algebra odd numbers

Question

A question states, using algebra, prove that when the square of any odd number is divided by four, the remainder is $1$

I managed to go up to $4(n^{2}+n)+1$, from $(2n+1)^{2}$ but I dont know how to prove it. Please help!

Answer 1

You say you managed to go from $(2n+1)^2$ to $4(n^2+n)+1$, but aren't sure how to continue from here.

The final step left (which depending on skill level of writer and reader can be omitted entirely) is to cite the quotient-remainder theorem which paraphrased states that for any integer $a$ and positive integer $b$ there exists a unique pair of integers $\color{red}q,\color{blue}r$ with $0\leq \color{blue}r<b$ such that $a=b\color{red}q+\color{blue}r$. Here $\color{red}q$ is called the "quotient" and $\color{blue}r$ is called the "remainder" for the division

So, we recognize that $(2n+1)^2 = 4\color{red}{(n^2+n)}+\color{blue}1$, so by the quotient-remainder theorem, since we could write $(2n+1)^2$ as above it follows from the uniqueness part of the theorem that $1$ is in fact the remainder.

Answer 2

When $4(n^2+n)+1$ is divided by $4$, the quotient is $n^2+n$ and the remainder is $1$.

Answer 3

(1) Let n be an odd number.

The goal is to show that n² is of the form : 4.k +1 , with k integer

(2) Since n is odd , n = m+1 ( with m even)

(3) n² = (m+1) ² = m² + 2m+1

(4) since m is even , m = 2 k’ ( with k’ integer)

(5) So, substitutiong 2k' for m in (3), we get :

     n² = (2k’)²+ 2(2k’) +1 = 4 (k’)²+4 k’ +1 = 4 [ (k’)²+k’ ] +1 

(6) since k’ is an integer, the number [ (k’)²+k’ ] is also an integer , and therefore, the number n² is of the form 4.k +1 , with k integer

Remark: the reason why the number : (k’)²+k’ is an integer is that (k')² is an integer ( since the set of integers is closed under the operation of multiplication) so that (k’)²+k’ is the sum of two integers; but the set of integers is also closed under the operation of addition