Editing Pythagorean triple (section)

==Generalizations==
There are several ways to generalize the concept of Pythagorean triples.

===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>

===Fermat's Last Theorem===
{{main|Fermat's Last Theorem}}

A generalization of the concept of Pythagorean triples is the search for triples of positive integers {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}, such that {{math|''a''<sup>''n''</sup> + ''b''<sup>''n''</sup> {{=}} ''c''<sup>''n''</sup>}}, for some {{math|''n''}} strictly greater than 2. [[Pierre de Fermat]] in 1637 claimed that no such triple exists, a claim that came to be known as [[Fermat's Last Theorem]] because it took longer than any other conjecture by Fermat to be proved or disproved. The first proof was given by [[Andrew Wiles]] in 1994.

==={{math|''n'' − 1}} or {{math|''n''}} {{math|''n''}}th powers summing to an {{math|''n''}}th power===
{{main|Euler's sum of powers conjecture}}
Another generalization is searching for sequences of {{math|1=''n'' + 1}} positive integers for which the {{math|''n''}}th power of the last is the sum of the {{math|''n''}}th powers of the previous terms. The smallest sequences for known values of {{math|''n''}} are:

* {{math|''n''}} = 3: {3, 4, 5; 6}.
* {{math|''n''}} = 4: {30, 120, 272, 315; 353}
* {{math|''n''}} = 5: {19, 43, 46, 47, 67; 72}
* {{math|''n''}} = 7: {127, 258, 266, 413, 430, 439, 525; 568}
* {{math|''n''}} = 8: {90, 223, 478, 524, 748, 1088, 1190, 1324; 1409}

For the {{math|1=''n'' = 3}} case, in which <math>x^3+y^3+z^3=w^3,</math> called the  [[Fermat cubic]], a general formula exists giving all solutions.

A slightly different generalization allows the sum of {{math|(''k'' + 1)}} {{math|''n''}}th powers to equal the sum of {{math|(''n'' − ''k'')}} {{math|''n''}}th powers. For example:
* ({{math|1=''n'' = 3}}): 1{{sup|3}} + 12{{sup|3}} = 9{{sup|3}} + 10{{sup|3}}, made famous by Hardy's recollection of a conversation with [[Ramanujan]] about the number [[Taxicab number|1729]] being the smallest number that can be expressed as a sum of two cubes in two distinct ways.

There can also exist {{math|''n'' − 1}} positive integers whose {{math|''n''}}th powers sum to an {{math|''n''}}th power (though, by [[Fermat's Last Theorem]], not for {{math|1=''n'' = 3)}}; these are counterexamples to [[Euler's sum of powers conjecture]].  The smallest known counterexamples are<ref>{{citation |first1=Scott |last1=Kim |author-link1=Scott Kim|title=Bogglers |journal=[[Discover (magazine)|Discover]] |page=82 |date=May 2002 |url=http://discovermagazine.com/2002/may/bogglers |quote=The equation w{{sup|4}} + x{{sup|4}} + y{{sup|4}} = z{{sup|4}} is harder. In 1988, after 200 years of mathematicians' attempts to prove it impossible, [[Noam Elkies]] of Harvard found the counterexample, 2,682,440{{sup|4}} + 15,365,639{{sup|4}} + 18,796,760{{sup|4}} = 20,615,673{{sup|4}}.}}</ref><ref>{{citation |author-link=Noam Elkies |last=Elkies |first=Noam |title=On A{{sup|4}} + B{{sup|4}} + C{{sup|4}} = D{{sup|4}} |journal=Mathematics of Computation |volume=51 |issue=184 |pages=825–835 |year=1988 |mr=930224 |url=https://www.ams.org/journals/mcom/1988-51-184/S0025-5718-1988-0930224-9/ | doi = 10.2307/2008781|jstor=2008781 }}</ref><ref name=MacHale>{{citation |last1=MacHale |first1=Des |author1-link=Des MacHale|last2=van den Bosch |first2=Christian |title=Generalising a result about Pythagorean triples |journal=[[Mathematical Gazette]] |volume=96 |pages=91–96 |date=March 2012 |doi=10.1017/S0025557200004010 |s2cid=124096076 |doi-access=free }}</ref>

* {{math|1=''n'' = 4}}: (95800, 217519, 414560; 422481)
* {{math|1=''n'' = 5}}: (27, 84, 110, 133; 144)

===Heronian triangle triples===
{{main|Heronian triangle}}

A '''Heronian triangle''' is commonly defined as one with integer sides whose area is also an integer. The lengths of the sides of such a triangle form a '''Heronian triple''' {{math|(''a, b, c'')}} for {{math|''a'' ≤ ''b'' ≤ ''c''}}.
Every Pythagorean triple is a Heronian triple, because at least one of the legs {{math|''a''}}, {{math|''b''}} must be even in a Pythagorean triple, so the area ''ab''/2 is an integer. Not every Heronian triple is a Pythagorean triple, however, as the example {{math|(4, 13, 15)}} with area 24 shows.

If {{math|(''a'', ''b'', ''c'')}} is a Heronian triple, so is {{math|(''ka'', ''kb'', ''kc'')}} where {{math|''k''}} is any positive integer; its area will be the integer that is {{math|''k''{{sup|2}}}} times the integer area of the {{math|(''a'', ''b'', ''c'')}} triangle.
The Heronian triple {{math|(''a'', ''b'', ''c'')}} is '''primitive''' provided ''a'', ''b'', ''c'' are [[coprime integers#Coprimality in sets|setwise coprime]].  (With primitive Pythagorean triples the stronger statement that they are ''pairwise'' coprime also applies, but with primitive Heronian triangles the stronger statement does not always hold true, such as with {{math|(7, 15, 20)}}.) Here are a few of the simplest primitive Heronian triples that are not Pythagorean triples:

: (4, 13, 15) with area 24
: (3, 25, 26) with area 36
: (7, 15, 20) with area 42
: (6, 25, 29) with area 60
: (11, 13, 20) with area 66
: (13, 14, 15) with area 84
: (13, 20, 21) with area 126

By [[Heron's formula]], the extra condition for a triple of positive integers {{math|(''a'', ''b'', ''c'')}} with {{math|''a'' < ''b'' < ''c''}} to be Heronian is that

: {{math|(''a''{{sup|2}} + ''b''{{sup|2}} + ''c''{{sup|2}}){{sup|2}} − 2(''a''{{sup|4}} + ''b''{{sup|4}} + ''c''{{sup|4}})}}
or equivalently
: {{math|2(''a''{{sup|2}}''b''{{sup|2}} + ''a''{{sup|2}}''c''{{sup|2}} + ''b''{{sup|2}}''c''{{sup|2}}) − (''a''{{sup|4}} + ''b''{{sup|4}} + ''c''{{sup|4}})}}

be a nonzero perfect square divisible by 16.

===Application to cryptography===
Primitive Pythagorean triples have been used in cryptography as random sequences and for the generation of keys.<ref>[[Subhash Kak|Kak, S.]] and Prabhu, M. Cryptographic applications of primitive Pythagorean triples. Cryptologia, 38:215–222, 2014. [http://www.tandfonline.com/doi/full/10.1080/01611194.2014.915257#.U5tvAvldXkg]</ref>