The discussion centers on finding an integer \( n \) such that both \( n+3 \) and \( n^2+3 \) are perfect cubes. The scope includes mathematical reasoning and exploration of the properties of cubes.
Participants generally disagree on the possibility of finding such an \( n \). While some believe it is impossible, others engage with the mathematical reasoning without reaching a consensus.
The discussion includes unresolved mathematical steps and assumptions regarding the properties of cubes and their relationships.
kaliprasad said:Find n such that both $n+3$ and $n^2+3$ are perfect cubes
[sp]So if $n+3$ and $n^2+3$ are cubes then so is $(n+3)(n^2+3) = n^3 + 3n^2 + 3n = (n+1)^3 - 1$. But two consecutive numbers cannot both be cubes unless they are $0$ and $1$. Since $n^2 + 3 \geqslant 3$, the pair $0$ and $1$ cannot arise in this way. Therefore the question is asking for something impossible (which I think is a bit sneaky! (Tongueout) ).[/sp]kaliprasad said:hint
product of 2 cubes is a cube
Opalg said:[sp]So if $n+3$ and $n^2+3$ are cubes then so is $(n+3)(n^2+3) = n^3 + 3n^2 + 3n = (n+1)^3 - 1$. But two consecutive numbers cannot both be cubes unless they are $0$ and $1$. Since $n^2 + 3 \geqslant 3$, the pair $0$ and $1$ cannot arise in this way. Therefore the question is asking for something impossible (which I think is a bit sneaky! (Tongueout) ).[/sp]
[sp]Oops, yes, of course it should be $(n+3)(n^2+3) = n^3 + 3n^2 + 3n + 9 = (n+1)^3 + 8$. But the only cubes that differ by $8$ are $0$ and $\pm8$. Since none of those numbers is of the form $n^2+3$, no such $n$ exists.kaliprasad said:Sorry Opalg,
product you evaluated is not correct
Opalg said:[sp]
I still think it is unfair to phrase the question in a way that implies it is possible to find $n$, when in fact it is not possible.[/sp]