Question 26

If n is an odd positive integer, show that $(n^2-1)$ is divisible by 8.

Solution:

To show that $$n^2-1$$ is divisible by 8 when $$n$$ is an odd positive integer, we can follow these steps:

Step 1: Express $$n$$ as an odd integer
An odd integer can be expressed in the form:
$$n=2k+1$$
where $$k$$ is an integer.

Step 2: Calculate $$n^2$$
Now, we calculate $$n^2$$:
$${\mathrm{n2}}^{}=(2k+1{)2}^{}=4{\mathrm{k2}}^{}+4k+1$$

Step 3: Calculate $$n^2-1$$
Next, we find $${\mathrm{n2}}^{}-1$$:
$${\mathrm{n2}}^{}-1=(4{\mathrm{k2}}^{}+4k+1)-1=4{\mathrm{k2}}^{}+4k$$

Step 4: Factor out the common terms
We can factor out $$4$$ from the expression:
$${\mathrm{n2}}^{}-1=4({\mathrm{k2}}^{}+k)$$

Step 5: Show that $$k^2+k$$ is even
Notice that $$k(k+1)$$ is the product of two consecutive integers, which is always even. Therefore, we can express it as:
$${\mathrm{k2}}^{}+k=2m\text{for some integer}m$$

Step 6: Substitute back into the equation
Substituting back, we get:
$${\mathrm{n2}}^{}-1=4\left(2m\right)=8m$$

Conclusion
Thus, $${\mathrm{n2}}^{}-1$$ is divisible by 8.