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
\[\begin{array} &0^2=0, &(\pm1)^2=1, &(\pm2)^2=4, &(\pm3)^2=9, &(\pm4)^2=16, \\ (\pm5)^2=25, &(\pm6)^2=36, &(\pm7)^2=49, &(\pm8)^2=64, &(\pm9)^2=81 .\end{array}\]
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
\[\begin{array} &1^2-0^2=1-0=1, &&&2^2-1^2=4-1=3, \\3^2-2^2=9-4=5, &&&4^2-3^2=16-9=7,\\ 5^2-4^2=25-16=9, &&&6^2-5^2=36-25=11, \\ 7^2-6^2=49-36=13, &&&8^2-7^2=64-49=15, \\ 9^2-8^2=81-64=17 .\end{array}\]
Thus, the answers are
\[1, 3, 5, 7, 9, 11, 13, 15, 17. \ _\square\]
Which of the following is not a perfect square?
\[\begin{array} &(a)~ 125 &&&(b)~ 144 &&&(c)~ 441 &&&(d)~ 225 \end{array}\]
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
\[\begin{array} &(\pm17)^2=289, &(\pm18)^2=324, &(\pm19)^2=361, &(\pm20)^2=400.\end{array}\]
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
\[\begin{array} 0^3 = 0, & 1^3 = 1, & 2^3 = 8, & 3^3 = 27, & 4^3 = 64, \\
5^3 = 125, & 6^3 = 216, & 7^3 = 343, & 8^3 = 512, & 9^3 = 729. \end{array}\]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
\[\begin{array} 2^8 = 512, & 3^6 = 729, & 7^4 = 2401, & 5^4 = 625, & 10^3 = 1000, &4^6 = 4096. \end{array}\]
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 \(64^{7}:\)
\[\begin{array} 60^7 = 6^7 \times 10000000 = 2799360000000, & 70^7 = 7^7 \times 10000000 = 8235430000000. \end{array}\]
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. \(_\square\)
