Can both
n + 3 and
n^2 + 3 be perfect cubes if n is an integer ?
One can easily prove that n cannot be odd:
Suppose that n = 2m + 1 for an integer m, and further suppose that q^3 = n^2 + 3 for an integer q. Then by substitution we would have that q^3 = 4m^2 + 4m +4 = 4(m^2 + m + 1).
By considering the prime factorization of q^3 we see that in order for the above equation to be true, m^2 + m + 1 must be
even (for else there will not be sufficient powers of 2 in q^3).
However, this is impossible as that expression always results in an odd output for integral entries.
How does that sound?
A preliminary thought for half of a solution
