Difference between the squares of two consecutive numbers is not divisible by 2

Hmm, let's prove that

consecutive integers are even then odd or odd then even

all even integers can be represented by 2n where n is an integer
all odd integers can be represetned by 2n+1 or 2n-1

all numbers divisible by 2 are even

alright

we wil start with even then odd
2n and 2n+1 are our even and odd numbes, the
their square are 4n² and 4n²+4n+1 respectively
their difference is 4n+1 or 2n+2n+1, an even+an odd, resulting in an odd difference

lets try odd then even
odd is 2n-1 and the even is 2n
their square are 4n²-4n+1 and 4n² respectively
their difference is 4n-1 which is 2n+2n-1, another even+odd combo, resulting in an odd difference

proven, the difference between the squares of 2 consecutive numbers is not divisible by 2

Ican’t sorry i would if i could