Can both
n + 3 and
n^2 + 3 be perfect cubes if n is an integer ?
As both (n+3) and (n^2+3) are perfect cubes, the product (n+3)(n^2+3) = n^3+3n^2+3n+9 = (n+1)^3 + 8 = (n+1)^3 + 2^3 will be a perfect cube.
However, by Fermat's Last Theorem, (n+1)^3+2^3 cannot correspond to a perfect cube.
This is a contradiction. This definitely proves that both n+3 and n^2+3 cannot correspond to a perfect cube for any integer n.
Edited on December 6, 2023, 2:04 am
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