Answered

Prove that the difference between the square of any odd integer and the integer itself is always an even integer.

Sagot :

Sophiecamilleviz

Answer:

So x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Since k is an integer, 2k2 + 2k is also an integer, so we can write x2 = 2l + 1, where l = 2k2 + 2k is an integer. Therefore, x2 is odd. Since this logic works for any odd number x, we have shown that the square of any odd number is odd.