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

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The discussion revolves around finding an integer n such that both n+3 and n^2+3 are perfect cubes. It is established that if both expressions are cubes, their product also forms a specific cubic expression. However, it is noted that two consecutive cubes can only be 0 and 1, which is not applicable here since n^2 + 3 is always greater than or equal to 3. The conclusion drawn is that no such integer n exists, making the problem inherently impossible. The phrasing of the question is criticized for suggesting that a solution might be possible when it is not.
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Find n such that both $n+3$ and $n^2+3$ are perfect cubes
 
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kaliprasad said:
Find n such that both $n+3$ and $n^2+3$ are perfect cubes

hint

product of 2 cubes is a cube
 
kaliprasad said:
hint

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]
 
Last edited:
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]

Sorry Opalg,

product you evaluated is not correct
 
kaliprasad said:
Sorry Opalg,

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]
 
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]

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

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