Question

Prove algebraically that the sum of the squares of two consecutive intgers is always an odd number

Answer 1

Answer:

See below.

Step-by-step explanation:

Let's let our first integer be n.

Then, our second, consecutive integer must be (n+1).

We want to prove that the sum of the square of two consecutive integers is always odd.

So, let's square our two expressions and add them up:

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

Square. Use the perfect square trinomial pattern. So:

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

Combine like terms:

$=2n^2+2n+1$

Notice that the first term 2n² has a 2 in front. Anything multiplied by 2 is even. So, regardless of what integer n is, 2n² is always even.

Our second term 2n also has a 2 in front. So, whatever n is, 2n is also always even.

So far, 2n² and 2n are both even. Therefore, the sum of 2n² and 2n is also even.

Lastly, we have the constant term 1. It doesn't matter what n is, we know that (2n² + 2n) is definitely and is always even. So, if we add 1 to this, we will get an odd number.

Q.E.D.