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This can never be true.
Claim:
n+3 and n² + 3 can never be perfect cubes.
Proof (by contradiction):
Suppose not; suppose that ∃ a, b, n ∈ Z ∋ a³=n+3 and b³=n²+3.
Then:
(a³)(b³) = (ab)³
= (n+3)(n²+3)
= n² + 3n² + 3n + 9
So, n² + 3n² + 3n + 9 = (ab)³ is a perfect cube.
Also, we can show that:
(n+1)³ = n³ + 3n² + 3n + 1
= (n² + 3n² + 3n + 9) - 8
= (n+3)(n²+3) - 8
and
(n+2)³ = n³ + 6n² + 12n + 8
= (n² + 3n² + 3n + 9) + 3n² + 9n - 1
= (n+3)(n²+3) + 3n² + 9n - 1
Also, (3n² + 9n - 1) is positive for all integers except for n∈{0,-1,-2,-3}
(the proof of this is relatively inconsequential and is left to the reader).
Thus, ∀ n ∈ Z, (n+1)³ < (n+3)(n²+3) < (n+2)³,
and (n+1) < 3√[(n+3)(n²+3)] < n+2 (assuming -3<n or n>0).
Since there are no integers between n+1 and n+2, there is no integer that when cubed equals (n+3)(n²+3) for any integer n (-3<n or n>0).
Now, for the remaining cases, it is simple enough to prove by exhaustion that 0²+3=3, (-1)²+3=4, (-2)²+3=7, and (-3)²+3=12, none of which are perfect cubes.
Thus, the statement that (n+1)(n²) is a perfect cube is false, so the assumption must be false, and the original claim must be true.//
Other methods for proving this can be found in the problem comments. |
