Editing Pythagorean triple (section)

===Pythagorean {{math|''n''}}-tuple===
{{see also|Pythagorean quadruple}}

The expression
:<math>\left(m_1^2 - m_2^2 - \ldots - m_n^2\right)^2 + \sum_{k=2}^n (2 m_1 m_k)^2 = \left(m_1^2 + \ldots + m_n^2\right)^2</math>
is a Pythagorean {{mvar|n}}-tuple for any tuple of positive integers {{math|(''m''{{sub|1}}, ..., ''m''{{sub|''n''}})}} with {{math|''m''{{supsub|2|1}} > ''m''{{supsub|2|2}} + ... + ''m''{{supsub|2|''n''}}}}.  The Pythagorean {{mvar|n}}-tuple can be made primitive by dividing out by the largest common divisor of its values.

Furthermore, any primitive Pythagorean {{mvar|n}}-tuple {{math|1=''a''{{supsub|2|1}} + ... + ''a''{{supsub|2|''n''}} = ''c''{{sup|2}}}} can be found by this approach.  Use {{math|1=(''m''{{sub|1}}, ..., ''m''{{sub|''n''}}) = (''c'' + ''a''{{sub|1}}, ''a''{{sub|2}}, ..., ''a''{{sub|''n''}})}} to get a Pythagorean {{mvar|n}}-tuple by the above formula and divide out by the largest common integer divisor, which is {{math|1=2''m''{{sub|1}} = 2(''c'' + ''a''{{sub|1}})}}.  Dividing out by the largest common divisor of these {{math|1=(''m''{{sub|1}}, ..., ''m''{{sub|''n''}})}} values gives the same primitive Pythagorean {{mvar|n}}-tuple; and there is a one-to-one correspondence between tuples of [[setwise coprime]] positive integers {{math|1=(''m''{{sub|1}}, ..., ''m''{{sub|''n''}})}} satisfying {{math|''m''{{supsub|2|1}} > ''m''{{supsub|2|2}} + ... + ''m''{{supsub|2|''n''}}}} and primitive Pythagorean {{mvar|n}}-tuples.

Examples of the relationship between setwise coprime values <math>\vec{m}</math> and primitive Pythagorean {{mvar|n}}-tuples include:<ref>{{Cite OEIS|A351061|Smallest positive integer whose square can be written as the sum of n positive perfect squares|mode=cs2}}</ref>

:<math>\begin{align}
                  \vec{m} = (1) & \leftrightarrow 1^2 = 1^2 \\
               \vec{m} = (2, 1) & \leftrightarrow 3^2 + 4^2 = 5^2 \\
            \vec{m} = (2, 1, 1) & \leftrightarrow 1^2 + 2^2 + 2^2 = 3^2 \\
         \vec{m} = (3, 1, 1, 1) & \leftrightarrow 1^2 + 1^2 + 1^2 + 1^2 = 2^2 \\
      \vec{m} = (5, 1, 1, 2, 3) & \leftrightarrow 1^2 + 1^2 + 1^2 + 2^2 + 3^2 = 4^2 \\
   \vec{m} = (4, 1, 1, 1, 1, 2) & \leftrightarrow 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 = 3^2 \\
\vec{m} = (5, 1, 1, 1, 2, 2, 2) & \leftrightarrow 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 = 4^2
\end{align}</math>

====Consecutive squares====
Since the sum {{math|''F''(''k'',''m'')}} of {{math|''k''}} consecutive squares beginning with {{math|''m''{{sup|2}}}} is given by the formula,<ref>{{citation|url= http://www.math.niu.edu/~rusin/known-math/97/cube.sum|title= Sum of consecutive cubes equal a cube|url-status= dead|archive-url= https://web.archive.org/web/20080515132209/http://www.math.niu.edu/~rusin/known-math/97/cube.sum|archive-date= 2008-05-15}}</ref>

:<math>F(k,m)=km(k-1+m)+\frac{k(k-1)(2k-1)}{6}</math>

one may find values {{math|(''k'', ''m'')}} so that {{math|''F''(''k'',''m'')}} is a square, such as one by Hirschhorn where the number of terms is itself a square,<ref>{{citation |last=Hirschhorn |first=Michael |title=When is the sum of consecutive squares a square? |journal=The Mathematical Gazette |volume=95 |pages=511–2 |date=November 2011 |doi=10.1017/S0025557200003636 |s2cid=118776198 |oclc=819659848 |issn=0025-5572|doi-access=free }}</ref>

:<math>m=\tfrac{v^4-24v^2-25}{48},\; k=v^2,\; F(m,k)=\tfrac{v^5+47v}{48}</math>

and {{math|''v'' ≥ 5}} is any integer not divisible by 2 or 3. For the smallest case {{math|1=''v'' = 5}}, hence {{math|1=''k'' = 25}}, this yields the well-known cannonball-stacking problem of [[Édouard Lucas|Lucas]],

:<math>0^2+1^2+2^2+\dots+24^2 = 70^2</math>

a fact which is connected to the [[Leech lattice#Using the Lorentzian lattice II25,1|Leech lattice]].

In addition, if in a Pythagorean {{math|''n''}}-tuple ({{math|''n'' ≥ 4}}) all [[addend]]s are consecutive except one, one can use the equation,<ref>{{citation |last=Goehl |first=John F. Jr. |title=Reader reflections |journal=Mathematics Teacher |volume=98 |issue=9 |page=580 |date=May 2005 |doi=10.5951/MT.98.9.0580 |url=http://www.nctm.org/publications/article.aspx?id=18884}}</ref>

:<math>F(k,m) + p^{2} = (p+1)^{2}</math>

Since the second power of {{math|''p''}} cancels out, this is only linear and easily solved for as <math>p=\tfrac{F(k,m)-1}{2}</math> though {{math|''k''}}, {{math|''m''}} should be chosen so that {{math|''p''}} is an integer,  with a small example being {{math|1=''k'' = 5}}, {{math|1=''m'' = 1}} yielding,

:<math>1^2+2^2+3^2+4^2+5^2+27^2=28^2</math>

Thus, one way of generating Pythagorean {{math|''n''}}-tuples is by using, for various {{math|''x''}},<ref>Goehl, John F., Jr., "Triples, quartets, pentads", ''Mathematics Teacher'' 98, May 2005, p. 580.</ref>

:<math>x^2+(x+1)^2+\cdots +(x+q)^2+p^2=(p+1)^2,</math>

where ''q = n''–2 and where

:<math>p=\frac{(q+1)x^2+q(q+1)x+\frac{q(q+1)(2q+1)}{6} -1}{2}.</math>