X2 1 is always divisible by 8 if x is a an

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Question: Show that n^2 -1 is divisible by 8, if n is an odd positive integer. Solution: If n is odd and positive, we define n=2k+1 where k is a non-negative integer. from which we substitute, expand and factor: n²-1 =(2k+1)²-1 =4k²+4k+1-1 =4k²+4k =4k(k+1) Since k is a non-negative integer, we have two possible cases: 1. k is odd, in which case (k+1) is even, and equal to 2q (q=non-negative integer) => n²-1=4k(2q)=8kq (where both k and q are non-negative integers) therefore 8 divides n²-1 2. k is even, then k=2q (q=non-negative integer) => n²-1=4(2q)(k+1)=8q(k+1) (where both k and q are non-negative integers) therefore 8 divides n²-1 Since in both cases, 8 divides n²-1, therefore it is proved that 8 divides n²-1 in for all positive values of n.

Alternatively, you could prove this by induction on $x$.

Base case: Consider the case where $x = 1$. Clearly, $(1)^2 - 1 = 0$ is divisible by 8. Hence, the base case is true

Inductive Hypothesis: Assume that the statement is true for some $x = k$ where $k$ is an odd integer greater than 1.

Inductive Step: Since $k$ is odd, we must show that the statement holds for the next odd integer, $k + 2$. Plugging $x = k + 2$ into $x^2 - 1$,

$$ (k + 2)^2 - 1 = k^2 + 4k + 3 = (k^2 - 1) + 4(k + 1) $$

$k^2 - 1$ was divisible by 8 by the inductive hypothesis. Thus, the statement holds for $k + 2$ if and only if $4(k + 1)$ is divisible by 8. Observe that since $k$ is odd, is can be written as $k = 2q + 1$ for some integer $q$. This gives us

$$ 4(k + 1) = 4((2q + 1) + 1) = 4(2q + 2) = 8q + 8 = 8(q + 1) $$

Thus, we have shown that 8 divides $4(k + 1)$ which means that it also divides $(k + 2)^2 - 1$, completing the proof.