The problem of finding an integer n such that both n+3 and n²+3 are perfect cubes is impossible. The analysis shows that if both expressions are cubes, their product (n+3)(n²+3) simplifies to (n+1)³ + 8. The only cubes that differ by 8 are 0 and ±8, which do not yield valid values for n²+3. Thus, no integer n satisfies the given conditions.
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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]