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I must prove, that when dividing the following by 6 it yields no remainder: $$n^2+11n $$ Is this done with mathematical induction method? and what other technique can I use?

Update: some may find this question as a potential duplicate of the correct case when we have $n^3$ instead of $n^2$. I want to clarify that this is not true.

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    As a first attempt, I don't see a reason for not trying induction. – 2017-01-29
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    Not true. For $n=2$, the expression is $26$. – 2017-01-29
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    $n^3+11n\, ???$ – 2017-01-29
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    Do you know that $$6|n^3-n$$ ? if yes It suffice to sum it $+12n$ – 2017-01-29
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    @Niki Are you sure that this is the correct expression? This is not true for $n^2+11n$, as OpenBall says. However, it is true for $n^3+11n$. – 2017-01-29
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    This is not the case of $n^3+11n$. How would I proceed to prove this with the induction method? – 2017-01-29
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    Without further information from the OP, since the claim $6\mid n^2 + 11n$ fails, whereas it **is true** that $6\mid (n^3 + 11n)$, I'm closing as a duplicate above. – 2017-01-29
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    What exactly are you trying to prove? For $n \geq ? ... n^2 + 11n$ is divisible by 6? – 2017-01-29
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    @Niki If it isn't the case of $n^3+11n$, there is no point of trying to prove it with mathematical induction, since the statement is wrong for $n=2$. – 2017-01-29
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    I'm asking to know where the induction method fails (as I'm practicing it), I already know this is not true. Thank you. – 2017-01-29
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    It is true that $n^3 + 11n$ is divisible by 6. However, the case of $n=2$ shows that $6\mid (n^2 + 11n)$ is false (you need only one counter-example to show that a statement does not hold for all $n\in \mathbb N$. – 2017-01-29
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    In general, induction can be used to *prove* many statements about divisibility, and so much more. But it is not sufficient to prove such a statement false. If you're stuck trying to prove by induction, your going to end up going in circles and not arriving at any successful proof. – 2017-01-29
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    If your goal is to find out why induction *fails* to prove this, the question is worded ... really weirdly! Because you only ask how to prove that it *is* true. I would like to reword, given your most recent comment, but it wouldn't make sense unless I completely rewrote the question (which I am averse to doing). – 2017-01-29
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    Niki - I'm sorry if I'm frustrating you; ultimately, I am trying to help you to find your next course of action. – 2017-01-29
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    Big thanks to @amWhy. although, what proving technique could I use in this case? – 2017-01-29
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    Big thanks to those who contributed – 2017-01-29
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    A proof by counter-example is almost, without question, the route to go. The claim that you were to prove, is a claim that it holds for all $n\in \mathbb N$. The almost guaranteed approach to disproving a *universal statement* about all elements in some set, is showing that there exists an element in the given set for which the claim fails. Hence, by giving that counterexample to the claim, you will have formally proven that the claim fails to hold for all elements in the given set. – 2017-01-29
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    Since you were, indeed, asked to prove (or disprove?) the following proposition: $$\forall n \in \mathbb N(6\mid n^2 + 11n)$$ $(1)$ We find that the claim fails when $n= 2$, because $(n = 2) \rightarrow (n^2 + 11n = 4 + 22 = 26)$, yet we know $6$ does **not** divide $26$. $(2)$ Therefore **the claim fails to hold** for all $n\in \mathbb N$. – 2017-01-29
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    The last two comments above, Niki, were going to be my answer to your last question to me. It would not post, as it seems the post was put on hold before I posted it. Please feel free to ask me about anything I have written that you find confusing. Thank you very much for being responsive to the users here trying to help. If I could upvote your post again, I would! – 2017-01-29
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    To know where the induction fails to show that $P_n=[6\mid n^2+11n]$ for every positive integer $n$, one can take $n=1$ as base case, then $P_1$ holds, then try to show that $P_n\implies P_{n+1}$ for every $n\geqslant1$. Then, write $$(n+1)^2+11(n+1)=(n^2+11n)+2n+12$$ hence, if $P_n$ holds, since $6\mid n^2+11n$ and $6\mid12$, one sees that $P_{n+1}$ holds if and only if $6\mid2n$ if and only if $3\mid n$, which is not true for every $n\geqslant1$, hence the proof by induction fails. (For example, $3$ does not divide $1$ hence $P_1\implies P_2$ is wrong.) – 2017-01-30

2 Answers 2

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If you really mean $n^2+11n$ then the remainder can be different to zero as @OpenBall point out, for $n=2$ we have $26$ is not congruent to zero modulo $6$.

If you mean $n^3+11n$ then it is true because

$$n^3+11n=n(n-1)(n+1)+12n$$

so the remainder is always zero.

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The number $(n-1)(n-2)(n-3)$ is always divisible by $6$ because this is a product of three consecutive integers. Now $$(n-1)(n-2)(n-3)=n^3-6n^2+11n-6,$$ whence $$n^3+11n=(n-1)(n-2)(n-3)+6(n^2-1)$$ is divisible by $6$.