Question

Provealgebraically that the square of any odd number is always 1more than a multiple of 8

Answer 1

The answer is 8

Explanation:

If n is "any integer", then 2n+1 is "any odd number."

The square of any odd number is then ...

(2n+1)² = 4n² +4n +1 = 4n(n+1) +1

Since n is any integer, one of n and n+1 will be an even integer, so the product 4n(n+1) will be divisible by 8.

Then the sum 4n(n+1) +1 is one more than a number divisible by 8, hence ...

the square of an odd number is 1 more than a multiple of 8.

Answer 2

Let's assume that we have an odd integer represented by the variable "n". Then, according to the definition of an odd integer, we can write it as follows:

n = 2k + 1

Where "k" is any integer.

Now, we need to find the square of "n". We can write it as follows:

n^2 = (2k + 1)^2

n^2 = 4k^2 + 4k + 1

Now, we can simplify the expression:

n^2 = 4k(k + 1) + 1

We can observe that the expression 4k(k + 1) is always an even number. This is because when we multiply two consecutive integers (k and k + 1), we always get an even result. Therefore, we can write:

4k(k + 1) = 8m

Where "m" is any integer. Substituting this expression into the previous equation, we get:

n^2 = 8m + 1

This shows that the square of any odd number is always 1 more than a multiple of 8, which is what we wanted to prove.