Question 3. $n^2-1$ is divisible by $8$, if $n$ is

Solution:

To determine the values of $$n$$ for which $n^2-1$ is divisible by $$8$$, we can analyze the expression step by step.

Step 1: Understand the expression
We start with the expression $n^2-1$. We can rewrite it as:
$$n2-1=(n-1)(n+1)$$
This shows that $n^2-1$ is the product of two consecutive integers, $$n-1$$ and $$n+1$$.

Step 2: Analyze the product of two consecutive integers
The product of two consecutive integers is always even. In fact, one of them is guaranteed to be divisible by $$2$$ and at least one of them will be divisible by $$4$$ if the integers are consecutive. Therefore, the product $$(n-1)(n+1)$$ is divisible by $$8$$ if one of these integers is divisible by $$4$$ and the other is even.

Step 3: Consider the cases for $$n$$
1. Case 1: $$n$$ is odd
– If $$n$$ is odd, we can express $$n$$ as $$n=2k+1$$ for some integer $$k$$.
– Then, $$n-1=2k$$ (even) and $$n+1=2k+2$$ (even).
– The product $$(n-1)(n+1)=2k(2k+2)=4k(k+1)$$.
– Since $$k(k+1)$$ is the product of two consecutive integers, it is always even. Thus, $$4k(k+1)$$ is divisible by $$8$$.

2. Case 2: $$n$$ is even
– If $$n$$ is even, we can express $$n$$ as $$n=2k$$ for some integer $$k$$.
– Then, $$n-1=2k-1$$ (odd) and $$n+1=2k+1$$ (odd).
– The product $$(n-1)(n+1)=(2k-1)(2k+1)$$ is the product of two odd numbers, which is odd and hence not divisible by $$8$$.

Conclusion
From the analysis, we conclude that $n^2-1$ is divisible by $$8$$ if and only if $$n$$ is an odd number.

Final Answer
Thus, $$n$$ must be an odd number.