Find n such that both n+3 and n^2+3 are perfect cubes

[kaliprasad]

Find n such that both $n+3$ and $n^2+3$ are perfect cubes

 

[kaliprasad]

Find n such that both $n+3$ and $n^2+3$ are perfect cubes

hint

Spoiler

product of 2 cubes is a cube

 

[Opalg]

hint

Spoiler

product of 2 cubes is a cube

[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]

[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]

Sorry Opalg,

Spoiler

product you evaluated is not correct

 

[Opalg]

Sorry Opalg,

Spoiler

product you evaluated is not correct

[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.

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]

 

[kaliprasad]

[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]

Spoiler

I do not think so. The answer is there is no such n. the set can be empty.