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Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.

I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.

What I did: $(n+2)(n+3)(n+7)=6P$ \begin{align*} ((k+2)+1)&((k+1)+3)((k+1)+7) = (k+3)(k+4)(k+8)\\ = &(k+3)[(k+2)+2][(k+7)+1]\\ = &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\ = &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\ = &6P+2k^2+20k+42+k^2+5k+6+2k+6\\ = &6P+3k^2+27k+54\\ = &6p+3(k^2+9k+18) \end{align*}

I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!

Especially Lime
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Killua
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    Welcome to math stack exchange ! Are you supposed to use induction ? If not, there is an easier way to show the claim. – Peter Oct 19 '20 at 09:20
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    $k^2+9k+18$ is always even, so you apparently succeeded. – Peter Oct 19 '20 at 09:22
  • You could simplify $n+7=n+1+6$ and the factor $6$ does not affect divisibility by $6$ – Mark Bennet Oct 19 '20 at 11:13
  • @Mark Yes, and that reduces it to showing $6$ divides $(n+1)(n+2)(n+2),,$ a product $3$ consecutive integers, so we can use an inductive proof that $k$ divides a product of $k$ consecutive integers to get a factor of $2$ and $3$ so also their lcm $= 6$. I was going to mention that but decided to stay closer to the OP's proof. – Bill Dubuque Oct 19 '20 at 11:22
  • Alternatively (for a non-inductive proof) one express it in terms of binomial coefficients $(n+2)(n+3)(n+7)=6\left(7+9n+5\binom{n}{2}+\binom{n}{3}\right)$. Of course not simpler than checking divisibility by $2$ and $3$ directly. – Sil Oct 19 '20 at 11:26
  • @BillDubuque. Indeed that still needs a proof, and it is no easier really than the original, but the three consecutive integers potentially gives a more secure sense of direction. It does, however, involve noticing an extra step. And just with parity (mod $2$) arguments, it is easy to forget that a mod $r$ (eg mod $3$) argument also contains an implicit induction, so there is a danger here. – Mark Bennet Oct 19 '20 at 12:56

5 Answers5

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Hint: You are done, because $k^2+9k+18$ is always even, so that you have divisibility by $2\cdot 3=6$.

Dietrich Burde
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This is fine so far. You have divisibility by $3$ and only need to check that $k^2+9k+18$ is even to get divisibility by $6$. This is true because $k^2$ and $k$ are either both even or both odd.

Especially Lime
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Easier: call it $f_n.\,$ Check $\,f_n-f_{n-2} = 6(n^2\!+\!8n\!+\!14) =: 6 b_n.$ Base cases $f_0,f_1 = 6(7),6(16)$. Assume for complete induction $\,f_n = 6 a_n\,$ for all naturals below $n\ge 2$. Since $\,f_n = \color{#0a0}{f_{n-2}}+6b_n\,$ and induction hypothesis $\,\Rightarrow\, \color{#0a0}{f_{n-2}}=6a_{n-2},\:$ we conclude with the sought: $\ f_n = \color{#c00}6a_{n-2}+\color{#c00}6b_n$.

Remark $ $ It's instructive to explain why $\,f_n-f_{n-2}\,$ has the sought factor of $2$ too. By the Factor Theorem if $\,f(n)\,$ is a polyomial in $\,n\,$ with integers coef's then $\,f(n)-f(n-2)\,$ is divisible by $\,n-(n-2) = 2,\,$ i.e. $\,f(n)-f(n\!-\!2) = 2g(n)\,$ for a polynomial $\,g(n)\,$ with integer coefs. The factor of $3$ you found in $\,f_n - f_{n-1}\,$ also remains in $\,f_n - f_{n-2} = (f_n-f_{n-1}) + (f_{n-1}-f_{n-2})$ since both braced terms have a factor of $3$. Hence $\,f_n-f_{n-2}\,$ has a factor of $\,{\rm lcm}(2,3) = 6$.

Bill Dubuque
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    Note that we don't need any parity argument this way. To be rigorous the parity arguments also require proof by induction (or they invoke lemma(s) that require such proof). – Bill Dubuque Oct 19 '20 at 11:01
  • Yes - parity arguments sometimes seem too obvious to require proof, but of course they do. – Mark Bennet Oct 19 '20 at 12:52
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Hint:

$k^2+9k+18=(k+3)(k+6)$

Now if k is odd then $k+3$ is even.

If k is even then $k+6$is even. So $k^2+9k+18$is always even.

sirous
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Let's look at 3 cases:

1) n is divisible by 3: (n+3) is divisible by 3 and one of (n+2) or (n+3) must be even and, therefore, divisible by 2. Being divisible by both 2 and 3, the expression is divisible by 6.
2) n is not divisible by 3 but (n-1) is: (n+2) is divisible by 3 and one of (n+2) or (n+3) must be even. 
3) n is not divisible by 3 but (n-2) is: (n+7) is divisible by 3 and one of (n+2) or (n+3) must be even.

That covers all possible scenarios for n. Thus, the expression is always divisible by 6.

Raeny von Haus
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