Answer:
True
Step-by-step explanation:
To prove by induction, first show that the statement is true at an initial value (in this case, x = 1).
8 | (1² − 1)
8 | 0
Next, assume the statement is true at x = k.
8 | (k² − 1)
Now show that it is true at the next value of x (x = k + 2).
(k + 2)² − 1
= k² + 4k + 4 − 1
= k² − 1 + 4(k + 1)
k + 1 is an even number, so 4(k + 1) is a multiple of 8. And since we're assuming that k² − 1 is a multiple of 8, that means the sum is also a multiple of 8.
8 | ((k + 2)² − 1)
So the statement is true.
Logically, here's another way of proving it:
x² − 1 = (x − 1) (x + 1)
Since x is an odd integer, both x − 1 and x + 1 are even numbers. Since the difference between them is 2, one of them is a multiple of 4. An even number times a multiple of 4 is a multiple of 8.
