Editing Square number (section)

==Properties==
The number ''m'' is a square number if and only if one can arrange ''m'' points in a square:
{| cellpadding="8"
|-
|{{bigmath|1=''m'' = 1<sup>2</sup> = 1}}
|[[Image:Square number 1.png]]
|-
|{{bigmath|1=''m'' = 2<sup>2</sup> = 4}}
|[[Image:Square number 4.png]]
|-
|{{bigmath|1=''m'' = 3<sup>2</sup> = 9}}
|[[Image:Square number 9.png]]
|-
|{{bigmath|1=''m'' = 4<sup>2</sup> = 16}}
|[[Image:Square number 16.png]]
|-
|{{bigmath|1=''m'' = 5<sup>2</sup> = 25}}
|[[Image:Square number 25.png]]
|}

The expression for the {{mvar|n}}th square number is {{math|''n''<sup>2</sup>}}. This is also equal to the sum of the first {{mvar|n}} [[parity (mathematics)|odd number]]s 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:<math display="block">n^2 = \sum_{k=1}^n (2k-1).</math>For example, {{math|1=5<sup>2</sup> = 25 = 1 + 3 + 5 + 7 + 9}}.
[[File:The sum of the first n odd integers is n². 1+3+5+...+(2n-1)=n²..gif|thumb|The sum of the first ''n'' odd integers is ''n''<sup>2</sup>. {{nowrap|1 + 3 + 5 + ... + (2''n'' − 1) {{=}} ''n''<sup>2</sup>}}. Animated 3D visualization on a tetrahedron.]]
There are several [[recursion#Recursion in mathematics|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}} + (2''n'' − 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=''n''<sup>2</sup> = 2(''n'' − 1)<sup>2</sup> − (''n'' − 2)<sup>2</sup> + 2}}. For example,
:{{math|1= 2 × 5<sup>2</sup> − 4<sup>2</sup> + 2 = 2 × 25 − 16 + 2 = 50 − 16 + 2 = 36 = 6<sup>2</sup>}}.
The square minus one of a number {{mvar|m}} is always the product of <math>m - 1</math> and <math>m + 1;</math> that is,<math display="block">m^2-1=(m-1)(m+1).</math>For example, since {{Math|1=7<sup>2</sup> = 49}}, one has <math>6 \times 8 = 48</math>. 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,<math display="block">a^2-b^2=(a+b)(a-b)</math>This is the [[difference of two squares|difference-of-squares formula]], which can be useful for mental arithmetic: for example, {{math|1=47 × 53}} can be easily computed as {{math|1=50<sup>2</sup> − 3<sup>2</sup> = 2500 − 9 = 2491}}.
A square number is also the sum of two consecutive [[triangular number]]s. 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 [[parity (mathematics)|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<sup>''k''</sup>(8''m'' + 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|4''k'' + 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 number|prime]]&nbsp;{{mvar|p}} divides a square number&nbsp;{{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 of a positive integer|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&nbsp;{{mvar|k}}, if {{math|''k''<sup>2</sup> − ''m''}} is the square of an integer&nbsp;{{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|100<sup>2</sup> − 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 <math>k \geq \sqrt{m}.</math>

A square number cannot be a [[perfect number]].

The sum of the ''n'' first square numbers is<math display="block">\sum_{n=0}^N n^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + \cdots + N^2 = \frac{N(N+1)(2N+1)}{6}.</math>The first values of these sums, the [[square pyramidal number]]s, are: {{OEIS|id=A000330}}
<blockquote>
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...
</blockquote>
[[File:Proofwithoutwords.svg|thumb|100px|Proof without words for the sum of odd numbers theorem]]
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 <math>s=ut+\tfrac{1}{2}at^2</math>, for {{Math|1=''u'' = 0}} and constant {{Math|''a''}} (acceleration due to gravity without air resistance); so {{Math|''s''}} is proportional to {{Math|''t''<sup>2</sup>}}, and the distance from the starting point are consecutive squares for integer values of time elapsed.<ref>{{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}}</ref>

The sum of the ''n'' first [[cube (algebra)|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 <math>T_n</math> is:<math display="block">(T_n)^2 + (T_{n+1})^2 = T_{(n+1)^2}</math>