# On the equation $m^3-m^2+1 = n^2$

(i) How can I find all positive integers $m$ such that $m\equiv 4 \pmod 7$ and $m^3-m^2+1$ is a perfect square?

(ii) Is there a method to solve this equation over positive integers: $$m^3-m^2+1 = n^2.$$

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Notice that if $m \equiv 4 \mod 7$ then $n^2 = m^3 - m^2 + 1 \equiv 0 \mod 7$. This means that $7$ divides $n$ and $7^2$ divides $m^3-m^2+1$. It might help... –  Joel Cohen May 29 '11 at 14:29
The equation describes an elliptic curve of rank 1 with trivial torsion subgroup. See my answer here on how to find integral points on elliptic curves in general: math.stackexchange.com/questions/30457/… –  Alex B. May 29 '11 at 14:59
Thanks... Can someone please help me with the cube version of the problem? I mean solving this equation: $m^3−m^2+1=n^{\color{red}3}$ in positive integers. –  Amir Hossein May 29 '11 at 15:29
A hint/comment for the cube version (new poster, not good enough reputation to comment, sorry): [Edit: moved this to become a comment now that I CAN comment] If $m$ and $n$ are positive integers, then you are left with relatively few alternatives given that whenever $m>1$ we have the inequalities $$(m-1)^3=m^3-3m^2+3m-1< m^3-m^2+1< m^3.$$ Thus your choices for $n$ are severely limited given that cubing is an increasing function :-) –  Jyrki Lahtonen Jun 8 '11 at 6:10