Junior vibes, people! Answer. The solutions $(x,y,p)$ are $(8,1,19)$, $(1,8,19)$, $(7,2,13)$, $(2,7,13)$, $(5, 4, 7)$ and $(4,5,7)$. Without loss of generality treat $x\geq y$. Writing $a = x + y \geq 2$, $b=xy$, we have $a^3 - 3ab = p(b+p)$, i.e. $b = \frac{a^3 - p^2}{3a+p}$. As $p+3a \mid p^3 + 27a^3$, we get $p+3a \mid p^3+27a^3 - 27(a^3 - p^2)$, i.e. $p+3a \mid p^2(p+27)$. If $p \nmid a$ and $p \neq 3$, then $p + 3a \mid p + 27$, hence $p+3a \mid 27 - 3a$, implying $a \leq 9$ (for $a> 9$ we would get $p+3a \leq 3a - 27 < 3a$, impossible). For $a \geq 4$ we get $p + 3a \geq 17 > 27 - 3a$ for $p\geq 5$ and for $p=2$ and $a=4$ we get $14 \mid 15$, nope. For $a=3$ we get $p+9 \mid 18$, impossible; for $a=2$ we get $x=y=1$ which gives (from the initial equation) $p=1$, nope. It remains only $a=9$ and now carefully checking all $(x,y)$, $x\geq y$ with $x+y = 9$ we get the solutions $(8,1, 19)$, $(7, 2, 13)$ and $(5, 4, 7)$ Now suppose $p\mid a$, then with $a=kp$ we get $b = \frac{k^3p^2 - p}{3k+1}$ and $b \leq \frac{a^2}{4} = \frac{k^2p^2}{4}$ (equivalent to $(x-y)^2 \geq 0$) implies $4k^3p^2 - 4p \leq 3k^3p^2 + k^2p^2$, i.e. $k^2(k-1)p \leq 4$, which is impossible for $k\geq 2$. If $k=1$, then $b=\frac{p^2-p}{4}$ and $(x-y)^2 = a^2 - 4b = p$, impossible for $p$ prime. Finally, treat $p=3$. Then $b = \frac{3(3k^3 - 1)}{3k+1}$, so $3k+1 \mid 3k^3 - 1$, so $3k+1 \mid 3k(k^2 + 1)$, i.e. $3k+1 \mid k^2 + 1$, so $3k+1 \mid k(k-3)$, i.e. $3k + 1 \mid k-3$. Hence none of $k\geq 4$ and $k=1,2$ are possible and for $k=3$ we get $a = kp = 9$, which was already treated above.

Solved with GoodMorning and cj13609517288. The only solutions are $\boxed{(x,y,p) = (8,1,19), (1,8,19), (7,2,13), (2,7,13), (5,4,7), (4,5,7)}$, which work. Now we prove they are the only solutions. If $p$ divided either $x$ or $y$, then $p$ would divide both $x$ and $y$. In this case, the LHS would be divisible by $p^3$, while the RHS wouldn't, contradiction. If another prime $q$ divided $x,y$, then taking mod $q$ gives $q\mid p^2$, so $q = p$, contradiction. Thus, $\gcd(x,y) = 1$. The equation becomes \[(x+y)(x^2 - xy + y^2) = p(xy+p)\]By Four Number Lemma, there exist positive integers $a,b,c,d$ such that \[ab = x+y, cd = x^2 - xy + y^2, ac = p, bd = xy +p \] Since $p$ is a prime, we have either $a=p, c=1$ or $a=1, c=p$. Claim: If $a=p$ and $c=1$, there are no solutions. Proof: Suppose $a=p$ and $c=1$. We get \[x + y = p b, x^2 - xy + y^2 = d, xy + p = bd\]By AM-GM, $x^2 + y^2 \ge 2xy$, so $d\ge xy$. Using the third equation, we get $p \ge (b-1)xy$. Therefore, \[x+y \ge b(b-1)xy,\]which is evidently impossible if $b>1$ (because there are no solutions when $x=y=1$). Thus, $b=1$, so $p = x + y$. Then, dividing both sides in the equation by $x+y$, we get $x^2 - xy + y^2 = xy + p$, so $p = (x-y)^2$, which is absurd since no prime is a perfect square. $\square$ Thus, $a=1$ and $c=p$. We get \[x + y = b, x^2 - xy + y^2 = pd, xy + p = bd\] Claim: $d=3$. Proof: Suppose this was not the case. Notice that $(x+y)^2 - 3xy = x^2 - xy + y^2$, so \[b^2 - 3(bd - p) = pd\implies b^2 - 3bd = pd - 3p,\]so $b\mid pd - 3p = p(d-3)$. If $p\mid b$, then taking the third equation mod $p$ gives $p\mid xy$, which we know is not true. Thus, $\gcd(b,p) = 1$, so $b\mid d-3$. If $d<3$, then $b=2$. This case leads to no solutions, so assume $d>3$. We have $d-3\ge b$, so $d>b = x + y $. Now we have \begin{align*} d = \frac{xy + p}{x+y} \\ \le \frac{xy + pd}{x+y} = \frac{x^2 + y^2}{x+y} \\ \le x + y = b, \\ \end{align*}which is absurd since $d>b$. $\square$ Now, we have $x^2 - xy + y^2 = b^2 - 3(3b - p) = b^2 - 9b + 3p$. Since this is equal to $3p$, $b^2 - 9b = 0\implies b= x+y=9$. Since $\gcd(x,y) = 1$, we must have $\{x,y\} = \{1,8\}, \{2,7\},$ or $ \{4,5\}$, which gives the desired solution set.

surprising The solutions are $(x,y,p)=(8,1,19),(7,2,13),(5,4,7)$, and the obvious $3$ others by swapping $x$ and $y$. WLOG let $x \geq y$. Note that $p \nmid x,y$, since if it divides one then it divides the other, but then $\nu_p(LHS)\geq 3$ while $\nu_p(RHS)=2$. Furthermore, no prime $q \neq p$ divides both $x$ and $y$, else it divides $p$, which is absurd. Now we observe that if $x>p$, then we have $$x^3+y^3<x(xy+x)=x^2y+x^2 \implies x^2(x-y-1)+y^3<0,$$which is also absurd, as this inequality is clearly false for $x=y=1$ and otherwise $x\geq y-1$ since $x \neq y$ is impossible. Thus $x \leq p$, so in fact $p>x>y$. Factor the LHS as $(x+y)(x^2-xy+y^2)$. If $p \mid x+y$, then $x+y=p$ for size reasons. But then $x^2-xy+y^2=xy+p \implies (x-y)^2=p$, which clearly never happens. Thus $p \mid x^2-xy+y^2$, so write $x^2-xy+y^2=kp$. We then have $$(x+y)k=xy+p \implies (x+y)k^2-kxy=(x+y)^2-3xy \implies (x+y)(k^2-x-y)=(k-3)xy.$$Clearly $x+y \mid (k-3)xy$. Because $\gcd(x,x+y)=\gcd(y,x+y)=1$, it follows that $x+y \mid k-3$. Suppose that $k \neq 3$. Because $x$ and $y$ are positive we can't have $k=2$, and if $k=1$ then $x=y=1$ which is clearly never a solution for any primes $p$. Thus $k-3$ is positive and thus at least $x+y$. Substituting this back into the original definition of $k$, we have $$x^2-xy+y^2\geq (x+y+3)p>x(x+y+3) \implies y^2>2xy+3x,$$which is preposterous. Hence $k=3$, so we should also have $x+y=k^2=9$. A case check finishes. $\blacksquare$ Remark: Although this solution does not appear to be heavily motivated by the equality cases, it was. One notices that the value of $k$ must always equal $3$, which makes it inevitable that the term will be used somehow. On the other hand, the mere existence of solutions implies that one cannot blindly go in applying size or some other method to obtain a contradiction. When it comes to finding solutions themselves, this is much easier once we reduce the problem to $p \mid x^2-xy+y^2$. For cyclotomic polynomial reasons this implies that only $p \equiv 1 \pmod{6}$ possibly work, and for those primes $p$ there is some integer $a$ such that any solution $(x,y)$ will have $x \equiv ay \pmod{p}$ up to switching $x$ and $y$ (namely $a$ such that $a^2-a+1 \equiv 0$), so our search space is greatly narrowed. I would be surprised to find a solution that relies on $p \equiv 1 \pmod{6}$ being true.

First, we claim that $p\nmid x$ and $p\nmid y$. Indeed, if $p$ divides $x$ or $y$, then obviously $p$ divides the other one, and then we get a contradiction by looking at the $v_p$ on both sides. This tells us that $p\nmid xy+p$, thus $p$ divides exactly one of $x+y$ and $x^2-xy+y^2$. Case 1 : $p\mid x+y$. In this case, we have $\gcd(p,x^2-xy+y^2)=1$, so $x^2-xy+y^2\mid xy+p$. Suppose WLOG that $x\ge y$, and let $x=y+\alpha$. The divisibility condition then becomes \[(y+\alpha)^2-y(y+\alpha)+y^2=y^2+y\cdot \alpha+\alpha^2\mid y(y+\alpha)+p=y^2+y\cdot \alpha+p\]\[\implies y^2+y\cdot \alpha+\alpha^2\mid p-\alpha^2\]Since $p$ is a prime number, the right hand side is non zero, so we must have $y^2+y\cdot \alpha+\alpha^2\le p-\alpha^2\le p$. Keeping in mind that $p$ divides $x+y$ yields $y^2+y\cdot \alpha+\alpha^2\le x+y=2y+\alpha$. We can easily check that this fails due to size issues whenever $y$ or $\alpha$ is greater than $1$, and when $y=\alpha=1$ we have $x=2$, so $p\mid x+y=3$ and thus $p=3$, but this fails in the given statement. Case 2 : $p\nmid x+y$, and so $p\mid x^2-xy+y^2$. In this case we have $x+y\mid xy+p$ and $p\mid x^2-xy+y^2$. Furthermore, we must have \[k:=\frac{xy+p}{x+y}=\frac{x^2-xy+y^2}{p}\]Note that $k$ must be at least $2$ for this to be possible. Now, note that $x+y\mid xy+p\implies x+y\mid (x+y)^2-3(xy+p)=x^2-xy+y^2-3p=p(k-3)$, thus $x+y\mid k-3$. If $k\ne 3$, then $k\ge 4$ because if $k=2$ we get $x+y\mid 1$, contradiction. Thus we have $x+y\le k-3<k=\frac{xy+p}{x+y}$, so $p>x^2+xy+y^2>x^2-xy+y^2$, contradiction. It follows that we must have $k=3$, so \[x^2-xy+y^2=3p=3(3(x+y)-xy)=9(x+y)-3xy\implies x+y=9\]The rest is casework, and we find that $(8,1,19),(7,2,13)$ and $(5,4,7)$ up to permutation are the only solutions, like @above. $\square$

Let $p\mid x+y$. If $x+y=p$ then we obtain $x^2-xy+y^2 = xy+x+y$, so $(x-y)^2=x+y=p$, a contradiction. Likewise if $x+y\ge 2p$, \[ \frac{x+y}{2} + xy \ge p+xy = \frac{x+y}{p}(x^2-xy+y^2)\ge 2xy, \]a clear contradiction. So, $p\nmid x+y$ and $p\mid x^2-xy+y^2$. In particular, $p\mid x^2-xy+y^2$ and $x+y\mid xy+p$. Setting $x^2-xy+y^2=\ell p$, we thus get $x+y\mid (x+y)^2 = \ell p + 3xy$, so that $x+y\mid (\ell-3)p$. As $(x+y,p)=1$, we get $x+y\mid \ell-3$. Clearly $\ell\ne 1,2$; whereas if $\ell=3$ then $p = (x^2-xy+y^2)/3$, which upon plugging yields $x+y=9$. Finally, we rule out $\ell\ge 4$. To that end, let $x\ge y$ without loss of generality. Note that \[ p=\frac{x^2-xy+y^2}{\ell} \le \frac{x^2-xy+y^2}{x+y+3}<x. \]Next, $x\equiv -y\pmod{x+y}\implies xy+p\equiv -y^2+p\pmod{x+y}$. So, $x+y\mid y^2-p$. If $p>y^2$ then $p-y^2\ge x+y$ so that $p>x$. Finally if $p<y^2$ then $x+y\mid y^2-p$ yields $y^2\ge p+y+x>x$, but on the other hand using $p<x$ we get $p(xy+p)<x(xy+x) =x^2(y+1)$ but $x^3+y^3\ge xy(x+y)$, forcing $x^2(y+1)>xy(x+y) \implies x>y^2$, another contradiction. So $\ell=3$ and $x+y=9$. From here, we immediately deduce the triplets $(8,1,19)$, $(1,8,19)$, $(7,2,13)$, $(2,7,13)$, $(5, 4, 7)$ and $(4,5,7)$.

A problem inspired by this: Determine all prime numbers $p(p\neq 3)$ and all positive integers ${}x$, $y(\gcd (x,y)=1)$ and ${}n$ satisfying $$x^3+y^3=p^n(xy+p^n).$$Proposed by Gong Gu

Taking $\pmod{\gcd(x,y)}$ we have $\gcd(x,y) \mid p^2$ but it cannot be a multiple of $p$ because $\nu_p$ comparision on both sides give a contradiction, so $\gcd(x,y)=1$. If $p\le \tfrac{x+y}{2}$ then \[(x+y)(x^2-xy+y^2) \le \tfrac{x+y}{2} \left(xy + \tfrac{x+y}{2}\right) \iff 2xy \le x+y\]which can be verified to have no solutions. So $p > \tfrac{x+y}{2}$ and if $p \mid x+y$ then $p = x+y$ but plugging this in original equation gives $(x-y)^2 = x+y=p$ which is not possible so $p \mid x^2 - xy + y^2$. Let $x^2 -xy + y^2 = kp$ then $x+y \mid xy + \tfrac{x^2 - xy + y^2}{k} \implies x^2 + y^2 + xy(k-1) \equiv 0 \pmod{x+y} \implies (k-1)xy \equiv 2xy \pmod{x+y} \implies k\equiv 3 \pmod{x+y}$ because $\gcd(x+y,xy)=1$. If $k\neq 3$ then $k \ge 3 + x + y$ so \[xy + p = k(x+y)\ge (x+y)(x+y+3) \implies p \ge x^2 + y^2 + xy + 3x + 3x\]which ofcourse makes the RHS a lot bigger, forcing a contradiction. So $k=3$, note that $3p = x^2 - xy + y^2 \ge xy \implies p \ge \tfrac{xy}{3}$ so \[4(x+y) > 3(x+y) = xy + p \ge xy + \tfrac{xy}{3} \iff (x-3)(y-3)<9\]and a finite casecheck finishes the problem.

Answer $(x,y,p)=(1,8,19),(2,7,13),(4,5,7),(5,4,7),(7,2,13),(8,1,19)$. Step 1 As $p|(x^3+y^3)=(x+y)(x^2-xy+y^2)$, we have $p|(x+y)$ or $p|(x^2-xy+y^2)$. Step 2 If $p|(x+y)$, let $x+y=ap$, $a\in\mathbb Z^+$, then $x^2-xy+y^2=\frac{xy+p}{a}$. If $a=1$, then $p=(x-y)^2$ which is not true. Let $a\geq 2$. Using $p=\frac{x+y}{a}$, we have $x+y=a^2(x^2-xy+y^2)-axy\geqslant 4(x^2-xy+y^2)-2xy\geqslant 2xy$. Therefore $x=y=1$, $a=2$, $p=1$ which is not true. Step 3 If $p|(x^2-xy+y^2)$, and $p\nmid (x+y)$, let $x^2-xy+y^2=ap$, $a\in\mathbb Z^+$, then $x+y=\frac{xy+p}{a}$. From $ap=x^2-xy+y^2=(x+y)^2-3xy=(x+y)^2-3[a(x+y)-p]$, we can get $p(a-3)=(x+y)(x+y-3a)\cdots (*)$. If $a>3$, we have $p|(x+y-3a)$, therefore $x+y-3a\geqslant p\Rightarrow x+y\geqslant p+3a>a-3$, so $(*)$ is not true. If $a=3$, then $x+y=3a=9$, we can get $(x,y,p)=(1,8,19),(2,7,13),(4,5,7),(5,4,7),(7,2,13),(8,1,19)$. If $a=2$, then from $(*)$, we have $p=(x+y)(6-x-y)$ which is not true. If $a=1$, then $x+y=xy+p>xy+1\geqslant x+y$ which is not true as well.

My first post! Let $x+y=a$ and $xy=b$. Then, the given equation can be written as $a(a^2-3b)=p(b+p)$. We proceed with two cases: Case 1. If $p \mid a$, let $a=pk$. Then, we have $b+p=(a^2-3b)k$ from the original equation. Since $a^2 \geq 4b$, we obtain the inequality $$bk^2 \leq (a^2-3b)k^2 = (b+p)k = bk+a \leq bk+b+1 \leq b(k+2).$$Thus, we must have $k=1$ or $k=2$. If $k=1$, we get $b+p=a^2-3b$. Hence, $p=a^2-4b=(x-y)^2$, which contradicts the given condition that $p$ is a prime. If $k=2$, then $b=1$, and setting $x=y=1$ gives no solution. Case 2. If $p \nmid a$, let $a^2-3b=pk$, and $b+p=ak$. Substituting these into the given equation, we get $a^2-3(ak-p)=pk$, which simplifies to $a(a-3k)=p(k-3)$. Since $p \nmid a$, we have $p \mid a-3k$. Let $a-3k=pt$ and $k-3=at$. Then, we have $a-3(at+3)=pt$, which implies $a=3at+pt+9$. If $t>0$, then $a \geq 3a+p+9$, which has no solution. If $t<0$, then $a \leq -3a-p+9$, which yields $8 \leq 4a \leq 9-p$, a contradiction. Therefore, the only possibility is $t=0$, which implies $a=9$ and $k=3$. Now, we consider the four possible cases: $(x,y)=(1,8), (2,7), (3,6),$ and $(4,5)$. From $a^2-3b=pk=3p$, we get $p=27-b$. For each case, we obtain $p=19, 13, 9,$ and $7$. Since $p$ is a prime, the solutions are $\boxed{(x, y, p)=(1, 8, 19), (8, 1, 19), (2, 7, 13), (7, 2, 13), (4, 5, 7), (5, 4, 7)}$.

If $xy > p^2$, then $p(xy+p) < 2pxy < 2(xy)^{\frac 32} \le x^3+y^3$ by AM-GM. Consequently, we can assume $xy \le p^2$. We let $x\le y$ since otherwise, we can flip the choices of $x$ and $y$ to get a working triple $(p,x,y)$. We claim that $y < p$. Otherwise, we have $y \ge p$ and $x \le p$. If $x = p$ and $y = p$ the equality fails, so we may assume $x < p$. The given equation rearranges to $(py - x^2)x = y^3 - p^2$. We then have $y^2(y - 1) \le y^3 - p^2 < pyx \le y^2(p - 1)$, but this is absurd. Accordingly, we have $0 < x\le y < p$. Since $x^3 + y^3 = (x+y)(x^2-xy+y^2)$, one of $p\mid x+y$ and $p\mid x^2 - xy + y^2$ is true. In the former case, we have $p = x+y$, so we get $x^2 - xy + y^2 = xy + p$. But this rearranges to $(x-y)^2 = p$, which cannot be true. Since we have $\left(\frac{x}{y}\right)^3 \equiv -1\pmod p$ and $\frac{x}{y} \not\equiv -1\pmod p$, $p$ must be congruent to 1 modulo 3. For fixed $y$, we have $x^2 - xy + y^2 = (x-y/2)^2 + 3y^2/4 \le y^2/4 + 3y^2/4 = y^2$. Let $x^2 - xy + y^2 = kp$, so we have $k < p$. The equation becomes $k(x+y) = xy + p$. Observe that $k\equiv \frac{xy+p}{x+y} \equiv \frac{1}{3}\cdot \frac{xy + xy + x^2+y^2}{x+y} \equiv \frac{x+y}{3}\pmod p$. For some integer $i$, we get $k = \frac{x+y+ip}{3}$. This transforms $k(x+y) = xy + p$ into $x^2+2xy+y^2 + ip(x+y) = (x+y+ip)(x+y) = 3xy + 3p$. This rearranges to $x^2 - xy + y^2 = p(3 - ix - iy)$. But this expression is also equal to $kp$, so $\frac{x+y+ip}{3} = 3 - i(x+y)$. We clearly cannot have $i \ge 1$ because then $x$ and $y$ are not both 1 and so the expression is not positive. Since $x+y < 2p$, $i$ is at least $-1$. We then have two cases: if $i = 0$, then $x+y=9$, and if $i = -1$, then $9+3(x+y) = x+y-p$, so $2(x+y) = -p-9$ which is absurd. Since we must have $x+y=9$ and $i=0$, we get $x^2 - xy + y^2 = 3p$. The left side may be rewritten as $x^2-x(9-x)+(9-x)^2 = x^2-9x+x^2 + 81 - 18x + x^2 = 3x^2 - 27x + 81$, so we get $x^2 - 9x + 27 = p$. The possible pairs $(x, p)$ are $(1, 19)$, $(2, 13)$, and $(4, 7)$. The resulting triples $(x, y, p)$ are $(1, 8, 19)$, $(8, 1, 19)$, $(2, 7, 13)$, $(7, 2, 13)$, $(4, 5, 7)$, and $(5, 4, 7)$.

Cute problem. I had seen this sort of substitution beforehand, so it was pretty easy for me. Solution: The answers are $(x,y,p) = (8,1,19), (7,2,13), (5,4,7)$ along with permutations of $x,y$. It is easy to see that all of them work. We will now move on showing they are the only possibilities. Without loss of generality, we may assume that $x \ge y$. From here on, it is easy to show that $\gcd(x,y) = 1$ and none of them are divisible by $p$. Set $a \coloneqq x + y$ and $b \coloneqq xy$. Upon re-writing the original equation and solving for $b$ we get \[a^3 - 3ab = bp + p^2 \iff b = \frac{a^3 - p^2}{3a + p}.\]Since $b$ is an integer, we must have $3a + p \mid a^3 -p^2$. Time to do some algebra with this to get a upper bound on the value of $a$. \begin{align*} 3a + p &\mid a^3 -p^2 \\ 3a + p &\mid 3(a^3 -p^2) - a^2(3a+p) \\ 3a + p &\mid pa^2 + 3p^2 \\ 3a + p &\mid a^2 + 3p \\ 3a + p &\mid 3(a^2 + 3p) - a(3a + p) \\ 3a + p &\mid p(9-a) \\ 3a + p &\mid a-9 \end{align*}By size reasons, one can conclude that $a = 9$. It is possible to case-work a handful of values to get that the claimed solutions are the only one. So, we're done! $\blacksquare$

We uploaded our solution https://calimath.org/pdf/RMM2023-1.pdf on youtube https://youtu.be/5t4Nlvvehz0.

Determine all prime $p$ and positive integers $(x,y)$ such that $$\boxed{x^3+y^3=p(xy+p)}$$ $\textbf{Proof:}$ Case 1: $p|x+y$ then there exist a positive integer $k$ such that $x+y=kp$, then we have, $(x^2+y^2-xy)k=xy+p$, Assume for the sake of contrary $k\geq 3$ then, $xy+p\geq 3(x^2+y^2-xy)\geq 3xy$, thus $p\geq 2xy$ thus we have, $kp\geq 6xy$ Hence, $x+y\geq 6xy$ thus $(6x-1)(6y-1)\leq 1$ which is a contradiction as $x, y\geq 1$. Thus,we have two case, as $2\geq k\geq 1$ Subcase 1: $x+y=p$ so we have, $x^2+y^2-2xy=p=(x-y)^2$ clearly as $p\geq 2$ so, $(x-y)^2>1$ but that is a contradiction as $p$ is prime it cannot be a perfect square. Subcase 2: $x+y=2p$ then $4(x^2+y^2-xy)=2xy+x+y\geq 4xy$ which further implies, $x+y\geq 2xy$ thus, $(2x-1)(2y-1)\leq 1$ which has solution $(x, y) =(1,1)$ but $2=x+y=2p$ which is a contradiction as $p$ is a prime number. Case 2: $p|x^2+y^2-xy$, then $x^2+y^2-xy=pa$, $x+y=s$ hence by original equation, $xy+p=as$ $axy+pa=a^2s$ meaning $axy+x^2+y^2-xy=a^2s$. So, $(a-3)xy+(x+y)^2=a^2s$ ,$(a-3)(as-p)=a^2s-s^2$. $s(s-3a)=p(a-3)... (*) $ Assume for the case of contrary $a\geq 4$ Subcase 1: $p|s$ , then as $xy+p=as$ the $p|xy... (i) $ also $x+y=s$ so $p|x+y... (ii) $ thus by $(i)$ and $(ii) $ if $p$ divides any one of $x, y$ then $p$ divides both $x, y$, Hence from the original equation given in the question, $p^2|p$ which is merely impossible. Subcase 2:$p|s-3a$ then $s\geq 3a+p$ but then by $(*)$ $s\leq a-3$. Thus $3a+p\leq a-3$ implying $p<0$ contradiction. Hence we have three choice $a=1,2,3$, If $a=1$, $xy+2\leq xy+p=x+y=s$ hence $(x-1)(y-1)\leq - 1$ contradiction as $x, y\geq 1$ If $a=2$ we have $s(6-s)=p$ Hence $s=5$ as $p$ is prime implying $p=5$,as $xy+p=as$ we have, $xy=5$ so no solution to the original given equation. If $a=3$ we have $s=9$ then $x+y=9$ and $xy+p=27$, $x^2+y^2-xy=3p$ So,$(x,y)=(8,1), (7,2), (5,4)$ and symmetry because equation is symmetric over $(x, y) $ Thus $p=19,13,7$ Note that all works. Done.

This problem is quite cancerous. Claim. $p \nmid x+y$. Proof. Suppose otherwise. Then $x^2+xy+y^2 \mid xy+p$, implying $x^2+y^2 \leq p$, which is clearly impossible. $\blacksquare$ Now let $a = x+y$ and $d = \frac{x^2-xy+y^2}p$ for convenience. As $a \mid a^2-3(xy+p) = p(d-3)$ and $p \nmid a$ by the claim, $a \mid d-3$. If $d > 3$, this is clearly impossible as $$\frac{xy+p}{x+y} - 3 \geq x+y \iff p \geq x^2+xy+y^2+3(x+y)$$which contradicts the premise. Hence we have $d=3$, and in particular $xy \leq 3x+3y$. Now set $x+y = a$ and $xy = b$. By the equations \begin{align*} 3p &= a^2 - 3b \\ b +p &= 3a \end{align*}we obtain $a(a-9) = 0$, hence $a=0$. Then a (needlessly long) case check yields the final solution set $(x, y, p) = (1, 8, 19), (2, 7, 13), (4, 5, 7)$ and equivalent permutations.

We claim the only solutions are \[(x, y, p) = \{(4, 5, 7), (5, 4, 7), (2, 7, 13), (7, 2, 13), (1, 8, 19), (8, 1, 19)\}\]all of which clearly work. Because $p \mid x^3 + y^3$, we know $p \mid x$ if and only if $p \mid y$. Now, suppose $p \mid x, y$ or $x = pa$ and $y = pb$ for some $a, b \in \mathbb{Z^+}$. This yields \[v_p(x^3 + y^3) = v_p(p^3(a^3 + b^3)) = 3 + v_p(a^3 + b^3) \ge 3\]and \[v_p(p(xy+p)) = 1 + v_p(p^2ab + p) = 2 + v_p(pab + 1) = 2\]which is absurd. Hence, we must have $p \nmid x, y$, which also yields \[v_p(p(xy+p)) = 1 + v_p(xy+p) = 1.\]Now, since $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, we have two cases. Suppose $p \mid x+y$ or $x+y = kp$ where $k \in \mathbb{Z^+}$. Then, \[p(xy+p) = (x+y)(x^2-xy+y^2) = kp((x-y)^2 + xy)\]\[xy+p = k((x-y)^2 + xy)\]\[xy(k-1) = p - k(x-y)^2.\]Now, if $k \ge 2$, we have \[xy(k-1) \ge xy \ge 1 \cdot (kp - 1) > p \ge p - k(x-y)^2.\]Hence, we must have $k = 1$, implying $x+y = p$ and \[x^2 - xy + y^2 = xy + p \implies (x-y)^2 = p.\]This gives \[(x, y) = \left\{ \left( \frac{p + \sqrt{p}}{2}, \frac{p - \sqrt{p}}{2} \right), \left( \frac{p - \sqrt{p}}{2}, \frac{p + \sqrt{p}}{2} \right) \right\} \]meaning there are no $x, y \in \mathbb{Z^+}$ for this case. Now, suppose $p \mid x^2 - xy + y^2$ or $x^2 - xy + y^2 = kp$ where $k \in \mathbb{Z^+}$. These conditions give $3xy = (x+y)^2 - kp$ and \begin{align} p(xy+p) = (x+y)(x^2-xy+y^2) = kp(x+y) \\ \nonumber 3k(x+y) = 3(xy + p) = ((x+y)^2 - kp) + 3p \\ p(k-3) = (x+y)(x+y - 3k). \end{align}Now, recall that \[v_p((x+y)(x^2 - xy + y^2)) = v_p(p(xy+p)) = 1 \]so $p \nmid x+y$. This implies $p \mid x+y - 3k$ or $x+y - 3k = mp$ for some $m \in \mathbb{Z}$, which simplifies $(2)$ to $k = m(x+y) + 3$. Because $k \in \mathbb{Z^+}$ and $x+y \ge 2$, we know the only possibilities are $(x, y, m) = (1, 1, -1)$ or $m \ge 0$. For the former case, noting $k = 1$ and using $x^2 - xy + y^2 = kp$ yields $p = 1$, which is invalid. Moreover, if $m \ge 1$, then \begin{align} \nonumber x+y - mp = 3k = 3m(x+y) + 9 \hspace{52.5mm} \text{(3)} \end{align}gives \[(x+y)(1 - 3m) = mp + 9 > 0\]which is a contradiction, so the only remaining possibility is $m = 0$. Indeed, $m = 0$ and $(3)$ immediately imply $k = 3$ and $x+y = 9$, and using $(1)$ gives \[xy + p = k(x+y) = 27.\]Thus, testing all possible pairs $(x, y)$ yields the $6$ aforementioned solutions, which finishes. $\blacksquare$

RMM 2023 p1 We claim the answers are: $(x,y,p)=(1,8,19),(2,7,13),(4,5,7),(5,4,7),(7,2,13),(8,1,19)$ We divide the problem into three parts. Part 1: Showing that $p$ doesn’t divide both $x$ and $y$, and that $v_p(x^3+y^3)=1$. Part 2: Showing that $p>x>y$, implying that $p$ doesn’t divide $x+y$ by size. Part 3: Solving the case $p | x^2+y^2-xy$.

Let $a=x+y$ and $b=x^2-xy+y^2$. We have \[ab=p\left(\frac{a^2-b}{3}+p\right)\]Since $p\mid ab$, either $p\mid a$ or $p\mid b$. $~$ In the first case, either $p=a$ or $p\le \tfrac{a}{2}$. If $p=a$ then $p=(x-y)^2$ which is a contradiction because perfect squares aren't prime. If $p\le \tfrac{a}{2}$ then \[\frac{a^2-b}{3b}+\frac{p}{b}=\frac{a}{p}\ge 2\]Note that $a^2-4b=-3x^2+6xy-3y^2\le 0$ so $a^2-b\le 3b$, meaning $\tfrac{p}{b}\ge 1\implies \tfrac{a}{2}\ge p\ge b\ge \tfrac{a^2}{4}$ so $a\le 2$. This implies $x=y=1$ which gives $p(p+1)=2$ which has no prime solutions. $~$ Therefore, $p\nmid a$, so $p\mid b$. Let $b=pb'$. We have \begin{align*}ab' &= \frac{a^2-pb'}{3}+p \\ 3ab'-a^2 &= 3p-pb' \\ a(3b'-a) &= p(3-b') \end{align*}If $b'=1$ then $p=b$ and $2p=a(3-a)$ which gives no solutions. If $b'=2$ then $p=a(6-a)$ which forces $a$ to be $1$ or $5$, giving $p=5$. Clearly $a\ge 2$ so $a=5$, $b=5$, which solves to give no solutions. If $b'=3$ then $3b'-a=0\implies a=9$. Testing all possibilities, we have the following cases: \begin{align*} (x,y)=(1,8)&:~513=p(p+8),~p=19 \\ (x,y)=(2,7)&:~351=p(p+14),~p=13 \\ (x,y)=(3,6)&:~243=p(p+18),~\text{no solutions}\\ (x,y)=(4,5)&:~189=p(p+20),~p=7 \end{align*}Now to consider the cases where $b'>3$, in which we arrange our equation as such: \[a(a-3b') = p(b'-3)\]Since already established that $p\nmid a$, we have $p\mid a-3b'$. We also already established that $p>\tfrac{a}{2}$ so $p=a-3b'$. This implies that $b'-3=a$, so $p=b'-3-3b'=-2b'-3$ which is not positive at all. Therefore, there are no more solutions.