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

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In summary, the conversation discusses the impossibility of finding an integer n such that both n+3 and n^2+3 are perfect cubes. This is due to the fact that two consecutive numbers cannot both be perfect cubes unless they are 0 and 1, which is not possible in this case. The question is deemed unfair for implying that it is possible to find such an n when it is not.
  • #1
kaliprasad
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Find n such that both $n+3$ and $n^2+3$ are perfect cubes
 
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  • #2
kaliprasad said:
Find n such that both $n+3$ and $n^2+3$ are perfect cubes

hint

product of 2 cubes is a cube
 
  • #3
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:
  • #4
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
 
  • #5
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]
 
  • #6
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.
 

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

What does it mean for a number to be a perfect cube?

A perfect cube is a number that can be expressed as the cube of an integer. In other words, it is the result of multiplying a number by itself three times.

How do we find a number that satisfies the given conditions?

To find a number that satisfies both n+3 and n^2+3 being perfect cubes, we can use trial and error or algebraic methods. Trial and error involves testing different values for n until we find one that works. Algebraic methods involve setting up and solving equations with n as the variable.

Can we use any number for n or are there restrictions?

There are restrictions for n in this problem. We are looking for positive integers that satisfy the conditions. Additionally, n must be a multiple of 3 in order for both n+3 and n^2+3 to be perfect cubes.

Are there multiple solutions for n?

Yes, there can be multiple solutions for n. For example, if n=3, then both n+3 and n^2+3 are perfect cubes. However, there can also be cases where there is only one solution or no solution at all.

What are the applications of solving problems like this?

Problems like this can be used to exercise critical thinking and problem-solving skills. They can also be applied in fields such as cryptography, where finding specific types of numbers with certain properties is important.

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