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.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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