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If I were to prove $2\mid n^2+n$ for all integers $n$, would the following proof by induction be valid?

Base: $2\mid 1^2+1 = 2$, so this is obviously true.

Hypothesis: $2 \mid (k^2+k) \to 2 \mid ((k+1)^2+(k+1))$

Induction: \begin{align} &2 \mid k^2 + k \to 2 \mid (k+1)^2 + (k+1)\\ &2 \mid k^2+k\\ &2 \mid 2 \quad \therefore 2 \mid 2(k+1) \quad \because 2 \mid a \to 2 \mid ac \quad \text{for all integers} \ c \ \text{(in this case} \ a=2, c=k+1)\\ &(2 \mid 2k+2) \wedge (2 \mid k^2+k)\\ &2 \mid k^2+k+2k+2 \quad \because (2 \mid a) \wedge (2 \mid b) \to (2 \mid a+b)\\ &2 \mid k^2+3k+2\\ &2 \mid k^2 +2k+1+(k+1)\\ &2 \mid (k+1)^2+(k+1)\\ \end{align} Therefore, by the principle of mathematical induction we conclude that $2 \mid n^2+n$.

I think it all makes sense, but I'm not sure if it properly qualifies as induction. Please let me know if I'm missing any steps/made any incorrect assumptions.

Ѕᴀᴀᴅ
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Omaro_IB
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    Looks about okay to me. – Prometheus Dec 02 '21 at 01:33
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    I would say that it's missing all the words: it's all symbols! But I guess some people think that makes it look more "mathematical" or something. – Misha Lavrov Dec 02 '21 at 01:36
  • It seems like you may have the correct idea but it is impossible to be sure since you omit the logical connectives necessary to uniquely parse your argument. Here's is a clearer way to do the inductive step. Let $,f_k = k^2+k.,$ Then $2\mid f_{k+1} = f_k + 2(k!+!1)$ by $,2\mid 2(k!+!1)$ and $,2\mid f_k,$ (by induction hypothesis). See here for how to view this more intuitively via "telescopic" cancellation. – Bill Dubuque Dec 11 '21 at 17:30

1 Answers1

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Your induction solution is correct.

By the way, here's my solution without induction -- it makes it a whole lot simpler, unless I'm being stupid.

Exactly one out of two consecutive integers is even, so $n(n+1)$ is even. But $n(n+1)=n^2+n$, so this proves that our desired expression is even as well. $\square$

I hope this helped.

Saksham Sethi
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  • Or alternatively, $n$ and $n^2$ give the same remainder when divided by 2. Therefore, adding $n^2$ and $n$ then taking remainder when divided by 2 is the same as dividing $2n$ by $2$ and finding remainder, which is $0$. – Prometheus Dec 02 '21 at 02:01
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    Also I am guessing that OP needs it to be done through induction, but is aware of alternative proofs. – Prometheus Dec 02 '21 at 02:01