Square triangular number
In mathematics, a square triangular number (or triangular square number) is a number which is both a triangular number and a perfect square. There are infinitely many square triangular numbers; the first few are 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in the OEIS).
Contents
Explicit formulas
Write N_{k} for the kth square triangular number, and write s_{k} and t_{k} for the sides of the corresponding square and triangle, so that
Define the triangular root of a triangular number to be . From this definition and the quadratic formula, Therefore, is triangular if and only if is square. Consequently, a number is square and triangular if and only if is square, i. e., there are numbers and such that . This is an instance of the Pell equation, with . All Pell equations have the trivial solution (1,0), for any n; this solution is called the zeroth, and indexed as . If denotes the k'th nontrivial solution to any Pell equation for a particular n, it can be shown by the method of descent that and . Hence there are an infinity of solutions to any Pell equation for which there is one nontrivial one, which holds whenever n is not a square. The first nontrivial solution when n=8 is easy to find: it is (3,1). A solution to the Pell equation for n=8 yields a square triangular number and its square and triangular roots as follows: and Hence, the first square triangular number, derived from (3,1), is 1, and the next, derived from (17,6) (=6×(3,1)(1,0)), is 36.
The sequences N_{k}, s_{k} and t_{k} are the OEIS sequences A001110, A001109, and A001108 respectively.
In 1778 Leonhard Euler determined the explicit formula^{[1]}^{[2]}^{:12–13}
Other equivalent formulas (obtained by expanding this formula) that may be convenient include
The corresponding explicit formulas for s_{k} and t_{k} are ^{[2]}^{:13}
and
Pell's equation
The problem of finding square triangular numbers reduces to Pell's equation in the following way.^{[3]} Every triangular number is of the form t(t + 1)/2. Therefore we seek integers t, s such that
With a bit of algebra this becomes
and then letting x = 2t + 1 and y = 2s, we get the Diophantine equation
which is an instance of Pell's equation. This particular equation is solved by the Pell numbers P_{k} as^{[4]}
and therefore all solutions are given by
There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers.
Recurrence relations
There are recurrence relations for the square triangular numbers, as well as for the sides of the square and triangle involved. We have^{[5]}^{:(12)}
We have^{[1]}^{[2]}^{:13}
Other characterizations
All square triangular numbers have the form b^{2}c^{2}, where b / c is a convergent to the continued fraction for the square root of 2.^{[6]}
A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers, to wit:^{[7]}
If the triangular number n(n+1)/2 is square, then so is the larger triangular number
We know this result has to be a square, because it is a product of three squares: 2^2 (by the exponent), (n(n+1))/2 (the n'th triangular number, by proof assumption), and the (2n+1)^2 (by the exponent). The product of any numbers that are squares is naturally going to result in another square. This can be seen from the fact that a necessary and sufficient condition for a number to be square is that there should be only even powers of primes in its prime factorisation, and multiplying two square numbers preserves this property in the product.
The triangular roots are alternately simultaneously one less than a square and twice a square, if k is even, and simultaneously a square and one less than twice a square, if k is odd. Thus, , and In each case, the two square roots involved multiply to give and ^{[citation needed]}
and In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.^{[citation needed]}
The generating function for the square triangular numbers is:^{[8]}
Numerical data
As becomes larger, the ratio approaches and the ratio of successive square triangular numbers approaches . The table below shows values of between 0 and 11, which comprehend all square triangular numbers up to .
See also
 Cannonball problem on numbers that are simultaneously square and square pyramidal
 Sixth power, numbers that are simultaneously square and cubical
Notes
 ^ ^{a} ^{b} Dickson, Leonard Eugene (1999) [1920]. History of the Theory of Numbers. 2. Providence: American Mathematical Society. p. 16. ISBN 9780821819357.

^ ^{a} ^{b} ^{c}
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 20090511.
According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.
 ^ Barbeau, Edward (2003). Pell's Equation. Problem Books in Mathematics. New York: Springer. pp. 16–17. ISBN 9780387955292. Retrieved 20090510.

^
Hardy, G. H.; Wright, E. M. (1979). An Introduction to the Theory of Numbers (5th ed.). Oxford University Press. p. 210. ISBN 0198531710.
Theorem 244
 ^ Weisstein, Eric W. "Square Triangular Number". MathWorld.
 ^ Ball, W. W. Rouse; Coxeter, H. S. M. (1987). Mathematical Recreations and Essays. New York: Dover Publications. p. 59. ISBN 9780486253572.
 ^ Pietenpol, J. L.; A. V. Sylwester; Erwin Just; R. M Warten (February 1962). "Elementary Problems and Solutions: E 1473, Square Triangular Numbers". American Mathematical Monthly. Mathematical Association of America. 69 (2): 168–169. doi:10.2307/2312558. ISSN 00029890. JSTOR 2312558.
 ^ Plouffe, Simon (August 1992). "1031 Generating Functions" (PDF). University of Quebec, Laboratoire de combinatoire et d'informatique mathématique. p. A.129. Retrieved 20090511.
External links
 Triangular numbers that are also square at cuttheknot
 Weisstein, Eric W. "Square Triangular Number". MathWorld.
 Michael Dummett's solution