Square triangular number

<h2 id="solution-as-a-pell-equation">Solution as a Pell equation</h2>
<p>Write 
  
    
      
        
          N
          
            k
          
        
      
    
    {\displaystyle N_{k}}
  
 for the 
  
    
      
        k
      
    
    {\displaystyle k}
  
th square triangular number, and write 
  
    
      
        
          s
          
            k
          
        
      
    
    {\displaystyle s_{k}}
  
 and 
  
    
      
        
          t
          
            k
          
        
      
    
    {\displaystyle t_{k}}
  
 for the sides of the corresponding square and triangle, so that 
</p>

N
            
              k
            
          
          =
          
            s
            
              k
            
            
              2
            
          
          =
          
            
              
                
                  t
                  
                    k
                  
                
                (
                
                  t
                  
                    k
                  
                
                +
                1
                )
              
              2
            
          
          .
        
      
    
    {\displaystyle \displaystyle N_{k}=s_{k}^{2}={\frac {t_{k}(t_{k}+1)}{2}}.}

<p>Define the <i>triangular root</i> of a triangular number 
  
    
      
        N
        =
        
          
            
              
                n
                (
                n
                +
                1
                )
              
              2
            
          
        
      
    
    {\displaystyle N={\tfrac {n(n+1)}{2}}}
  
 to be 
  
    
      
        n
      
    
    {\displaystyle n}
  
. From this definition and the quadratic formula,
</p>

n
          =
          
            
              
                
                  
                    8
                    N
                    +
                    1
                  
                
                −
                1
              
              2
            
          
          .
        
      
    
    {\displaystyle \displaystyle n={\frac {{\sqrt {8N+1}}-1}{2}}.}

<p>Therefore, 
  
    
      
        N
      
    
    {\displaystyle N}
  
 is triangular (
  
    
      
        n
      
    
    {\displaystyle n}
  
 is an integer) <a href="/facts/If_and_only_if/bYSxGJ66">if and only if</a> 
  
    
      
        8
        N
        +
        1
      
    
    {\displaystyle 8N+1}
  
 is square. Consequently, a square number 
  
    
      
        
          M
          
            2
          
        
      
    
    {\displaystyle M^{2}}
  
 is also triangular if and only if 
  
    
      
        8
        
          M
          
            2
          
        
        +
        1
      
    
    {\displaystyle 8M^{2}+1}
  
 is square, that is, there are numbers 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
 such that 
  
    
      
        
          x
          
            2
          
        
        −
        8
        
          y
          
            2
          
        
        =
        1
      
    
    {\displaystyle x^{2}-8y^{2}=1}
  
. This is an instance of the <a href="/facts/Pell_equation/n0ul61eQ">Pell equation</a> 
  
    
      
        
          x
          
            2
          
        
        −
        n
        
          y
          
            2
          
        
        =
        1
      
    
    {\displaystyle x^{2}-ny^{2}=1}
  
 with 
  
    
      
        n
        =
        8
      
    
    {\displaystyle n=8}
  
. All Pell equations have the trivial solution 
  
    
      
        x
        =
        1
        ,
        y
        =
        0
      
    
    {\displaystyle x=1,y=0}
  
 for any 
  
    
      
        n
      
    
    {\displaystyle n}
  
; this is called the zeroth solution, and indexed as 
  
    
      
        (
        
          x
          
            0
          
        
        ,
        
          y
          
            0
          
        
        )
        =
        (
        1
        ,
        0
        )
      
    
    {\displaystyle (x_{0},y_{0})=(1,0)}
  
. If 
  
    
      
        (
        
          x
          
            k
          
        
        ,
        
          y
          
            k
          
        
        )
      
    
    {\displaystyle (x_{k},y_{k})}
  
 denotes the 
  
    
      
        k
      
    
    {\displaystyle k}
  
th nontrivial solution to any Pell equation for a particular 
  
    
      
        n
      
    
    {\displaystyle n}
  
, it can be shown by the method of descent that the next solution is
</p>

x
                    
                      k
                      +
                      1
                    
                  
                
                
                  
                  =
                  2
                  
                    x
                    
                      k
                    
                  
                  
                    x
                    
                      1
                    
                  
                  −
                  
                    x
                    
                      k
                      −
                      1
                    
                  
                  ,
                
              
              
                
                  
                    y
                    
                      k
                      +
                      1
                    
                  
                
                
                  
                  =
                  2
                  
                    y
                    
                      k
                    
                  
                  
                    x
                    
                      1
                    
                  
                  −
                  
                    y
                    
                      k
                      −
                      1
                    
                  
                  .
                
              
            
          
        
      
    
    {\displaystyle \displaystyle {\begin{aligned}x_{k+1}&=2x_{k}x_{1}-x_{k-1},\\y_{k+1}&=2y_{k}x_{1}-y_{k-1}.\end{aligned}}}

<p>Hence there are infinitely many solutions to any Pell equation for which there is one non-trivial one, which is true whenever 
  
    
      
        n
      
    
    {\displaystyle n}
  
 is not a square. The first non-trivial solution when 
  
    
      
        n
        =
        8
      
    
    {\displaystyle n=8}
  
 is easy to find: it is 
  
    
      
        (
        3
        ,
        1
        )
      
    
    {\displaystyle (3,1)}
  
. A solution 
  
    
      
        (
        
          x
          
            k
          
        
        ,
        
          y
          
            k
          
        
        )
      
    
    {\displaystyle (x_{k},y_{k})}
  
 to the Pell equation for 
  
    
      
        n
        =
        8
      
    
    {\displaystyle n=8}
  
 yields a square triangular number and its square and triangular roots as follows:
</p>

s
            
              k
            
          
          =
          
            y
            
              k
            
          
          ,
          
          
            t
            
              k
            
          
          =
          
            
              
                
                  x
                  
                    k
                  
                
                −
                1
              
              2
            
          
          ,
          
          
            N
            
              k
            
          
          =
          
            y
            
              k
            
            
              2
            
          
          .
        
      
    
    {\displaystyle \displaystyle s_{k}=y_{k},\quad t_{k}={\frac {x_{k}-1}{2}},\quad N_{k}=y_{k}^{2}.}

<p>Hence, the first square triangular number, derived from 
  
    
      
        (
        3
        ,
        1
        )
      
    
    {\displaystyle (3,1)}
  
, is 
  
    
      
        1
      
    
    {\displaystyle 1}
  
, and the next, derived from 
  
    
      
        6
        ⋅
        (
        3
        ,
        1
        )
        −
        (
        1
        ,
        0
        )
        −
        (
        17
        ,
        6
        )
      
    
    {\displaystyle 6\cdot (3,1)-(1,0)-(17,6)}
  
, is 
  
    
      
        36
      
    
    {\displaystyle 36}
  
.
</p><p>The sequences 
  
    
      
        
          N
          
            k
          
        
      
    
    {\displaystyle N_{k}}
  
, 
  
    
      
        
          s
          
            k
          
        
      
    
    {\displaystyle s_{k}}
  
 and 
  
    
      
        
          t
          
            k
          
        
      
    
    {\displaystyle t_{k}}
  
 are the <a href="/facts/OEIS/yow6cVsU">OEIS</a> sequences <a href="/facts/On-Line_Encyclopedia_of_Integer_Sequences/yow6cVsU">OEIS</a>: A001110, <a href="/facts/On-Line_Encyclopedia_of_Integer_Sequences/yow6cVsU">OEIS</a>: A001109, and <a href="/facts/On-Line_Encyclopedia_of_Integer_Sequences/yow6cVsU">OEIS</a>: A001108 respectively.
</p>
<h2 id="explicit-formula">Explicit formula</h2>
<p>In 1778 <a href="/facts/Leonhard_Euler/z2fsbjgj">Leonhard Euler</a> determined the explicit formula<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a><a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a>: 12–13 
</p>

N
            
              k
            
          
          =
          
            
              (
              
                
                  
                    
                      
                        (
                        
                          3
                          +
                          2
                          
                            
                              2
                            
                          
                        
                        )
                      
                      
                        k
                      
                    
                    −
                    
                      
                        (
                        
                          3
                          −
                          2
                          
                            
                              2
                            
                          
                        
                        )
                      
                      
                        k
                      
                    
                  
                  
                    4
                    
                      
                        2
                      
                    
                  
                
              
              )
            
            
              2
            
          
          .
        
      
    
    {\displaystyle \displaystyle N_{k}=\left({\frac {\left(3+2{\sqrt {2}}\right)^{k}-\left(3-2{\sqrt {2}}\right)^{k}}{4{\sqrt {2}}}}\right)^{2}.}

<p>Other equivalent formulas (obtained by expanding this formula) that may be convenient include
</p>

N
                    
                      k
                    
                  
                
                
                  
                  =
                  
                    
                      
                        1
                        32
                      
                    
                  
                  
                    
                      (
                      
                        
                          
                            (
                            
                              1
                              +
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            2
                            k
                          
                        
                        −
                        
                          
                            (
                            
                              1
                              −
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            2
                            k
                          
                        
                      
                      )
                    
                    
                      2
                    
                  
                
              
              
                
                
                  
                  =
                  
                    
                      
                        1
                        32
                      
                    
                  
                  
                    (
                    
                      
                        
                          (
                          
                            1
                            +
                            
                              
                                2
                              
                            
                          
                          )
                        
                        
                          4
                          k
                        
                      
                      −
                      2
                      +
                      
                        
                          (
                          
                            1
                            −
                            
                              
                                2
                              
                            
                          
                          )
                        
                        
                          4
                          k
                        
                      
                    
                    )
                  
                
              
              
                
                
                  
                  =
                  
                    
                      
                        1
                        32
                      
                    
                  
                  
                    (
                    
                      
                        
                          (
                          
                            17
                            +
                            12
                            
                              
                                2
                              
                            
                          
                          )
                        
                        
                          k
                        
                      
                      −
                      2
                      +
                      
                        
                          (
                          
                            17
                            −
                            12
                            
                              
                                2
                              
                            
                          
                          )
                        
                        
                          k
                        
                      
                    
                    )
                  
                  .
                
              
            
          
        
      
    
    {\displaystyle \displaystyle {\begin{aligned}N_{k}&={\tfrac {1}{32}}\left(\left(1+{\sqrt {2}}\right)^{2k}-\left(1-{\sqrt {2}}\right)^{2k}\right)^{2}\\&={\tfrac {1}{32}}\left(\left(1+{\sqrt {2}}\right)^{4k}-2+\left(1-{\sqrt {2}}\right)^{4k}\right)\\&={\tfrac {1}{32}}\left(\left(17+12{\sqrt {2}}\right)^{k}-2+\left(17-12{\sqrt {2}}\right)^{k}\right).\end{aligned}}}

<p>The corresponding explicit formulas for 
  
    
      
        
          s
          
            k
          
        
      
    
    {\displaystyle s_{k}}
  
 and 
  
    
      
        
          t
          
            k
          
        
      
    
    {\displaystyle t_{k}}
  
 are:<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a>: 13 
</p>

s
                    
                      k
                    
                  
                
                
                  
                  =
                  
                    
                      
                        
                          
                            (
                            
                              3
                              +
                              2
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            k
                          
                        
                        −
                        
                          
                            (
                            
                              3
                              −
                              2
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            k
                          
                        
                      
                      
                        4
                        
                          
                            2
                          
                        
                      
                    
                  
                  ,
                
              
              
                
                  
                    t
                    
                      k
                    
                  
                
                
                  
                  =
                  
                    
                      
                        
                          
                            (
                            
                              3
                              +
                              2
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            k
                          
                        
                        +
                        
                          
                            (
                            
                              3
                              −
                              2
                              
                                
                                  2
                                
                              
                            
                            )
                          
                          
                            k
                          
                        
                        −
                        2
                      
                      4
                    
                  
                  .
                
              
            
          
        
      
    
    {\displaystyle \displaystyle {\begin{aligned}s_{k}&={\frac {\left(3+2{\sqrt {2}}\right)^{k}-\left(3-2{\sqrt {2}}\right)^{k}}{4{\sqrt {2}}}},\\t_{k}&={\frac {\left(3+2{\sqrt {2}}\right)^{k}+\left(3-2{\sqrt {2}}\right)^{k}-2}{4}}.\end{aligned}}}

<h2 id="recurrence-relations">Recurrence relations</h2>
<p>The solution to the Pell equation can be expressed as a <a href="/facts/Recurrence_relation/JolXC71M">recurrence relation</a> for the equation's solutions. This can be translated into recurrence equations that directly express the square triangular numbers, as well as the sides of the square and triangle involved.  We have<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a>: (12) 
</p>

N
                    
                      k
                    
                  
                
                
                  
                  =
                  34
                  
                    N
                    
                      k
                      −
                      1
                    
                  
                  −
                  
                    N
                    
                      k
                      −
                      2
                    
                  
                  +
                  2
                  ,
                
                
                  
                    with 
                  
                  
                    N
                    
                      0
                    
                  
                
                
                  
                  =
                  0
                  
                     and 
                  
                  
                    N
                    
                      1
                    
                  
                  =
                  1
                  ;
                
              
              
                
                  
                    N
                    
                      k
                    
                  
                
                
                  
                  =
                  
                    
                      (
                      
                        6
                        
                          
                            
                              N
                              
                                k
                                −
                                1
                              
                            
                          
                        
                        −
                        
                          
                            
                              N
                              
                                k
                                −
                                2
                              
                            
                          
                        
                      
                      )
                    
                    
                      2
                    
                  
                  ,
                
                
                  
                    with 
                  
                  
                    N
                    
                      0
                    
                  
                
                
                  
                  =
                  0
                  
                     and 
                  
                  
                    N
                    
                      1
                    
                  
                  =
                  1.
                
              
            
          
        
      
    
    {\displaystyle \displaystyle {\begin{aligned}N_{k}&=34N_{k-1}-N_{k-2}+2,&{\text{with }}N_{0}&=0{\text{ and }}N_{1}=1;\\N_{k}&=\left(6{\sqrt {N_{k-1}}}-{\sqrt {N_{k-2}}}\right)^{2},&{\text{with }}N_{0}&=0{\text{ and }}N_{1}=1.\end{aligned}}}

s
                    
                      k
                    
                  
                
                
                  
                  =
                  6
                  
                    s
                    
                      k
                      −
                      1
                    
                  
                  −
                  
                    s
                    
                      k
                      −
                      2
                    
                  
                  ,
                
                
                  
                    with 
                  
                  
                    s
                    
                      0
                    
                  
                
                
                  
                  =
                  0
                  
                     and 
                  
                  
                    s
                    
                      1
                    
                  
                  =
                  1
                  ;
                
              
              
                
                  
                    t
                    
                      k
                    
                  
                
                
                  
                  =
                  6
                  
                    t
                    
                      k
                      −
                      1
                    
                  
                  −
                  
                    t
                    
                      k
                      −
                      2
                    
                  
                  +
                  2
                  ,
                
                
                  
                    with 
                  
                  
                    t
                    
                      0
                    
                  
                
                
                  
                  =
                  0
                  
                     and 
                  
                  
                    t
                    
                      1
                    
                  
                  =
                  1.
                
              
            
          
        
      
    
    {\displaystyle \displaystyle {\begin{aligned}s_{k}&=6s_{k-1}-s_{k-2},&{\text{with }}s_{0}&=0{\text{ and }}s_{1}=1;\\t_{k}&=6t_{k-1}-t_{k-2}+2,&{\text{with }}t_{0}&=0{\text{ and }}t_{1}=1.\end{aligned}}}

<h2 id="other-characterizations">Other characterizations</h2>
<p>All square triangular numbers have the form 
  
    
      
        
          b
          
            2
          
        
        
          c
          
            2
          
        
      
    
    {\displaystyle b^{2}c^{2}}
  
, where 
  
    
      
        
          
            
              b
              c
            
          
        
      
    
    {\displaystyle {\tfrac {b}{c}}}
  
 is a <a href="/facts/Convergent_(continued_fraction)/Wg1cMPxN">convergent</a> to the <a href="/facts/Simple_continued_fraction/Wg1cMPxN">continued fraction expansion</a> of 
  
    
      
        
          
            2
          
        
      
    
    {\displaystyle {\sqrt {2}}}
  
, the <a href="/facts/Square_root_of_2/SD0sag94">square root of 2</a>.<a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a>
</p><p>A. V. Sylwester gave a short proof that there are infinitely many square triangular numbers: If the 
  
    
      
        n
      
    
    {\displaystyle n}
  
th triangular number 
  
    
      
        
          
            
              
                n
                (
                n
                +
                1
                )
              
              2
            
          
        
      
    
    {\displaystyle {\tfrac {n(n+1)}{2}}}
  
 is square, then so is the larger 
  
    
      
        4
        n
        (
        n
        +
        1
        )
      
    
    {\displaystyle 4n(n+1)}
  
th triangular number, since:
</p>

(
                  
                
                4
                n
                (
                n
                +
                1
                )
                
                  
                    )
                  
                
                
                  
                    (
                  
                
                4
                n
                (
                n
                +
                1
                )
                +
                1
                
                  
                    )
                  
                
              
              2
            
          
          =
          4
          
          
            
              
                n
                (
                n
                +
                1
                )
              
              2
            
          
          
          
            
              (
              
                2
                n
                +
                1
              
              )
            
            
              2
            
          
          .
        
      
    
    {\displaystyle \displaystyle {\frac {{\bigl (}4n(n+1){\bigr )}{\bigl (}4n(n+1)+1{\bigr )}}{2}}=4\,{\frac {n(n+1)}{2}}\,\left(2n+1\right)^{2}.}

<p>The left hand side of this equation is in the form of a triangular number, and as the product of three squares, the right hand side is square.<a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a>
</p><p>The <a href="/facts/Generating_function/K7UVjnjL">generating function</a> for the square triangular numbers is:<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a>
</p>

1
              +
              z
            
            
              (
              1
              −
              z
              )
              
                (
                
                  
                    z
                    
                      2
                    
                  
                  −
                  34
                  z
                  +
                  1
                
                )
              
            
          
        
        =
        1
        +
        36
        z
        +
        1225
        
          z
          
            2
          
        
        +
        ⋯
      
    
    {\displaystyle {\frac {1+z}{(1-z)\left(z^{2}-34z+1\right)}}=1+36z+1225z^{2}+\cdots }

<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Cannonball_problem/CJlRFKU4">Cannonball problem</a>, on numbers that are simultaneously square and square pyramidal</li>
<li><a href="/facts/Sixth_power/j99ENzSr">Sixth power</a>, numbers that are simultaneously square and cubical</li></ul>
<h2 id="notes">Notes</h2>

<h2 id="external-links">External links</h2>
<ul><li><a href="http://www.cut-the-knot.org/do_you_know/triSquare.shtml">Triangular numbers that are also square</a> at <a href="/facts/Cut-the-knot/WLgV43jd">cut-the-knot</a></li>
<li><a href="/facts/Eric_W._Weisstein/FbMuVDep">Weisstein, Eric W.</a> <a href="https://mathworld.wolfram.com/SquareTriangularNumber.html">"Square Triangular Number"</a>. <i><a href="/facts/MathWorld/yc1jodbq">MathWorld</a></i>.</li>
<li><a href="https://opinionator.blogs.nytimes.com/2012/01/04/remembering-michael-dummett/">Michael Dummett's solution</a></li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>Dickson, Leonard Eugene (1999) [1920]. History of the Theory of Numbers. Vol. 2. Providence: American Mathematical Society. p. 16. ISBN 978-0-8218-1935-7. <a href="978-0-8218-1935-7" target="_blank">978-0-8218-1935-7</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
<li id="fn:2"><p>Euler, Leonhard (1813). "Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers)". Mémoires de l'Académie des Sciences de St.-Pétersbourg (in Latin). 4: 3–17. Retrieved 2009-05-11. According to the records, it was presented to the St. Petersburg Academy on May 4, 1778. <a href="/wiki/Leonhard_Euler" target="_blank">/wiki/Leonhard_Euler</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li>
<li id="fn:3"><p>Euler, Leonhard (1813). "Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers)". Mémoires de l'Académie des Sciences de St.-Pétersbourg (in Latin). 4: 3–17. Retrieved 2009-05-11. According to the records, it was presented to the St. Petersburg Academy on May 4, 1778. <a href="/wiki/Leonhard_Euler" target="_blank">/wiki/Leonhard_Euler</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li>
<li id="fn:4"><p>Weisstein, Eric W. "Square Triangular Number". MathWorld. <a href="/wiki/Eric_W._Weisstein" target="_blank">/wiki/Eric_W._Weisstein</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li>
<li id="fn:5"><p>Dickson, Leonard Eugene (1999) [1920]. History of the Theory of Numbers. Vol. 2. Providence: American Mathematical Society. p. 16. ISBN 978-0-8218-1935-7. <a href="978-0-8218-1935-7" target="_blank">978-0-8218-1935-7</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li>
<li id="fn:6"><p>Euler, Leonhard (1813). "Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers)". Mémoires de l'Académie des Sciences de St.-Pétersbourg (in Latin). 4: 3–17. Retrieved 2009-05-11. According to the records, it was presented to the St. Petersburg Academy on May 4, 1778. <a href="/wiki/Leonhard_Euler" target="_blank">/wiki/Leonhard_Euler</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li>
<li id="fn:7"><p>Ball, W. W. Rouse; Coxeter, H. S. M. (1987). Mathematical Recreations and Essays. New York: Dover Publications. p. 59. ISBN 978-0-486-25357-2. <a href="978-0-486-25357-2" target="_blank">978-0-486-25357-2</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li>
<li id="fn:8"><p>Pietenpol, J. L.; Sylwester, A. V.; Just, Erwin; Warten, R. M. (February 1962). "Elementary Problems and Solutions: E 1473, Square Triangular Numbers". American Mathematical Monthly. 69 (2). Mathematical Association of America: 168–169. doi:10.2307/2312558. ISSN 0002-9890. JSTOR 2312558. <a href="/wiki/Doi_(identifier)" target="_blank">/wiki/Doi_(identifier)</a> <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li>
<li id="fn:9"><p>Plouffe, Simon (August 1992). "1031 Generating Functions" (PDF). University of Quebec, Laboratoire de combinatoire et d'informatique mathématique. p. A.129. Archived from the original (PDF) on 2012-08-20. Retrieved 2009-05-11. <a href="/wiki/Simon_Plouffe" target="_blank">/wiki/Simon_Plouffe</a> <a href="#fnref:9" class="footnote-back-ref">↩</a></p></li>
</ol>

Square triangular number open-in-new

Square triangular number