Square Pyramidal Number

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A figurate number of the form

 P_n^((4))=1/6n(n+1)(2n+1),
(1)

corresponding to a configuration of points which form a square pyramid, is called a square pyramidal number (or sometimes, simply a pyramidal number). The first few are 1, 5, 14, 30, 55, 91, 140, 204, ... (OEIS A000330). The generating function for square pyramidal numbers is

 (x(x+1))/((x-1)^4)=x+5x^2+14x^3+30x^4+....
(2)

The square pyramidal numbers are sums of consecutive pairs of tetrahedral numbers and satisfy

 P_n=1/3(2n+1)T_n,
(3)

where T_n is the nth triangular number.

The only numbers which are simultaneously square S_m=m^2 and square pyramidal P_n=n(n+1)(2n+1)/6 (the cannonball problem) are P_1=1 and P_(24)=4900, corresponding to S_1=1 and S_(70)=4900 (Ball and Coxeter 1987, p. 59; Ogilvy 1988; Dickson 2005, p. 25), as conjectured by Lucas (1875), partially proved by Moret-Blanc (1876) and Lucas (1877), and proved by Watson (1918). The problem requires solving the Diophantine equation

 m^2=1/6n(n+1)(2n+1)
(4)

(Guy 1994, p. 147). Watson (1918) gave an almost elementary proof, disposing of most cases by elementary means, but resorting to the use of elliptic functions for one pesky case. Entirely elementary proofs have been given by Ma (1985) and Anglin (1990).

Numbers which are simultaneously triangular T_m=m(m+1)/2 and square pyramidal P_n=n(n+1)(2n+1)/6 satisfy the Diophantine equation

 1/2m(m+1)=1/6n(n+1)(2n+1).
(5)

Completing the square gives

 1/2(m+1/2)^2-1/8=1/6(2n^3+3n^2+n)
(6)
 1/8(2m+1)^2=1/6(2n^3+3n^2+n)+1/8
(7)
 3(2m+1)^2=8n^3+12n^2+4n+3.
(8)

The only solutions are (n,m)=(-1,0), (0, 0), (1, 1), (5, 10), (6, 13), and (85, 645) (Guy 1994, p. 147), corresponding to the nontrivial triangular square pyramidal numbers 1, 55, 91, 208335.

Numbers which are simultaneously tetrahedral Te_m=m(m+1)(m+2)/6 and square pyramidal P_n=n(n+1)(2n+1)/6 satisfy the Diophantine equation

 m(m+1)(m+2)=n(n+1)(2n+1).
(9)

Beukers (1988) has studied the problem of finding solutions via integral points on an elliptic curve and found that the only solution is the trivial Te_1=P_1=1.

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