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Using Induction proof makes sense to me and know how to do, but I am having a problem in using a direct proof for practice problem that was given to us.

The problem is:

For all natural numbers $n$, $2n^3 + 6n^2 + 4n$ is divisible by 4.

We are to use direct proof as a way proving it. I have no clue where to start.

scrappedcola
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user3321427
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7 Answers7

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hints

Note that $2n^3+6n^2+4n = 2n(n^2+3n+2) = 2n(n+1)(n+2)$.

Can you argue this is always divisible by 4?

gt6989b
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  • I do understand that it is divisible by 4. I'm confused on what direct proof techniques can be used instead of proving by induction – user3321427 May 20 '15 at 18:38
  • @user3321427 direct argument would be like "this has a factor of 2, hence it must be even" -- another similar one to argue it is divisible by 4 – gt6989b May 20 '15 at 18:39
  • @user3321427 You already have a $2$ here. Can you show that $n(n+1)(n+2)$ is even as well? Note that this amounts to showing that at least one of $n,n+1,n+2$ is even. – Wojowu May 20 '15 at 18:40
  • Mathematically I understand the problem I am just not aware on how to write a proof properly in the format of "Let n be a natural number. Then assume.......thus.....therefore....in conclusion." Something along those lines. – user3321427 May 20 '15 at 18:50
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We shall show that $4|2n^3+6n^2+4n$ for this it suffices to show that it's divisible by 2 twice. We have that $$2n^3+6n^2+4n=2n(n^2+3n+2)$$ Which shows it is divisible by two atleast once, what remains to show is that it is again divisible by two, we know $n$ cannot always be divisible by two as if it is odd it is not. So the only candidate is $(n^2+3n+2)$. Assume $n$ is even, then we have that $n^2$ and $3n$ is even, and the sum of even numbers is even, so $n^2+3n+2$ is even and we have that $2|n^2+3n+2$.

Now assume $n$ is odd, at this point we have that $n^2$ is odd and $3n$ is odd, as square of odd numbers and odd numbers times odd numbers are both odd. The sum of two odd numbers is even so the sum of these two is even. Then we have the last 2 in $n^2+3n+2$, which as the first two terms are odd, their sum is even, we have a sum of two even numbers, so it must be even too. Therefore $n^2+3n+2$ is even and $2|n^2+3n+2$

How is this one then?

Zelos Malum
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Here is my take, We have $$2n^3+6n^2+4n=2n(n^2+3n+2)$$ So we know it's divisible by 2, as $4=2\cdot 2$ all we need to do now is to check that $(n^2+3n+2)$ is divisible by two also. This in turn is simple as we have two cases, either $n$ is even, or it's odd. If $n$ is even, then $n^2$, $3n$ are divisible by 2 so $2|(n^2+3n+2)$, if $n$ is odd we have that $n^2$ is odd, we also have that $3n$ is odd then. this gives us in the case of $n^2+3n$ that we have odd plus odd number, which is even, then we have $n^2+3n+2$ which is even number plus even number, this is even, so it is divisible by 2 also. So no matter what it is divisible by two. As the entire expression is divisible by 2 twice, it means it is divisible by 4.

Zelos Malum
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  • My main question is more on how I write that on a proper proof form. Such as Let n be a natural number. Then assume.......thus.....therefore....in conclusion. Mathematically I understand the problem I am just not aware on how to write a proof properly in that format. – user3321427 May 20 '15 at 18:49
  • I am fairly certain that constitutes a proof, you can rewrite it to be more "formal" but that is a fairly easy way to write it with basal arithmetic – Zelos Malum May 20 '15 at 18:50
  • In arithmetic yes it does seem to be easier, but trying to learn formal proof techniques and how problems can be written in a "formal" proof just like how Daniel J Velleman shows examples in his book "How to Prove It" – user3321427 May 20 '15 at 18:52
  • Fix up the grammar, some if and thens and then it is a perfectly legitimate proof. The line of reasoning is more important and there it's fairly spot on. If you want I can give you another where it is more "formal" but the line of reasoning I gave – Zelos Malum May 20 '15 at 18:54
  • If it's fine with you I would love a formal one. I just want to be able to learn how I can do that with different problems such as this one. And yes I do agree that it is a legitimate proof but when your in a class based on proof, it's all about formal ways lol. – user3321427 May 20 '15 at 18:57
  • @user3321427 +1 The parity idea generalizes - see the link to the Parity Root Test in my answer. – Bill Dubuque May 20 '15 at 19:05
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We have$$2n^3+6n^2+4n=2n(n^2+3n+2)=2n(n+1)(n+2)$$ Here, note that either $n$ or $n+1$ is even.

mathlove
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Note that $$\sum_{k=1}^n 6(k+1)(k+2) = 2n^3+6n^2+4n$$ Since each term on the left in the summation is divisible by $12$ (since $2$ divides $(k+1)(k+2)$), $2n^3+6n^2+4n$ is also divisible by $12$.

Adhvaitha
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It is enough to check $n^3+3n^2=n^2(n+1)$ is even, which is trivial.

Bernard
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For all $\,n\!:\,$ $\,4\mid 2f(n)$ $\!\iff\! 2\mid f(n)$ $\!\iff\! f(0)\equiv 0\equiv f(1)\pmod{\!2}$ $\!\iff 2\mid f(0),f(1)$

for any polynomial $\,f(x)\,$ with integer coefficients (see also the Parity Root Test).

This applies to $ $ OP $\ f(x) = n^3+3n^2+2n\,$ since $\ 2\mid f(0) = 0,\ $ and $\ 2\mid f(1) = 6$.

Community
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Bill Dubuque
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