Why does this happen? Do you think it will happen forever?

Question

Homi · Accepted Answer

Understand the Question

Adding consecutive odd numbers starting from $1$ $1$ always results in a perfect square:
- $1 = 1^2$
- $1 + 3 = 4 = 2^2$
- $1 + 3 + 5 = 9 = 3^2$
- $1 + 3 + 5 + 7 = 16 = 4^2$
Visually, adding the next odd number $(2N + 1)$ to an existing $N \times N$ square of blocks wraps around two adjacent sides plus one corner block to form the next $(N + 1) \times (N + 1)$ square.
Since every positive integer $N$ has a unique next odd number $2N + 1$ , this geometric and algebraic pattern continues forever.

Step 1 · Identify the Pattern of Consecutive Odd Numbers

Observing the sum of consecutive odd numbers:

\begin{aligned} 1 &= 1 = 1 \times 1 = 1^2 \\ 1 + 3 &= 4 = 2 \times 2 = 2^2 \\ 1 + 3 + 5 &= 9 = 3 \times 3 = 3^2 \\ 1 + 3 + 5 + 7 &= 16 = 4 \times 4 = 4^2 \end{aligned}

In general, the sum of the first $N$ odd numbers is: $\text{Sum of } N \text{ odd numbers} = N^2$ Diagram 1

Step 2 · Geometric Reason for the Square Pattern

Diagram 2

Representing squares using unit blocks:

First odd number ( $1$ ): Forms a $1 \times 1$ square ( $1$ block).
Second odd number ( $3$ ): Adding $3$ blocks around the $1 \times 1$ square forms a $2 \times 2$ square ( $1 + 3 = 4$ blocks).
Third odd number ( $5$ ): Adding $5$ blocks around the $2 \times 2$ square forms a $3 \times 3$ square ( $4 + 5 = 9$ blocks).
Fourth odd number ( $7$ ): Adding $7$ blocks around the $3 \times 3$ square forms a $4 \times 4$ square ( $9 + 7 = 16$ blocks).

To expand an $N \times N$ square to an $(N+1) \times (N+1)$ square, we add $N$ blocks along the top, $N$ blocks along the side, and $1$ block at the corner. The total number of blocks added is always an odd number ( $2N + 1$ ).

Step 3 · Verify That the Pattern Continues Forever

For any square of size $N \times N$ , adding the next odd number $(2N + 1)$ gives:

\begin{aligned} N^2 + (2N + 1) &= (N + 1)^2 \end{aligned}

Since this algebraic identity holds for every positive integer $N$ , the pattern never ends and will continue forever.

Answer

(i) It happens because adding the next odd number $(2N + 1)$ creates an L-shaped layer that perfectly completes an $(N + 1) \times (N + 1)$ square. (ii) Yes, this will happen forever because $N^2 + (2N + 1) = (N + 1)^2$ holds for all positive integers $N$ .

Common Mistakes

Starting from an Even Number or Skipping: The pattern only holds when adding consecutive odd numbers starting strictly from $1$ (e.g., $3 + 5 = 8$ , which is not a square number).
Missing the Corner Block: Forgetting that expanding an $N \times N$ grid requires $N$ blocks on each of the two sides plus $1$ corner block, totaling $2N + 1$ blocks.

Question 1

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