Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers.

Let any integer be represented by x. 
Then, the consecutive integer (the number that follows) should be x+1. 

The statement wants us to prove; 

(x+1)²-x²=x+(x+1) <-- solve left hand side 

x²+x+x+1²-x²
2x+1 (solution for left hand side) 

Now solve for right hand side. 

x+(x+1)= 2x+1 

As noticed, the LHS=RHS (left hand side= right hand side), therefore, the difference of squared consecutive numbers subtracted is equal to the sum of the two integers. 

Hope I helped :)