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:
[tex](n)^2+(n+1)^2[/tex]
Square. Use the perfect square trinomial pattern. So:
[tex]=n^2+(n^2+2n+1)[/tex]
Combine like terms:
[tex]=2n^2+2n+1[/tex]
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.
