Perfect Squares

A perfect square is an integer that results from squaring another integer. In mathematical terms, a number $ n $ is a perfect square if there exists an integer $ k $ such that: \[n = k^2\]

Here are some common perfect squares:

$$ 1 = 1^2 \\ 4 = 2^2 \\ 9 = 3^2 \\ 16 = 4^2 \\ 25 = 5^2 \\ ... $$

Properties of Perfect Squares

Perfect squares have several interesting properties:

They always have an odd number of divisors
Every perfect square has an odd number of divisors. For any positive integer $ n $, its divisors come in pairs: if $ d $ is a divisor of $ n $, then so is $ \frac{n}{d} $, since $ d \times \frac{n}{d} = n $. However, if $ n $ is a perfect square, meaning $ n = a^2 $, one of its divisors—$ a $, the square root of $ n $—pairs with itself rather than with a distinct number. Since this divisor is counted only once, the total number of divisors is odd.
For example, take $ n = 16 $, which is $ 4^2 $. Its divisors are: $ 1, 2, 4, 8, 16 $. Notice that $ 1 $ pairs with $ 16 $ ($ 1 \times 16 = 16 \times 1 = 16 $), and $ 2 $ pairs with $ 8 $ ($ 2 \times 8 = 8 \times 2 = 16 $), following the commutative property. But $ 4 $, the square root of $ 16 $, does not have a distinct pair—it pairs with itself ($ 4 \times 4 = 16 $) and is counted only once. $$ 1 \times 16 = 16 \\ 2 \times 8 = 16 \\ \color{red}{4 \times 4 = 16} \\ 8 \times 2 = 16 \\ 16 \times 1 = 16 $$ This pattern holds for all perfect squares, ensuring they always have an odd number of divisors.
Their square roots are always integers
A number is NOT a perfect square if its square root is not a whole number.

For example, $ 10 $ is not a perfect square because its square root is $ \sqrt{10} \approx 3.162 $, which is not an integer.
The last digit follows a pattern
The last digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. If a number ends in 2, 3, 7, or 8, it is NOT a perfect square (e.g., 23, 57, 98, etc.).
The sum of the first $ n $ odd numbers is always a perfect square
The sum of the first $ n $ odd numbers always results in a perfect square: $$ 1+ 2 + 3+ ... +(2n-1) = n^2 $$ Examples: $$ 1 = 1^2 \\ 1 + 3 = 4 = 2^2 \\ 1 + 3 + 5 = 9 = 3^2 \\ 1 + 3 + 5 + 7 = 16 = 4^2 \\ 1 + 3 + 5 + 7 + 9 = 25 = 5^2 \\ ... $$ It's worth noting that this sum includes both prime and composite odd numbers. In other words, it consists of the first $ n $ odd numbers, starting from 1.

Proof. Consider the sum of the first $ n $ odd numbers: $$ 1 + 3 + 5 + \dots + (2n-1) = n^2. $$ We can prove this using mathematical induction. The base case for $ n = 1 $ is: $$ 1 = 1^2 $$ So the property holds for $ n = 1 $. Now, assume the formula holds for some $ n $, meaning: $$ 1 + 3 + 5 + \dots + (2n-1) = n^2. $$ We must now prove that it also holds for $ n+1 $, meaning: $$ 1 + 3 + 5 + \dots + (2n-1) + [2(n+1)-1] = (n+1)^2 $$ Using the induction hypothesis: $$ n^2 + [2(n+1)-1] = n^2 + (2n+2-1) = n^2 + 2n + 1 $$ Since $ n^2 + 2n + 1 = (n+1)^2 $, the property holds for $ n+1 $ as well. $$ n^2 + [2(n+1)-1] = n^2 + (2n+2-1) = (n+1)^2 $$
Perfect squares are always positive
This follows from the fact that any real number multiplied by itself is always non-negative.
Remainders when dividing a perfect square
When divided by 3 or 4, a perfect square always leaves a remainder of either 0 or 1.
Difference between consecutive perfect squares
The difference between two consecutive perfect squares follows a simple rule: $$ (n+1)^2 - n^2 = 2n + 1 $$ This means the difference is always an increasing odd number.
For example: $$ 4^2 - 3^2 = 16 - 9 = 7 $$ $$ 5^2 - 4^2 = 25 - 16 = 9 $$ $$ 6^2 - 5^2 = 36 - 25 = 11 $$ $$ 7^2 - 6^2 = 49 - 36 = 13 $$ In each case, the result is $ 2n+1 $, where $ n $ is the smaller of the two squares. For instance, if $ n=3 $ and $ n+1= 4 $, then $ 2n+1 = 2 \cdot 3 + 1 = 7 $, which matches the difference between their squares: $ 4^2-3^2=16-9=7 $.
Perfect squares and powers of 2
A power of 2 can only be a perfect square if its exponent is even, meaning it must be of the form $2^{2k}$.
For example: $$ 4 = 2^2 \\ 16 = 2^4 \\ 64 = 2^6 $$ However, $ 2^5 = 32 $ is NOT a perfect square.

Proof. Consider a number of the form $2^n$, where $n$ is a non-negative integer. For $2^n$ to be a perfect square, all exponents in its prime factorization must be even. Thus, $ n $ must be even, meaning $ n=2k $ for some integer $ k $: $$ 2^n = 2^{2k} $$ Using exponent rules, we can rewrite this as: $$ 2^n = 2^{2k} = (2^k)^2 $$ This confirms that a power of 2 can only be a perfect square if its exponent is even.

And so on.