# Homework Help: Product of two consecutive integers

1. Jan 4, 2010

### nobahar

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$n^2+n$$

3. 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?

2. Jan 4, 2010

### rochfor1

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

3. Jan 4, 2010

### 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.

4. Jan 5, 2010

### nobahar

Many thanks rochfor1 and HallsofIvy!

5. Jan 5, 2010

### 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.

6. Jan 5, 2010

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7. Jan 5, 2010

### Dick

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

8. Jan 6, 2010

### ideasrule

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

9. Jan 6, 2010

### Staff: Mentor

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.

10. Jan 7, 2010

### icystrike

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