Find all integers m such that m + 3 and m^2 + 3m + 3 are perfect cubes?
let me give you an algorithm
let m+3=a^3
then m=a^3-3
now put values of perfect cubes (its easy) in place of a^3.for each such value you will get an integral value of m.
Do the same for m^2 + 3m + 3
that is m^2 + 3m + 3-a^3=0
solve this and insert different value of a^3.this time every solution may not be integral.
Did you find a solution to this? My 4 month old daughter is tired of waiting while I try to crack it!
I have made a couple of observations.
1. It is easy to show that m = -2 is a solution. This comes from the special case when $\displaystyle m+3=a^3=b^3=m^2+3m+3$. Solving for m gives $\displaystyle m^2+2m=0$ and consequently m= -2 or m = 0.
2. If we let $\displaystyle m+3=a^3$ then the second equation can be written as $\displaystyle m*a^3+3=b^3$ which looks very much like the first equation.