Starting with S(1) = 1, S(2) = 36, and so on. Proof is fairly
straightforward. Well, it seems curious that a thing can be both a
triangle and a square, doesn't it?
Yes, of course. How can something be square and triangular simultaneously?
Well, I had to look through several books in my library till I discovered a discussion of these numbers in A. H. Beiler's Recreations in the Theory of Numbers. Beiler writes: "Eight triangles increased by unity produce a square," because
8·r(r + 1)/2 + 1 = 4r2 + 4r + 1 = (2r + 1)2.
(2) proves the converse as well: a square decreased by unity gives eight triangles.
This simple rule explains the recurrence relation (1). However, (1) does not enumerate all such numbers. From (1), we get S(3) = 41616, whereas there is another number, 1225, that is both triangular and square: 1225 = 49·50/2 = 352.
The question was in fact posted as the problem E1473 (AMM, 1962, p. 168) by J.L. Pietepol and solved by A.V. Sylvester [Mathematical Morsels, p. 145, Mathematical Quickies, #186]. Let Tn = n(n + 1)/2 denote the nth triangular number. The solution follows from the observation that if Tn is a perfect square, so is T4n(n+1). Indeed, let n(n + 1)/2 = k2. Then 4n(n + 1) = 8k2. Continue:
|T4n(n+1) = T8k2
|| = 8k2(8k2 + 1)/2 = 4k2(8k2 + 1)|
| = 4k2[4n(n + 1) + 1] = 4k2[4n2 + 4n + 1]|
| = 4k2(2n + 1)2.|
All such numbers can be found by solving a Pell equation: 8x2 + 1 = y2 which plays a prominent role in number theory.
|Numbers Simultaneously Square and Triangular|
|8x2 + 1 = ||y2||Triangular number||Side of triangle||Side of square|
|8·12+1 = ||32||1||1||1|
|8·62+1 = ||172||36||8||6|
|8·352+1 = ||992||1225||49||35|
|8·2042+1 = ||5772||41616||288||204|
|8·11892+1 = ||33632||1413721||1681||1189|
|8·69302+1 = ||196012||48024900||9800||6930|
|8·403912+1 = ||1142432||1631432881||57121||40391|
All such numbers can be obtained by squaring the members of the sequence generated from the recurrence relation
un = 6un-1 - un-2, n > 1, u0 = 0, u1 = 1,
There is also an explicit formula
un = (((1 + √2)2n - (1 - √2)2n)/(42))2
which can also be written as [Wells, p. 93]
un = [(17 + 12√2)n + (17 - 12√2)n - 2]/32.
([Wells, p. 93] also mentions that no triangular number may be a cube, or fourth or fifth power.)
Armando Guarnaschelli from Argentina found a very simple recurrence relation that generates all triangular numbers that are also square.
References I came across are listed below.
- A. H. Beiler's Recreations in the Theory of Numbers, Dover, 1966
- J. H. Conway, R. K. Guy, The Book of Numbers, Springer, 1996
- R. Honsberger's Mathematical Morsels, MAA, 1978
- G. M. Phillips, Mathematics Is Not a Spectator Sport, Springer, 2005
- C. W. Trigg, Mathematical Quickies, Dover, 1985
- D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987
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Copyright © 1996-2017 Alexander Bogomolny