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how could i prove that a cubic number $ u^3 $ is always of the form $ 9k+n $ where $n=0,1,2,3,4,\ldots,8$

can a similar proof be made to prove that a power of n $ n^k $ is of the form

$ am+b$ where k,m,a,b,u and n are integers $a=a(m) $ and $ b=0,1,2,3,\ldots,a-1 $

what congruence should i solve ? apparently i should study the congruence

$ u^3=b \pmod 9$ but i have no idea how to solve this.

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What about $27 = 3^3$? – Lord_Farin Oct 13 '12 at 22:15
$ 27=9*3 $ yes i forgot the case $n=0$ – Jose Garcia Oct 13 '12 at 22:16
up vote 3 down vote accepted

Every integer has form $\rm\: 9\,q+r\:$ with remainder $\rm\:0\le r< 9\:$ by the division algorithm.

Less trivially, one may show that cubes are either $\,0\,$ or $\,\pm1\!\pmod 9$.

$\rm {\bf Hint}\ \ \ mod\ 9\!:\ (j + 3k)^3 =\, j^3\! + 9(\cdots)\,\equiv\, j^3\in \{0,1,2\}^3 \equiv \{0,1,-1\}$

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