TITLE: For how many positive integers $m$ the number $m^3+5m^2+3$ is a perfect cube?
QUESTION [10 upvotes]: For how many positive integers $m$ the number $m^3+5m^2+3$ is a perfect cube?
What if $m$ is a negative integer?
It intuitively seems to me that no positive integer can satisfy the given condition, but I can't prove it mathematically.

REPLY [2 votes]: Assume, for a contradiction,that it's a perfect cube and notice : $m^3 < m^3 +5m^2+3 < (m+2)^3$ so we must have $m^3+5m^2+3=(m+1)^3$ but that equation has no integer solutions.