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Find all natural numbers $b>3$ such that the number $1331_b$ is the cube of some natural number.

In the notation above, "$1331_b$" means the number $1331$ is in base $b$.

I found that $1b^3+3b^2+3b^1+1b^0$ is $(b+1)^3$, does it means that the thing is a cube for all b>3?

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    $\begingroup$ Your work shows that the thing is a cube for all $b\gt 3$. $\endgroup$ Commented Oct 24, 2015 at 21:43
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    $\begingroup$ and that this number is b+1. $\endgroup$ Commented Oct 24, 2015 at 21:49
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    $\begingroup$ Similarly, $(b+1)^4 = b^4+4b^3+6b^2+4b+1$, so $14641_b$ is a fourth power for any base $b > 6$. $\endgroup$ Commented Oct 24, 2015 at 21:50
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    $\begingroup$ Or put another way, the number is $11_b$. $\endgroup$ Commented Oct 24, 2015 at 22:09
  • $\begingroup$ Yes, I figured out. Thanks! $\endgroup$
    – dreamer
    Commented Nov 21, 2015 at 19:46

1 Answer 1

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With any b, $(b+1)^3= 1*b^3+3*b^2+3*b^1+1*b^0$ so 1331 in base b is always equal to (b+1)^3 for b>3

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