The difference of the squares of two consecutive even integers is always divisible by 4.
Explanation
Let the two consecutive even integers be 2n2n2n and 2n+22n+22n+2, where nnn is an integer.
- The square of the first integer is (2n)2=4n2(2n)^2=4n^2(2n)2=4n2.
- The square of the second integer is (2n+2)2=4n2+8n+4(2n+2)^2=4n^2+8n+4(2n+2)2=4n2+8n+4.
The difference of their squares is:
(2n+2)2−(2n)2=(4n2+8n+4)−4n2=8n+4(2n+2)^2-(2n)^2=(4n^2+8n+4)-4n^2=8n+4(2n+2)2−(2n)2=(4n2+8n+4)−4n2=8n+4
Factoring out 4, the expression becomes:
8n+4=4(2n+1)8n+4=4(2n+1)8n+4=4(2n+1)
Since 4(2n+1)4(2n+1)4(2n+1) is clearly divisible by 4, the difference of the squares of two consecutive even integers is divisible by 4. This holds for all integers nnn. Thus, the difference is guaranteed to be divisible by 4 but not necessarily by any other integers such as 3, 6, or 7.
