There are no solutions for $p = 2$ so assume $p$ is odd. Note that $n < p$ or else $p^2(p^3+1) = n^2(n^6-1) \geq p^2(n^3+1)(n^3-1) > p^2(p^3+1)$. Thus, $p^2 | n^6 - 1 = n^2(n+1)(n-1)(n^2+n+1)(n^2-n+1)$. If $n = p-1$, we can easily show $n^2(n^6-1) > p^2(p^3+1)$ unless $p =3$ in which case we have the solution $(n,p) = (2,3)$. Now, $p$ divides exactly one of $n^2 +n+1$ and $n^2-n+1$ so in fact $p^2$ is a divisor. We have $p^2 < n^2 + n$ Now, $n^2(n+1)(n-1) | p^3+1$ so $n^4 - n^2 < p^3$. Thus, we have $(n^4 - n^2)^2 < (n^2+n)^3$ or $n^4(n^2-1)^2 < n^3(n+1)^3$ $\iff n(n^4-2n^2+1) < n^3+3n^2+3n+1$ $\iff n^5 - 3n^3 - 3n^2 - 2n - 1 < 0$ which cannot hold for large enough $n$, say $n \geq 3$ Thus the only solution is $(n,p) = (2,3)$.

$p=2$ leads to $n^8-n^2=36$ having no solution.If $p=2,$$n^8-n^2=252=2^2(2^6-1)$,thus $n=2$ We claim that this is the only solution. We are left with $p$ odd,if $n$ even,$n^8-p^5\equiv 0-1\mod 4$ as $\phi (8)=4$.But every factor of $n^2+p^2$ must be of the form $4k+1$,otherwise $4k-1|p$ leading to a contradiction.Because then $n^8-p^5>p$ almost trivially.So $n^8-p^5$ cant divide $p$.Then both $n,p$ odd. $n^2+p^2\equiv 2\mod 8$ $\implies n^8-p^5\equiv 2$ $\implies 1-p\equiv 2$ $\implies p\equiv 1$ $\implies n^8-p^5\equiv 1-1\equiv 0\mod 8,$contradiction!

mathmdmb wrote: We are left with $p$ odd,if $n$ even,$n^8-p^5\equiv 0-1\mod 4$ as $\phi (8)=4$. I think it is wrong. We cannot say that $p^5 \equiv 1 \pmod 4$ for odd prime $p.$

Valentin Vornicu wrote: Find all positive integers $n$ and $p$ if $p$ is prime and \[ n^8 - p^5 = n^2+p^2 . \] Adrian Stoica Lemma 1 : If $p$ be a prime and $k$ be a positive integer such that $p\mid k^2+1$, we have $p\equiv 1\;\pmod 4$. Lemma 2 : Given $x$ be a positive integer such that $p\mid x^2+x+1$ or $p\mid x^2-x+1$, we have $p\equiv 1\;\pmod 3$. Back to problem : Easily, there are no solutions for $p=2$. Suppose $p$ is odd. If $p=3$, we find out $n=2$. Considering $p>3$. We have : \[p^2(p^3+1)=n^8-n^2\equiv 0\;\pmod 4\]. Since $p$ odd : \[p^3+1\equiv 0\;\pmod 4\Rightarrow p\equiv 3\;\pmod 4\] We also have : \[p^2(p^3+1)=n^8-n^2\equiv 0\;\pmod 3\Rightarrow 3\mid p^3+1\Rightarrow p\equiv 2\;\pmod 3\] From the given equation, we get : \[n^3+1 \mid p^2(p^3+1)\] So : \[p^5+p^2\geq n^3+1\;\;\;(1)\] The given equation can be written as : \[n^2(n^3-1)(n^3+1)=p^2(p^3+1)\] Implies : \[p^2\mid n^2(n^3-1)(n^3+1)\] If $p\mid n$, let $n=pk$ : \[(kp)^8-(kp)^2=p^5+p^2\Leftrightarrow k^8.p^6-k^2=p^3+1\Leftrightarrow k^2+1=p^3(k^8p^3-1)\] Hence $p\mid k^2+1$ and apply lemma 1, we get $p\equiv 1\;\pmod 4$, a contradiction ! Thus : \[p^2\mid (n^3-1)(n^3+1)\] But $\gcd(n^3-1,n^3+1)\mid 2$, we divides two cases : Case 1 : If $p^2\mid n^3-1$, or $p^2\mid (n-1)(n^2+n+1)$. But $\gcd(n-1,n^2+n+1)\mid 3$ and $p>3$ : If $p\mid p ^2 \mid n^2+n+1$, use lemma 2, we imply $p\equiv 1\;\pmod 3$. A contradiction ! Hence $p^2\mid n-1$. So : \[n-1\geq p^2\;\;\;(2)\] From $(1)(2)$ we deduce : \[p^5+p^2\geq n^3+1\geq (p^2+1)^3+1\] Contradiction ! Case 2 : If $p^2\mid n^3+1$ or $p^2\mid (n+1)(n^2-n+1$. Similar to case 1, we get $p^2\mid n+1$, hence $n\geq p^2-1\;\;(3)$. From $(1)(3)$ : \[p^5+p^2\geq n^3+1\geq (p^2-1)^3+1\Leftrightarrow p^4+2\leq p^3+3p\] But $p^4+2> 3p^3+2> p^3+3p$. Thus, $(n,p)=(2,3)$

We first manipulate the equation to get \[ n^2(n^3+1)(n^3-1)=p^2(p^3+1) \]. Clearly there's no solution for $p=2$, assume from now that $p$ is odd. Note that the factors on the left cannot pair-wisely share any factors larger than $2$, therefore $p^2$ divide exactly one of the factors, which leads to the following three cases: Case 1: $p^2|n^2$ or $p|n\Rightarrow n=pk$ for $k\in \mathbb {N}$. However since $p^3+1\ge n^3+1$, therefore $k=1\Rightarrow n^3-1=1\Rightarrow $no solution. Case2: $p^2|n^3-1$, let $n^3-1=p^2k$ for $k\in \mathbb {N}$. We have \[n^2k(p^2k+2)=p^3+1\], this means $p^2k+2|p^3+1\Rightarrow p^2k+2|2p-k$. Since $k<p$, therefore $2p-k\ge p^2k+2\ge p^2+2\Rightarrow 2p-1\ge p^2+2 \Rightarrow$ no solution Case3: $p^2|n^3+1$, let $n^3+1=p^2k$ for $k\in \mathbb {N}$. We have \[n^2k(p^2k-2)=p^3+1\], this means $k<p+1,p^2k-2|p^3+1\Rightarrow p^2k-2|2p+k\Rightarrow 2p+k\ge p^2k-2$ $\ge p^2-2\Rightarrow 3p+1>p^2-2\Rightarrow p=2,3$ which gives the solution $(p,n)=(3,2)$ In conclusion after covering all possible cases, $(p,n)=(3,2)$ is the only solution. $\Box$

Nice probem. I used some ineqalities(number theory, divisibility) and got that n cannot be greater than 2.