$\left(m,n\right)=\left(k^{2}+1,k\left(k^{2}+1\right)\right)$ satisfies the equation for any integer $k$. So $N$ is infinite.

N is infinite. Take $m=k^2+1$ edit: sniped

$m^2+n^2=m^3 \rightarrow n^2=m^2(m-1)$ If $K$ is positive integer then plugging the value $m=K^2+1$ gives infinite solution for this equation . So,$N$ is infinte.

Sorry for bumping, but I have a rather idiotic question about this. Rearranging this equation, we see that: $$m^3 - m^2 = n^2$$$$m^2(m - 1) = n^2$$Where does $k^2 + 1$ come from though, I didn't understand that part.

ATGY wrote: Sorry for bumping, but I have a rather idiotic question about this. Rearranging this equation, we see that: $$m^3 - m^2 = n^2$$$$m^2(m - 1) = n^2$$Where does $k^2 + 1$ come from though, I didn't understand that part. $gcd(m^2, m-1) =1 $ if $m-1$ has a prime divisor such that $v_p(m-1) = 1\pmod2$ then $$v_p(m^2(m-1))=v_p(m-1)=2l+1=v_p(n^2)$$which is absurd since $v_p(n^2)$ always even. And for every $p$ that divides $m-1$ we must have $v_p(m-1)=0 \pmod2 \implies m-1 = k^2$ for some $k \in\mathbb{N}.$

What do you mean by $V_p$... Sorry for the dumb questions

$n^2=m^2(m-1)$ $\frac{n}{m}=\pm\sqrt{m-1}$ if $m-1$ is not a perfect square, then $\frac{n}{m}$ becomes irrational which is not possible since $n,m \in \mathbb{Z}$ so $m-1$ must be a perfect square, for any solution to exist (which does) $m-1=k^2 \implies m = k^2+1, n= \pm k(k^2+1)$ Also @above, it's something related to LTE I think.

ATGY wrote: What do you mean by $V_p$... Sorry for the dumb questions $v_p(n)$ is defined as the largest integer $k$ such that $p^k \mid n$, where $p, n$ are positive integers and $p$ prime. Or it's simply the exponent of $p$ in the prime factorization of $n$. If $n$ is a perfect square, then $v_p(n)$ must be even.