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I have a question regarding natural numbers that can be divided by $3$.

If $3|m^2 + n^2$, then $3|m$ and $3|n$ where $m$ and $n$ are natural numbers, does this provide a valid proof that every natural number can be written as $3k$, $3k + 1$ or $3k + 2$ with $k ∈ N$.

If so, I'd like to see.

Thanks!

Edit: Citation as asked for. Sorry, it's in German.

  1. Der folgende Satz wird betrachtet: Seien $m$, $n ∈ N$. Ist $m^2 + n^2$ durch $3$ teilbar, so sind $m$ und $n$ durch $3$ teilbar. Entscheiden Sie jeweils, ob die angegebene Argumentation einen gültigen Beweis des Satzes liefert.

(c) Jede natürliche Zahl lässt sich als $3k$, $3k + 1$ oder $3k + 2$ schreiben mit $k ∈ N$. Die folgenden Berechnungen begründen den Satz: $(3j)^2 + (3k + 1)^2 = 3(3j^2 + 3k^2 + 2k) + 1,(3j)^2 + (3k + 2)^2 = 3(3j^2 + 3k^2 + 4k) + 4, (3j + 1) + (3k + 1)^2 = 3(3j^2+3k^2+2j+2k) + 2,(3j + 1) + (3k + 2)^2 = 3(3j^2+3k^2+2j+4k) + 5, (3j + 2)^2 + (3k + 2)^2 = 3(3j^2 + 3k^2 + 4j + 4k) + 8$

CG7
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  • I read that as "$\exists m, n \in \mathbb{N}$ such that if $3 \mid m^2 + n^2$ then $3 \mid m$ and $3 \mid n$. So that doesn't mean all natural numbers can be written as $3k$, $3k +1$ or$3k + 2$ with $k \in \mathbb{N}$ – ewokx Sep 28 '22 at 08:57
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    The reverse implication is true (and is proved here many times), i.e. we can prove $3\mid m^2+n^2\Rightarrow 3\mid m,$ or $,3\mid n,,$ by checking all possible cases $,0,1,2,$ for the remainders of $,m,n,$ on division by $,3.,$ Is this direction what you really intended to ask about (and if not, why the other direction?). Perhaps you mixed up the intended implication direction in an exercise. – Bill Dubuque Sep 28 '22 at 09:08
  • We can also use that the only quadratic residues mod $3$ are $0$ and $1$. Only $0+0$ gives $0$ modulo $3$ , so we can conclude $3\mid m^2+n^2\implies m\equiv n\equiv 0\mod 3$ – Peter Sep 28 '22 at 09:13
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    Technically , the mentioned implication holds (something true can always "prove" something else also true) , but the other direction makes much more sense , namely that we start with that every number has the form $3k,3k+1,3k+2$. – Peter Sep 28 '22 at 09:17
  • @BillDubuque The question asks in my exercise: Given $3 | m^2 + n^2$ then $3 | m$ and $3 | n$, does this provide a vailid proof that every natural number can be written as $3k$, $3k + 1$ or $3k + 2$ with $k ∈ N$. I guess it's more of a yes or no type question. – CG7 Sep 28 '22 at 09:18
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    Could you please cite the source of the exercise. – Bill Dubuque Sep 28 '22 at 09:28
  • @BillDubuque citation added. Is this what you meant? – CG7 Sep 28 '22 at 09:44
  • We can just prove that every number is of the form $3k$ , $3k+1$ or $3k+2$ ,ignoring the implication $3\mid m^2+n^2\implies m\equiv n\equiv 0\mod 3$ , but it would be strange to consider this as a derivation. – Peter Sep 28 '22 at 09:45
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    What you added is exactly the proof by cases in the correct direction. – Peter Sep 28 '22 at 09:47
  • The German text gives the statement "If $3|m^2+n^2$ then $3|m$ and $3|n$". That is the only statement being proved, and all occurrences of "Satz" in the text are referring to that statement. – Jaap Scherphuis Sep 28 '22 at 12:30
  • Your added text shows that - as I surmised above - the exercise is the usual direction of induction by modular cases, which is discussed here in many prior answers, e.g. here and here in the linked dupes. – Bill Dubuque Sep 28 '22 at 16:45

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