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: 

Square. Use the perfect square trinomial pattern. So: 

Combine like terms: 

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 <em>always</em> even. 
Our second term 2n also has a 2 in front. So, whatever n is, 2n is also <em>always</em> even. 
So far, 2n² and 2n are both even. Therefore, the sum of 2n² and 2n is <em>also </em>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.