# Prove that n^4 - m^4 - n^2 - m^2 + 2 is always even

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### 1 Answer

To prove that `n^4 - m^4-n^2 - m^2 + 2` is always even, group the terms containing the same variables together:

`n^4 - m^4 - n^2 - m^2 + 2 = n^4 - n^2 - m^4 - m^2 + 2`

Consider each group separately:

`n^4 - n^2 = n^2(n^2 - 1)` This is a product of a number ` ` `n^2` and a number

`n^2 - 1` , which is 1 less than `n^2` . Since they are two consecutive numbers, one of them must be even and one odd. The product of an odd number and an even number is always an even number (since one factor is divisible by 2, the product is also divisible by 2.)

`-m^4 - m^2 = -m^2(m^2 + 1)` Similarly, this is a product of two consecutive numbers, so it must be even, as discussed above.

2 is even because it is divisible by 2.

So the original expression consists of a sum and difference of three even numbers, which means it is also an even number. (Sum or difference of even numbers is always even as well.)