Let $x^2=p^3-4p+9$. Since $x^2 \equiv 9 \pmod p,$ we have $x=kp\pm 3$ where $k$ is an integer. Then $(kp \pm 3)^2=p^3-4p+9 \: \Longrightarrow \: k^2p \pm 6k =p^2-4,$ and we obtain $ p \mid 6k \pm 4.$ If $p \neq 2,$ then $p \mid 3k \pm 2 \: \Longrightarrow \: p \leq 3k+2 \Longrightarrow \: \frac{p-2}{3} \leq k \: \Longrightarrow \: \frac{p^2-2p-9}{3} \leq pk-3 \leq x.$ Case-1: If $x \leq \frac{p^2}{4},$ then we have $\frac{p^2-2p-9}{3} \leq \frac{p^2}{4} \: \Longrightarrow \: p \leq 8+\frac{36}{p} \: \Longrightarrow \: p \leq 11.$ Case-2: If, on the other hand, $x> \frac{p^2}{4},$ then $x^2=p^3-4p+9 \: \Longrightarrow \: \frac{p^4}{16}<p^3-4p+9 \: \Longrightarrow$ $p<16-\frac{16(4p-9)}{p^3} \: \Longrightarrow \: p \leq 13.$ Finally, for $p \leq 13, \: (p,x)=(2,3), \: (7,18),$ and $(11,36)$ are the only solutions.

Writing $p^3-4p+9 = a^2$, thus $p(p^2-4) = (a-3)(a+3)$, we need $p\mid a\pm 3$, so $a = kp \mp 3$. Then $p^2 -k^2p \pm 6k - 4 = 0$. Its discriminant is $k^4 \mp 24k + 16$ needs be a perfect square, but $(k^2)^2 < k^4 + 24k + 16 < (k^2+1)^2$ for $k\geq 13$ and $(k^2-1)^2 < k^4 - 24k + 16 < (k^2)^2$ for $k\geq 12$. Thus there are only few cases to check, and although we could devise some methods to reduce the computations, it's easy enough to get that the only "good" value is $k=3$, leading to $p\in \{2,7,11\}$.

Mathematicalx wrote: Find all prime $p$ numbers such that $p^3-4p+9$ is perfect square. Hello Mathematicalx! Doing a search would be better before posting problems, you see.

YESMAths wrote: Mathematicalx wrote: Find all prime $p$ numbers such that $p^3-4p+9$ is perfect square. Hello Mathematicalx! Doing a search would be better before posting problems, you see. What is the point of bumping a 5 months old thread to advise someone to search?

So, that the OP doesn't keep posting problems without searching and realises the importance of search.

Setting the expression equal to $n^2$ and rearranging, we get $p(p-2)(p+2) = (n-3)(n+3)$. We split cases now. $\textbf{Case 1:}$ $p$ divides $n-3$. Write $n = mp + 3$ for some positive integer $m$. Then we have, upon substitution, that\[p^2 - 4 = m(n + 3) = m(mp + 6) \implies p^2 - m^2p - 6m - 4 = 0.\]Solving using the quadratic formula for $p$ we see that the discriminant is\[\Delta = m^4 + 4(6m + 4) = m^4 + 24m + 16\]which is squeezed between $(m^2)^2$ and $(m^2 + 1)^2$ for $m \geq 13$. $\textbf{Case 2:}$ $p$ divides $n+3$. Write $n = mp - 3$ for some positive integer $m$. Then we have, upon substitution, that\[p^2 - 4 = m(n - 3) = m(mp - 6) \implies p^2 - m^2p + 6m - 4 = 0.\]Solving using the quadratic formula for $p$ we see that the discriminant is\[\Delta = m^4 - 4(6m - 4) = m^4 - 24m + 16\]which is squeezed between $(m^2 - 1)^2$ and $(m^2)^2$ for $m \geq 12$. So we just need to check $m < 12$. This is slightly painful, but slogging through computation yields that $m = 3$ is the only one that results in any discriminant being a perfect square. Plugging this back in we see that $\boxed{p = 2, 7, 11}$ are our answers.