For an integer x, x squared minus x will always produce an even value

Step-by-step explanation:

Not exactly sure what your question is - I am assuming that it is something like:

Show/prove that for any integer x, x^2 - x is even.

Suppose that x is an even integer. The product of an even integer and any other integer is always even (x = 2n, so x * y = 2 n * y which is even. Therefore x^2 is even. An even minus an even is even. (The definition of an even number is that it is divisible by 2 or has a factor of 2. So the difference of even numbers could be written as 2 * (the difference of the two numbers divided by 2); therefore the difference is even)

Suppose that x is an odd integer. The product of 2 odd numbers is odd - each odd number can be written as the sum of an even number and 1; multiplying the even parts with each other and 1 will produce even; multiplying the 1's will produce 1, so the product can be written as the sum of an even number and 1 - which is an odd number. The difference between two odd numbers is even - the difference between the even parts is even (argument above), the difference between 1 and 1 is zero, so the result of the difference is even.

x^2 is therefore even if x is even and odd if x is odd; The difference x^2 - x is even by the arguments above.