Does there exist an integer such that its cube is equal to $3n^2 + 3n + 7,$ where $n$ is an integer.
Problem
Source: IMO Longlist 1967, Poland 4
Tags: modular arithmetic, number theory, perfect cube, Diophantine equation, IMO Shortlist, IMO Longlist
orl
14.10.2005 20:17
Please post your solutions. Use $\LaTeX$ please! Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:
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Happy Problem-Solving!
Arne
14.10.2005 20:43
We have \[ 3n^2 + 3n + 7 \equiv 4\ \textrm{ or }\ 7\ (\textrm{mod}\ 9) \] while perfect cubes are always congruent to 0, 1 or 8 (mod 9). So, there are no solutions. Solution by grobber: Assume such a number $k$ exists. We have $k^3-1=3(n^2+n+2)$, so $k\equiv 1\pmod 3$. Let $k=3a+1$. We then get $27a^3+27a^2+9a=3(n^2+n+2)\iff 9a^3+9a^2+3a=n^2+n+2$, but $n^2+n+2$ cannot be divisible by $3$. We have a contradiction, so no such $k$ exists.
DensSv
28.11.2024 19:43
Assume there exists such $k\in \mathbb{Z}$ for whom $3n^{2}+3n+7=k^{3}$. Now we should add up $n^{3}$ and the equation becomes $(n+1)^{3}+6=k^{3}+n^{3}$.
Observe that $(n+1)^{3}+6\equiv 5,6,7 \pmod{9}$ whilst $k^{3}+n^{3}\equiv 0,1,2,7 \pmod{9}$, so $(n+1)^{3}\equiv 1\pmod{9}$ and $n^{3}\equiv k^{3}\equiv 8\pmod{9}$. In order for $n^{3}\equiv 8 \pmod{9}$, $n$ must be congruent to one of these $2,5,8 \pmod{9}$ (just check) thus, $n+1\equiv 3;6;0 \pmod{9}$ hence $(n+1)^{3}\equiv0\pmod{9}$ contradiction.
AshAuktober
29.11.2024 05:20
$3n^2+3n+7 \equiv 4, 7 \pmod{9}$, contradiction.