square number

The squares {{OEIS|id=A000290}} smaller than 60² = 3600 are:

{{Clear}}

The difference between any perfect square and its predecessor is given by the identity {{math|n{{sup|2}} − (n − 1){{sup|2}} {{=}} 2n − 1}}. Equivalently, it is possible to count square numbers by adding together the last square, the last square's root, and the current root, that is, {{math|n{{sup|2}} {{=}} (n − 1){{sup|2}} + (n − 1) + n}}.

The number m is a square number if and only if one can arrange m points in a square:

The expression for the {{mvar|n}}th square number is {{math|n²}}. This is also equal to the sum of the first {{mvar|n}} odd numbers as can be seen in the above pictures, where a square results from the previous one by adding an odd number of points (shown in magenta). The formula follows:$$n^2\; =\; \backslash sum\_\{k=1\}^n\; (2k-1).$$For example, {{math|1=5² = 25 = 1 + 3 + 5 + 7 + 9}}.

File:The sum of the first n odd integers is n². 1+3+5+...+(2n-1)=n²..gif

There are several recursive methods for computing square numbers. For example, the {{mvar|n}}th square number can be computed from the previous square by {{math|n{{sup|2}} {{=}} (n − 1){{sup|2}} + (n − 1) + n {{=}} (n − 1){{sup|2}} + (2n − 1)}}. Alternatively, the {{mvar|n}}th square number can be calculated from the previous two by doubling the {{math|(n − 1)}}th square, subtracting the {{math|(n − 2)}}th square number, and adding 2, because {{math|1=n² = 2(n − 1)² − (n − 2)² + 2}}. For example,

:{{math|1= 2 × 5² − 4² + 2 = 2 × 25 − 16 + 2 = 50 − 16 + 2 = 36 = 6²}}.

The square minus one of a number {{mvar|m}} is always the product of $m\; -\; 1$ and $m\; +\; 1;$ that is,$$m^2-1=(m-1)(m+1).$$For example, since {{Math|1=7² = 49}}, one has $6\; \backslash times\; 8\; =\; 48$. Since a prime number has factors of only {{Math|1}} and itself, and since {{Math|1=m = 2}} is the only non-zero value of {{Math|m}} to give a factor of {{Math|1}} on the right side of the equation above, it follows that {{math|3}} is the only prime number one less than a square ({{math|1= 3 = 2{{sup|2}} − 1}}).

More generally, the difference of the squares of two numbers is the product of their sum and their difference. That is,$$a^2-b^2=(a+b)(a-b)$$This is the difference-of-squares formula, which can be useful for mental arithmetic: for example, {{math|1=47 × 53}} can be easily computed as {{math|1=50² − 3² = 2500 − 9 = 2491}}.

A square number is also the sum of two consecutive triangular numbers. The sum of two consecutive square numbers is a centered square number. Every odd square is also a centered octagonal number.

Another property of a square number is that (except 0) it has an odd number of positive divisors, while other natural numbers have an even number of positive divisors. An integer root is the only divisor that pairs up with itself to yield the square number, while other divisors come in pairs.

Lagrange's four-square theorem states that any positive integer can be written as the sum of four or fewer perfect squares. Three squares are not sufficient for numbers of the form {{math|4^k(8m + 7)}}. A positive integer can be represented as a sum of two squares precisely if its prime factorization contains no odd powers of primes of the form {{math|4k + 3}}. This is generalized by Waring's problem.

In base 10, a square number can end only with digits 0, 1, 4, 5, 6 or 9, as follows:

• if the last digit of a number is 0, its square ends in 00;
• if the last digit of a number is 1 or 9, its square ends in an even digit followed by a 1;
• if the last digit of a number is 2 or 8, its square ends in an even digit followed by a 4;
• if the last digit of a number is 3 or 7, its square ends in an even digit followed by a 9;
• if the last digit of a number is 4 or 6, its square ends in an odd digit followed by a 6; and
• if the last digit of a number is 5, its square ends in 25.

In base 12, a square number can end only with square digits (like in base 12, a prime number can end only with prime digits or 1), that is, 0, 1, 4 or 9, as follows:

• if a number is divisible both by 2 and by 3 (that is, divisible by 6), its square ends in 0, and its preceding digit must be 0 or 3;
• if a number is divisible neither by 2 nor by 3, its square ends in 1, and its preceding digit must be even;
• if a number is divisible by 2, but not by 3, its square ends in 4, and its preceding digit must be 0, 1, 4, 5, 8, or 9; and
• if a number is not divisible by 2, but by 3, its square ends in 9, and its preceding digit must be 0 or 6.

Similar rules can be given for other bases, or for earlier digits (the tens instead of the units digit, for example).{{citation needed|date=March 2016}} All such rules can be proved by checking a fixed number of cases and using modular arithmetic.

In general, if a prime {{mvar|p}} divides a square number {{mvar|m}} then the square of {{mvar|p}} must also divide {{mvar|m}}; if {{mvar|p}} fails to divide {{math|{{sfrac|m|p}}}}, then {{mvar|m}} is definitely not square. Repeating the divisions of the previous sentence, one concludes that every prime must divide a given perfect square an even number of times (including possibly 0 times). Thus, the number {{mvar|m}} is a square number if and only if, in its canonical representation, all exponents are even.

Squarity testing can be used as alternative way in factorization of large numbers. Instead of testing for divisibility, test for squarity: for given {{mvar|m}} and some number {{mvar|k}}, if {{math|k² − m}} is the square of an integer {{mvar|n}} then {{math|k − n}} divides {{mvar|m}}. (This is an application of the factorization of a difference of two squares.) For example, {{math|100² − 9991}} is the square of 3, so consequently {{math|100 − 3}} divides 9991. This test is deterministic for odd divisors in the range from {{math|k − n}} to {{math|k + n}} where {{mvar|k}} covers some range of natural numbers $k\; \backslash geq\; \backslash sqrt\{m\}.$

A square number cannot be a perfect number.

The sum of the n first square numbers is$$\backslash sum\_\{n=0\}^N\; n^2\; =\; 0^2\; +\; 1^2\; +\; 2^2\; +\; 3^2\; +\; 4^2\; +\; \backslash cdots\; +\; N^2\; =\; \backslash frac\{N(N+1)(2N+1)\}\{6\}.$$The first values of these sums, the square pyramidal numbers, are: {{OEIS|id=A000330}}

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201...

File:Proofwithoutwords.svg

The sum of the first odd integers, beginning with one, is a perfect square: 1, 1 + 3, 1 + 3 + 5, 1 + 3 + 5 + 7, etc. This explains Galileo's law of odd numbers: if a body falling from rest covers one unit of distance in the first arbitrary time interval, it covers 3, 5, 7, etc., units of distance in subsequent time intervals of the same length. From $s=ut+\backslash tfrac\{1\}\{2\}at^2$, for {{Math|1=u = 0}} and constant {{Math|a}} (acceleration due to gravity without air resistance); so {{Math|s}} is proportional to {{Math|t²}}, and the distance from the starting point are consecutive squares for integer values of time elapsed.{{Cite book| last1=Olenick|first1=Richard P.| url=https://books.google.com/books?id=xMWwTpn53KsC&pg=PA18| title=The Mechanical Universe: Introduction to Mechanics and Heat| last2=Apostol|first2=Tom M.| last3=Goodstein|first3=David L.| date=2008-01-14| publisher=Cambridge University Press| isbn=978-0-521-71592-8| pages=18| language=en}}

The sum of the n first cubes is the square of the sum of the n first positive integers; this is Nicomachus's theorem.

All fourth powers, sixth powers, eighth powers and so on are perfect squares.

A unique relationship with triangular numbers $T\_n$ is:$$(T\_n)^2\; +\; (T\_\{n+1\})^2\; =\; T\_\{(n+1)^2\}$$

==Odd and even square numbers==

File:visual_proof_centered_octagonal_numbers_are_odd_squares.svg that all centered octagonal numbers are odd squares]]

Squares of even numbers are even, and are divisible by 4, since {{Math|1=(2n)² = 4n²}}. Squares of odd numbers are odd, and are congruent to 1 modulo 8, since {{Math|1=(2n + 1)² = 4n(n + 1) + 1}}, and {{Math|n(n + 1)}} is always even. In other words, all odd square numbers have a remainder of 1 when divided by 8.

Every odd perfect square is a centered octagonal number. The difference between any two odd perfect squares is a multiple of 8. The difference between 1 and any higher odd perfect square always is eight times a triangular number, while the difference between 9 and any higher odd perfect square is eight times a triangular number minus eight. Since all triangular numbers have an odd factor, but no two values of {{math|2ⁿ}} differ by an amount containing an odd factor, the only perfect square of the form {{math|2ⁿ − 1}} is 1, and the only perfect square of the form {{math|2ⁿ + 1}} is 9.

• If the number is of the form {{math|m5}} where {{math|m}} represents the preceding digits, its square is {{math|n25}} where {{math|1=n = m(m + 1)}} and represents digits before 25. For example, the square of 65 can be calculated by {{math|1=n = 6 × (6 + 1) = 42}} which makes the square equal to 4225.
• If the number is of the form {{math|m0}} where {{math|m}} represents the preceding digits, its square is {{math|n00}} where {{math|1=n = m²}}. For example, the square of 70 is 4900.
• If the number has two digits and is of the form {{math|5m}} where {{math|m}} represents the units digit, its square is {{math|aabb}} where {{math|1=aa = 25 + m}} and {{math|1=bb = m²}}. For example, to calculate the square of 57, {{Math|1=m = 7}} and {{Math|1=25 + 7 = 32}} and {{Math|1=7² = 49}}, so {{Math|1=57² = 3249}}.
• If the number ends in 5, its square will end in 5; similarly for ending in 25, 625, 0625, 90625, ... 8212890625, etc. If the number ends in 6, its square will end in 6, similarly for ending in 76, 376, 9376, 09376, ... 1787109376. For example, the square of 55376 is 3066501376, both ending in 376. (The numbers 5, 6, 25, 76, etc. are called automorphic numbers. They are sequence A003226 in the OEIS.{{Cite OEIS|1=A003226|2=Automorphic numbers: n^2 ends with n.}})
• In base 10, the last two digits of square numbers follow a repeating pattern mirrored symmetrical around multiples of 25. In the example of 24 and 26, both 1 off from 25, {{Math|1=24² = 576}} and {{Math|1=26² = 676}}, both ending in 76. In general, $(25n+x)^2-(25n-x)^2=100nx$. An analogous pattern applies for the last 3 digits around multiples of 250, and so on. As a consequence, of the 100 possible last 2 digits, only 22 of them occur among square numbers (since 00 and 25 are repeated).

{{div col}}

• {{annotated link|Brahmagupta–Fibonacci identity}}
• {{annotated link|cube (algebra)|Cubic number}}
• {{annotated link|Euler's four-square identity}}
• {{annotated link|Fermat's theorem on sums of two squares}}
• Some identities involving several squares
• {{annotated link|Integer square root}}
• {{annotated link|Methods of computing square roots}}
• {{annotated link|Power of two}}
• {{annotated link|Pythagorean triple}}
• {{annotated link|Quadratic residue}}
• {{annotated link|Quadratic function}}
• {{annotated link|Square triangular number}}

{{div col end}}

{{reflist}}

• Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 30–32, 1996. {{isbn|0-387-97993-X}}
• Kiran Parulekar. Amazing Properties of Squares and Their Calculations. Kiran Anil Parulekar, 2012 https://books.google.com/books?id=njEtt7rfexEC

{{Figurate numbers}}

{{Classes of natural numbers}}

Category:Elementary arithmetic

Category:Figurate numbers

Category:Integer sequences

Category:Integers

Category:Number theory

Category:Quadrilaterals

Category:Squares in number theory