Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

Question

Homi · Accepted Answer

We will count the whole numbers that are not perfect squares, found between two perfect squares that are next to each other.

Step 1 — Understanding Consecutive Perfect Squares

A perfect square is a number we get by multiplying an integer by itself. For example, $1 \times 1 = 1$ , $2 \times 2 = 4$ , $3 \times 3 = 9$ . So, 1, 4, 9 are perfect squares. Consecutive perfect squares are perfect squares that follow each other. Let us consider some pairs of consecutive perfect squares. These are $n^2$ and $(n+1)^2$ for any positive integer $n$ .

Step 2 — Counting for Specific Examples

Let us find the numbers between $1^2$ and $2^2$ . $1^2 = 1$ . $2^2 = 4$ . The numbers between 1 and 4 are 2, 3. We count these numbers. $4 - 1 - 1$ $= 3 - 1$

$\boxed{\text{2 numbers}}$

Let us find the numbers between $2^2$ and $3^2$ . $2^2 = 4$ . $3^2 = 9$ . The numbers between 4 and 9 are 5, 6, 7, 8. We count these numbers. $9 - 4 - 1$ $= 5 - 1$

$\boxed{\text{4 numbers}}$

Let us find the numbers between $3^2$ and $4^2$ . $3^2 = 9$ . $4^2 = 16$ . The numbers between 9 and 16 are 10, 11, 12, 13, 14, 15. We count these numbers. $16 - 9 - 1$ $= 7 - 1$

$\boxed{\text{6 numbers}}$

Diagram 1

Step 3 — Noticing a Pattern

Between $1^2$ and $2^2$ , we found 2 numbers. The smaller base number in this pair is 1. Notice that $2 \times 1 = 2$ .

Between $2^2$ and $3^2$ , we found 4 numbers. The smaller base number in this pair is 2. Notice that $2 \times 2 = 4$ .

Between $3^2$ and $4^2$ , we found 6 numbers. The smaller base number in this pair is 3. Notice that $2 \times 3 = 6$ .

We observe a pattern here. The number of integers between $n^2$ and $(n+1)^2$ is $2n$ .

Step 4 — Generalizing the Pattern Algebraically

Let $n$ be any positive integer. The two consecutive perfect squares are $n^2$ and $(n+1)^2$ . We want to find the count of integers greater than $n^2$ and less than $(n+1)^2$ . The number of integers between two integers $A$ and $B$ (where $A < B$ ) is $B - A - 1$ . Here, $A = n^2$ and $B = (n+1)^2$ . So, the number of integers is $(n+1)^2 - n^2 - 1$ . Let us expand $(n+1)^2$ . We know that $(a+b)^2 = a^2 + 2ab + b^2$ . So, $(n+1)^2 = n^2 + 2n(1) + 1^2$ . $= n^2 + 2n + 1$ Now, substitute this back into our count formula. The number of integers is $(n^2 + 2n + 1) - n^2 - 1$ . $= n^2 + 2n + 1 - n^2 - 1$ $= 2n$ This confirms our observed pattern. The number of integers between $n^2$ and $(n+1)^2$ is $2n$ .

Answer

(i) The number of integers that lie between two consecutive perfect squares, $n^2$ and $(n+1)^2$ , is $2n$ . (ii) The pattern is that the count of numbers is always twice the smaller base number (the $n$ in $n^2$ ).

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