Product of two consecutive integers

[nobahar]

Homework Statement

Prove that $$n^2+n$$ is even. Where n is a positive integer.

Homework Equations

$$n^2+n$$

The Attempt at a Solution

$$n^2+n = n(n+1)$$

One of which must be even, and therefore the product of 2 and an integer k.

$$n = 2k, \left \left 2*(k(n+1))$$

or

$$n+1 = 2k, \left \left 2*(n*k)$$

Is there a better way of doing this? I read this is not an inductive proof; what would this entail?
Many thanks in advance.

 

[rochfor1]

This looks absolutely fine. No need for induction at all.

 

[HallsofIvy]

As rochfor1 said, that is a perfectly good proof and simpler than "proof by induction".

But since you ask, here goes:

If n= 1, then $n^2+ n= 1^2+ 1= 2$ which is even.

Now, suppose that $k^2+ k$ is even and look at $(k+1)^2+ (k+1)$
$(k+1)^2+ (k+1)= k^2+ 2k+ 1+ k+ 1$$= (k^2+ 2k)+ 2k+ 2$. By the induction hypothesis, $k^2+ k$ is even and so $k^2+ k= 2m$ for some integer m (that is the definition of "even") so $(k+1)^2+ (k+1)= (k^2+ 2k)+ 2k+ 2$$= 2m+ 2k+ 2= 2(m+k+1)$. Since that is "2 times an integer", it is even.

Having proved that the statement is true for 1 and that "if it is true for k, it is true for k+1", by induction, it is true for all positive integers.

Yet a third way: If n is a positive integer, it is either even or odd.

case 1: n is even. Then n= 2m for some integer m. $n^2= 4m^2$ so $n^2+ n= 4m^2+ 2m= 2(m^2+ m)$. Since that is 2 times an integer, it is even.

case 2: n is odd. Then n= 2m+ 1 for some integer m. $n^2= (2m+1)^2= 4m^2+ 4m+ 1$ so $n^2+ n= 4m^2+ 4m+ 1+ 2m+ 1$$= 4m^2+ 6m+ 2= 2(2m^2+ 3m+ 1)$. Again that is 2 times an integer and so is even.

 

[nobahar]

Many thanks rochfor1 and HallsofIvy!

 

[icystrike]

Simple reasoning is needed.
For two consecutive integer , one will be even and another will be odd.
Thus;
$$n\equiv0(mod 2)$$
and
$$n+1\equiv1(mod 2)$$

We may now conclude that $$n(n+1)\equiv0(mod 2)$$ and means that n²+n is even.

 

[icystrike]

 

[Dick]

If n is odd, n^2 is odd. If n is even, n^2 is even. odd+odd=even and even+even=even.

 

[ideasrule]

OP: Why do you want a proof by induction? The proof you gave is simple and beautiful.

 

[Mark44]

Simple reasoning is needed.
For two consecutive integer , one will be even and another will be odd.

What you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.

Thus;
$$n\equiv0(mod 2)$$
and
$$n+1\equiv1(mod 2)$$

We may now conclude that $$n(n+1)\equiv0(mod 2)$$ and means that n²+n is even.

 

[icystrike]

What you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.

yes that is true. for the other case , it can be proved analogously .