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In doing unrelated research, I conjectured that $n^3+(n+3m)^2$ is divisible by $3$ for all n in $[2+k,3+k]$ for all $k$ in $\mathbb{N}$ for all m in $\mathbb{N}$.

I'd like to prove this algebraically, but don't know how to frame the function for $n$.

Siong Thye Goh
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Calogero
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  • Try $k=2, m=3$, $n=4$. – markvs Oct 31 '21 at 12:59
  • Firstly, your $m$ plays no role, since it is multiplied by $3$ and this product is always divisible by $3$. Further, what if $k=1,n=4$ (and, whatever, say $m=0$)? – Zuy Oct 31 '21 at 13:01
  • you're right, n=4 fails. and you're right the setting k=2 makes 2+k =4. So even the setting of 2+k, 3+k is wrong. What I need is a function that gives 2, 3, 5, 6, 8, 9, ... – Calogero Oct 31 '21 at 13:06
  • m does play a role, since the n is added to 3m, n+3m is not always divisible by 3 – Calogero Oct 31 '21 at 13:11
  • You can easily prove that $n^3+(n+3m)^2$ is divisible by $3$ if and only if $n^3+n^2$ is (for instance using modular arithmetic). – Zuy Oct 31 '21 at 13:22
  • I tried m=0, which reduces the expression to n^3+n^2 and found that that is always divisible by 3 for any n in 2, 3, 5, 6, 8, 9, ... – Calogero Oct 31 '21 at 13:23
  • I don't know why the comment that gave me what I wanted was deleted, but thanks to whomever gave me the solution – Calogero Oct 31 '21 at 13:30

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I think that you wanted to say "for all $n$ in $[2 + 3k, 3 + 3k]$", since I can't see what purpose $k$ would have otherwise.

I will use congruence relations to prove your conjecture :

You have two cases :

  • $n \equiv 0 \mod 3$ (Meaning $n = 3 + 3k$):

    Then $n^3 \equiv 0 \mod 3$.

    Also, since $3m \equiv 0 \mod 3$, then $n + 3m \equiv 0 \mod 3$. Therefore, $(n + 3m)^2 \equiv 0 \mod 3$.

    Summing our results, we get that $n^3 + (n + 3m)^2 \equiv 0 \mod 3$, which means that your conjecture is true for $n = 3 + 3k$.

  • $n \equiv 2 \mod 3$ (Meaning $n = 2 + 3k$):

    Then $n^3 \equiv 8 \mod 3 \equiv 2 \mod 3$.

    Also, since $3m \equiv 0 \mod 3$, then $n + 3m \equiv 2 \mod 3$. Therefore, $(n + 3m)^2 \equiv 4 \mod 3 \equiv 1 \mod 3$.

    Summing up, we get that $n^3 + (n + 3m)^2 \equiv 2 + 1 \mod 3 \equiv 0 \mod 3$, which means that your conjecture is true for $n = 2 + 3k$.

In conclusion, your conjecture is true for all $n$ in $[2 + 3k,\,3 + 3k],\,k \in \mathbb{N}$.

Pankkake
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