$p^{3}+m^{2}$ is square of a number. Well i thought it is a nice problem so i will post it here.
1) Prove that for every natural numbers $m$, There is at most two primes $p$ where $p^{3}+m^{2}$ is a perfect square.
2) Find all natural numbers $m$ so that $p^{3}+m^{2}$ is the square of a number for exactly $2$ prime numbers $p$.
 A: I have my own answer so tell me if you had a better answer :)
Suppose that $p^{3}+m^{2}=b^{2}$ for some natural numbers $b$. So, $(b-m)(b+m)=p^{3}$ We know that the number $p^{3}$ can only be written as $(1)(p^{3}),(p)(p^{2}),(p^{2})(p),(p^{3})(1)$. Knowing $b+m>b-m$ makes the only answers of $(b-m,b+m)$ be $(1,p^{3}),(p,p^{2})$.

Case 1: $b-m=1$ and $b+m=p^{3}$
This results in $p^{3}-1=2m$ and so $p=\sqrt[3]{2m+1}$
Case 2: $b-m=p$ and  $b+m=p^2$
This one results in $p^2+p=2m$ and so $p=\frac{1+\sqrt{8m+1}}{2}$ (since the negative one is unacceptable.)

This itself solves the first part since for every $m$, $p$ can only be first one, second one, both, or non.
Now we have to find all numbers $m$ so that both $\frac{1+\sqrt{8m+1}}{2}$ and $\sqrt[3]{2m+1}$ are prime. We first find all values of $m$ which $8m+1$ is a square. Suppose $8m+1=n^2$ for some numbers $n$. Then, $(4m+1)^2=n^2+(4m)^2$. Using pythagorean theorem, we find that all possible values of $m$ are in the form of $\frac{x(x+1)}{2}$. Putting $m$ in the first case results in $2*\frac{x(x+1)}{2} +1=p^{3}$ and so $x^2+x+1=p^3$. Now using the solutions which are given in  here, the only possible values of $(x,p)$ becomes $(18,7)$, and so $m=\frac{x(x+1)}{2}=\frac{18*19}{2}=171$.Checking this answer in both cases, we obtain $p=19$ and $p=17$. So the only possible $m$ becomes $171$. 
