},x+y^{p-1}=p^{y_0},x_0,y_0\ge p-1$. Now observe that $0\equiv y^{p-1}+x\equiv x\pmod {p^{y_0}}$, which is infinite "inscent" in the sense that x is now divisible by p-1th power, which henceforth increases the $\nu_p$ in y from the first equation.) If both of them are relatively prime, then mod p we find$$1+y\equiv 0\equiv 1+x\implies x\equiv y\implies x(x^{p-2}+1)\equiv 0\pmod p\implies ord_{\{x,y\}}|\text{gcd}(2p-4,2p-2)=2\implies \{x,y\}\equiv\{-1,1\}\pmod p.$$Checking, none of these permutations work mod p, unless p=3; this is the most bashy case yet. Prepare yourself! (x,y)=(2,5) works, and we prove it's the only one. Indeed, it's obvious taking mod 3 that x,y are 2 mod 3; mod 9, $x^2$ or $y^2$ can be 1,4,7 mod 9, so $(x,y)\equiv(2,5)\pmod 9$ is the only pair that works. Then, it's obvious if (at least) one of the expressions in the original problem=9 or 27, (2,5) is only solution. If one of them is divisible by 81, $(x^2,y^2)\equiv(4,25)\pmod 81$; it's obvious that this doesn't work because we would need $(x,y)\equiv(56,77)$, but plugging this in mod 81 doesn't actually work. In conclusion, the only solutions are $(p,x,y)=(2,k,2^n-k),(3,2,5)$ with switching x,y allowed.

Isn't it easy for N5? Answer: $(x, y, p) = (x, 2^k - x, 2), (2, 5, 3)$. If $p = 2$, then $x + y$ and $y + x$ are power of 2, so $y = 2^k - x$ for some $k$. Now assume $p \ge 3$. Note that $\gcd(p, x) = \gcd(p, y) = 1$, so $x^{p-1} + y \neq x + y^{p-1}$. Let $p^a = x^{p-1} + y$ and $p^b = x + y^{p-1}$.By Fermat's little theorem, we have $x + 1 \equiv y + 1 \equiv 0 (p)$. WLOG assume $a < b$. Then since $p^a \mid p^b$, so $x^{p-1} + y \mid x + y^{p-1}$,so $p^a \mid x^{p^2 - 2p} + 1$, so by lifting the exponent we have $\nu_p(x + 1) + \nu_p(p^2 - 2p) \ge a$, so $\nu_p(x + 1) \ge a - 1$. So by lifting the exponent we have $\nu_p(x^{p-1} - 1) = a - 1$, therefore $\nu_p(y + 1) = a - 1$. So $p^a = x^{p-1} + y \ge (p^{a-1} - 1)^{p-1} + (p^{a-1} - 1)$. Assume $a \ge 3$. Then $p^a > (p^{a-1} - 1)^{p-1} \ge (p^{a-1} - 1)^2 > (p^{a-1} - 1)(p + 1) = p^a + p^{a-1} - p - 1 > p^a$, a contradiction. Note that $a \neq 1$, since $x^{p-1} + y > p$. So $a = 2$. Then we have $(p-1)^{p-1} + (p-1) \le p^2$, so $(p-1)^{p-1} < p^2 - 1$. Assume $p \ge 5$. Then $(p-1)^3 \le (p-1)^{p-2} < p + 1$, a contradiction. So $p = 3$ and $a = 2$. It follows that $x = 2$ and $y = 5$, and it's not hard to check that $(x, y, p) = (2, 5, 3)$ is answer. This completes proof. $\blacksquare$

The only answers are $(x, 2^k-x)$ when $p=2$ and $(2, 5)$ when $p=3$. Note that when $p=2$ the problem is tautological. Hence assume $p \geq 3$. Note that if $p \mid x$ and $p \mid y$, then we have an obvious contradiction when $\nu_p(x) \geq \nu_p(y)$. It follows that $x, y \equiv -1 \pmod p$. Notice that when $x=y$, $x(x^{p-2} + 1)$ must be a power of $p$, which is impossible as $p \nmid x$. Thus assume $y > x$ and set $\nu_p(x+1) = k$. It follows that $$x^{p-1} + y \mid x^{(p-1)^2}-y^{p-1} + y^{p-1} + x = x(x^{(p-1)^2-1}+1).$$However, by LTE $$\nu_p(x^{(p-1)^2-1}+1) = \nu_p(x+1) + 1 = k+1.$$On the other hand, because $x^{p-1}+y$ is a power of $p$, it follows that $$(p^k-1)^{p-1} + y \leq x^{p-1} + y \leq p^{k+1}.$$The only solutions to this are found when $p=3$, which gives the desired pair $(2, 5)$.

Nice Question! took some time to solve it though If $p=2$ then $x+y=2^k$ for some $k\in \mathbb{N}$. So the triple would be $\boxed{(2,l,2^k-l)}$. For $p\geq 3$ let $x^{p-1}+y=p^a$ and $y^{p-1}+x=p^b$. If $p\mid x\implies p\mid y$ so let $\nu_p(x)=r$ and $\nu_{p}(y)=s$. From this we get $(p-1)s=r$ and $(p-1)r=s$ which is obviously a contradiction. Now by FLT we get $x,y\equiv -1 \pmod{p}$. WLOG assume $x>y$, then we have $$y^{p-1}+x \mid x^{p-1} +y \implies y^{p-1}+x \mid \left(-y^{p-1}\right)^{p-1}+y\implies y^{p-1}+x \mid y^{p(p-2)}+1$$By applying LTE we have $v_p(y^{p-1}+y)<\nu_{p}(y^{p-1}+x)\leq \underbrace{\nu_{p}(p(p-2))}_{1}+\nu_{p}(y+1)$, so we have $y^{p-1}+x \mid p(y+1)\implies y^{p-1}+y<p(y+1)$, because $y\equiv -1 \pmod{p}$ we have $y\geq p-1$, so we get $p=3,5$ so the triples are $\boxed{(3,5,2), (3,2,5)}$