Prove that the sum of two consecutive odd numbers is divisible by 4.

Question

Homi · Accepted Answer

Understand the Question

Any odd number can be algebraically expressed in the form $2n + 1$ , where $n$ is an integer.
Consecutive odd numbers differ by $2$ , so the next odd number is $(2n + 1) + 2 = 2n + 3$ .
Adding these two expressions and factoring out $4$ shows that their sum is always a multiple of $4$ , and therefore divisible by $4$ .

Step 1 · Represent Consecutive Odd Numbers

Let the first odd number be $2n + 1$ , where $n$ is an integer. Diagram 1

The next consecutive odd number is $(2n + 1) + 2 = 2n + 3$

Step 2 · Find the Sum and Factorise

Add the two consecutive odd numbers

\begin{aligned} (2n + 1) + (2n + 3) &= 2n + 1 + 2n + 3 \\ &= 4n + 4 \\ &= 4(n + 1) \end{aligned}

Since $n$ is an integer, $(n + 1)$ is also an integer. Hence, $4(n + 1)$ is a multiple of $4$ and is divisible by $4$ .

Answer

Hence proved, the sum of two consecutive odd numbers is divisible by $4$ .

Common Mistakes

Consecutive Integer vs. Consecutive Odd Integer: Adding $1$ instead of $2$ to get the next odd number (i.e. writing $(2n + 1) + 1 = 2n + 2$ , which is an even number).
Proof by Example: Showing that the statement works only for specific numbers (e.g. $3 + 5 = 8$ ) instead of providing a general algebraic proof valid for all integers $n$ .

Question 1