# Perfect Squares, Cubes, and Powers

A **perfect square** is an integer that can be expressed as the product of two equal integers. For example, $100$ is a perfect square because it is equal to $10\times 10$. If $N$ is an integer, then $N^2$ is a perfect square. Because of this definition, perfect squares are always non-negative.

Similarly, a **perfect cube** is an integer that can be expressed as the product of three equal integers. For example, $27$ is a perfect cube because it is equal to $3\times 3 \times 3.$ Determining if a number is a perfect square, cube, or higher power can be determined from the prime factorization of the number.

## Perfect Squares

Make a list of $10$ perfect squares from the smallest.

We have

$\begin{array}{c}&0^2=0, &(\pm1)^2=1, &(\pm2)^2=4, &(\pm3)^2=9, &(\pm4)^2=16, \\ (\pm5)^2=25, &(\pm6)^2=36, &(\pm7)^2=49, &(\pm8)^2=64, &(\pm9)^2=81 .\end{array}$

Thus, the answer is

$0, 1, 4, 9, 16, 25, 36, 49, 64, 81. \ _\square$

Find the differences between two adjacent perfect squares, 9 of them from the smallest. For example, start from

$1^2-0^2=1-0=1.$

We have

$\begin{array}{c}&1^2-0^2=1-0=1, &&&2^2-1^2=4-1=3, \\3^2-2^2=9-4=5, &&&4^2-3^2=16-9=7,\\ 5^2-4^2=25-16=9, &&&6^2-5^2=36-25=11, \\ 7^2-6^2=49-36=13, &&&8^2-7^2=64-49=15, \\ 9^2-8^2=81-64=17 .\end{array}$

Thus, the answers are

$1, 3, 5, 7, 9, 11, 13, 15, 17. \ _\square$

Which of the following is

nota perfect square?$\begin{array}{c}&(a)~ 125 &&&(b)~ 144 &&&(c)~ 441 &&&(d)~ 225 \end{array}$

Since $144=12 \times 12, 441=21 \times 21,$ and $225=15 \times 15,$ none of $(b), (c)$ and $(d)$ is the answer. Now, $125=5 \times 5 \times 5=25 \times 5,$ which is not a perfect square but a perfect cube. So, the answer is $(a).$ $_\square$

In the following equation, $a, b$ and $c$ are all distinct positive integers (This is part of the Pythagorean Theorem):

$a^2+b^2=c^2.$

What is the smallest possible value of $c?$

Observe that

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

Then the answer is $c=5.$ $_\square$

What is the positive number $a$ in the following equation:

$5^2+12^2=a^2?$

Observe that

$5^2+12^2=25+144=169=13^2.$

Then the answer is $a=13.$ $_\square$

What are the perfect squares between $301$ and $399?$

Observe that

$\begin{array}{c}&(\pm17)^2=289, &(\pm18)^2=324, &(\pm19)^2=361, &(\pm20)^2=400.\end{array}$

Then the answers are $324$ and $361.$ $_\square$

Some properties regarding perfect squares are as follows (their proofs are omitted here):

- Perfect squares
**cannot**have a units digit of 2, 3, or 7. (You can check them out for yourself.) - The square of an even number is even and the square of an odd number is odd.
- All odd squares are of the form $4n + 1$, hence all odd numbers of the form $4n+3$, where $n$ is a positive integer, are not perfect squares. For instance, 361 can be written as $4 \times 90 + 1,$ and we know $361 = 19^2.$ However, 843 is not a perfect square since $29^{2}=841$ and $30^{2} = 900;$ it can be expressed as $4 \times 210 + 3.$
- All even numbers of the form $4n + 2$, where $n$ is a positive integer, are not perfect squares
- All even squares are divisible by 4. (You can take any even square and check this.)
- The difference of 2 odd squares is a multiple of 8. For example, $15^{2} - 11^{2} =104,$ which is $8 \times 13.$
- The sum of the first $n$ odd numbers is in fact $n^2.$ For example, $1+3+5+7+9+11= 36.$ Here, there are 6 odd numbers, so we can find the sum as just $n^2=6^2.$ Similarly, $1+3+5+\cdots+57 = 784,$ as $n = 28$ here.
- The sum of the first $n$ perfect squares $1^{2} + 2^{2} + 3^{2} +\cdots+n^{2}$ is given by $\frac{n(n+1)(2n+1)}{6}.$
- If $p$ divides $a^{2},$ then $p$ divides $a$ as well (Euclid's theorem). From this, we can say that a number is a perfect square if its prime factorization contains all primes raised to some even power.
- Given two positive integers $K$ and $m,$ if $K^2-m$ is the square of an integer $n,$ then $K-n$ divides $m.$

Ending digits for squared numbers (we consider decimal system):

- If a number has units digit 1 or 9, its square will have units digit 1.
- If a number has units digit 2 or 8, its square will have units digit 4.
- If a number has units digit 3 or 7, its square will have units digit 9.
- If a number has units digit 4 or 6, its square will have units digit 6.
- If a number has units digit 5, its square will have units digit 5.
- If a number has units digit 0, its square will have units digit 0.

The proof for this is left for the reader.

## Perfect Cubes

When you cube something, you multiply it by itself three times. For example, $5^3 = 5 \times 5 \times 5 = 125$. Cubing a positive number will result in a positive number while cubing a negative number will result in a negative number.

Make a list of 10 perfect non-negative cubes starting from the smallest.

We have

$\begin{array}{c}0^3 = 0, & 1^3 = 1, & 2^3 = 8, & 3^3 = 27, & 4^3 = 64, \\ 5^3 = 125, & 6^3 = 216, & 7^3 = 343, & 8^3 = 512, & 9^3 = 729. \end{array}$

Thus the answer is

$0, 1, 8, 27, 64, 125, 216, 343, 512, 729. \ _\square$

Some simple properties of perfect cubes are given below. The proofs for them are omitted here.

- Every perfect cube has digital root 1, 8, or 9. By "digital root" we mean the sum of digits that is done until we get a single digit. For instance, the digital root of 1234 can be obtained as follows: We have $1+2+3+4=10.$ Since we got a 2-digit number, we add the digits again to get $1+0= 1,$ which is the digital root of 1234. The number 54 has digital root $5+4 = 9.$ Note that if a number has digital root 1, 8, or 9, it does not necessarily mean that it is a perfect cube (as is the case with 54, which has digital root 9 but is not a perfect cube). This can be proven with modular arithmetic.
- Perfect cubes can have any number from 0 to 9 as their units digit.
- The sum of the first $n$ perfect cubes $1^{3} + 2^{3} + 3^{3} + 4^{3} +\cdots+ n^{3}$ is $\left(\frac{n(n+1)}{2}\right)^{2}.$ This is equivalent to the square of the sum of the first $n$ natural numbers.
- Every positive rational number can be expressed as the sum of three cubes of rational numbers.
- It is possible to express any perfect cube as the sum of four odd numbers. For example, $64= 13+15+17+19.$

Units digits of perfect cubes:

- If a number ends in 0, its cube ends in 0.
- If a number ends in 2, its cube ends in 8.
- If a number ends in 3, its cube ends in 7.
- If a number ends in 4, its cube ends in 4.
- If a number ends in 5, its cube ends in 5.
- If a number ends in 6, its cube ends in 6.
- If a number ends in 7, its cube ends in 3.
- If a number ends in 8, its cube ends in 2.
- If a number ends in 9, its cube ends in 9.

The proof for this is left for the reader.

Only 6 of the following 7 numbers are perfect cubes. Which one is

not?$8000,\ 15625,\ 40323,\ 132651,\ 941192,\ 103823,\ 42875$

By applying the $1^\text{st}$ property, the digital root of 40323 is $4+0+3+2+3= 12 \longrightarrow 1+2=3.$ Thus, 40323 is not a perfect cube. $_\square$

## Perfect Powers

A perfect power is the more general form of squares and cubes. Specifically, it is any number that can be written as the product of some non-negative integer multiplied by itself at least twice. In other words, it is of the form $n^m$ for some integers $n\ge 0$ and $m > 1.$

The set of perfect powers is the union of the sets of perfect squares, perfect cubes, perfect fourth powers, and so on. The perfect powers less than or equal to $100$ are

$0,1,4,8,9,16,25,27,32,36,49,64,81,100.$

A few simple results are given below. The proofs for these are omitted.

- The $n^\text{th}$ power of a number with units digit 5 will have units digit 5 again.
- The $n^\text{th}$ power of a number with units digit 1 will have units digit 1 again.
- The $n^\text{th}$ power of a number with units digit 0 will have units digit 0.
- The $n^\text{th}$ power of a number with units digit 6 will have units digit 6.
- The units digit of a number raised to 5 is the units digit of the original number. In fact, $a^5=10{m} + a,$ where $a$ and $m$ are positive numbers (which comes from a famous theorem by Euler).
- 0 to the $n^\text{th}$ power is 0, where $n$ is not equal to zero. 0 to the zeroth power is undefined.
- 1 to the $n^\text{th}$ power is 1.
- The number of digits in 10 to the power $n$ is $n+1,$ where there will be $n$ zeroes.
- Every $4^\text{th}, 6^\text{th}, 8^\text{th}, ...., (2n)^\text{th}$ power of a positive integer is a perfect square.

Which is the greatest of the following?

$2^8,\ 3^6,\ 7^4,\ 5^4,\ 10^3,\ 4^6?$

We have

$\begin{array}{c}2^8 = 512, & 3^6 = 729, & 7^4 = 2401, & 5^4 = 625, & 10^3 = 1000, &4^6 = 4096. \end{array}$

Thus $4^6$ is the greatest among these numbers. Do note that there are better ways to determine which is the greatest or least given a set of numbers; the above powers can all be easily computed.

Find the number of digits in $64^{7}$, given $6^{7} = 279936$ and $7^{7}= 823543.$

Since we are given $6^{7}$ and $7^{7}$, we can easily see that the number of digits of any positive integer between 60 and 70 is in fact 13. Here's how: If $n$ is greater than $m,$ then $n$ raised to $k$ is greater than $m$ raised to $k,$ where $n$ and $m$ are real numbers and $k$ is a positive integer. Knowing this, we can find the number of digits in $64^{7}:$

$\begin{array}{c}60^7 = 6^7 \times 10000000 = 2799360000000, & 70^7 = 7^7 \times 10000000 = 8235430000000. \end{array}$

Since both $60^{7}$ and $70^{7}$ have an equal number of digits and $60^{7}$ is less than $64^{7}$ but $64^{7}$ is less than $70^{7}$, $64^{7}$ must also have an equal number of digits. Hence $64^{7}$ has 13 digits. $_\square$

## See Also

**Cite as:**Perfect Squares, Cubes, and Powers.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/perfect-squares/