Can both
n + 3 and
n^2 + 3 be perfect cubes if n is an integer ?
Three cheers for analytic number theory:
If both (n + 3) and (n^2 + 3) were perfect cubes, then their product would be a perfect cube as well.
However, (n + 3)(n^2 + 3) = n^3 + 3n^2 +3n + 9 is bounded between the functions (n + 1)^3 and (n + 2)^3 for all values of n less than -3 and greater than zero, and thus cannot be a perfect cube in those instances. One then verifies that for the remaining four cases the above fact does not hold.
Proof of the above claims:
(n + 1)^3 = n^3 + 3n^2 + 3n + 1
This is always 8 less than the polynomial in question, and thus forms a lower bound.
(n + 2)^3 = n^3 + 6n^2 + 12n + 8
subtracting our polynomial from this we get:
3n^2 + 9n - 1 which is greater than zero for values of n in the range described above.
Furthermore, n^2 + 3 takes the values 12, 7, 4, and 3 for the remaining cases, none of which are perfect cubes.
QED.
Cheers! Let me know what you think.
Solution
