Direct Proof on Divisibilty

Question

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.

Answer 1

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?

Answer 2

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?

Answer 3

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.

Answer 4

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 $12$.

Answer 5

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

Answer 6

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.

Answer 7

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 $\ 2\mid f(1) = 6$.