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As the title says, I have to prove this. Since n just ranges from 1 to 5, I wondered if it is okay to just prove it like the following?

For n=1: $5|{1}^3+k$, when $k \in${…,-11,-6,-1,4,9,14,…}

For n=2: $5|{2}^3+k$, when $k \in${…,-13,-8,-3,2,7,12,…}

For n=3: $5|{3}^3+k$, when $k \in${…,-12,-7,-2,3,8,13,…}

For n=4: $5|{4}^3+k$, when $k \in${…,-14,-9,-4,1,6,11,…}

For n=5: $5|{5}^3+k$, when $k \in${…,-15,-10,-5,0,5,10,15,…}

Labeling the different sets of k as ${k}_1,{k}_2,…,{k}_5$,

we see that ${k}_1 \cup {k}_2 \cup {k}_3 \cup {k}_4 \cup {k}_5 = \mathbb{Z}$

Bill Dubuque
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J3ck_Budl7y
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  • This proof would work if you correct the congruence classes for $n=2,3$ and the union over the $k_i$ still produces $\Bbb Z$... (Hint: those two are swapped.) – abiessu Sep 13 '22 at 05:04
  • Thanks for the hint @abiessu, mistyped that. – J3ck_Budl7y Sep 13 '22 at 05:15
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    Note $,5\mid n^3+k\iff \bmod 5!:\ n^3\equiv -k\iff -k,$ has a $\rm\color{#c00}{cube}$ root $!\bmod \color{#0a0}5,,$ which holds true by a simple Theorem proved in the dupe linked above, because $\color{#c00}3$ is coprime to $,\phi(\color{#0a0}5)=4.,$ It yields for $,a\not\equiv 0, $ that $n^3 \equiv a !\iff! n\equiv a^3,,$ e.g. $,n^3\equiv 2\iff n\equiv 2^3\equiv 3.\ \ $ – Bill Dubuque Sep 13 '22 at 08:15

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