There exist triangular numbers that are also square

Wonder of wonders. I was very pleasantly surprised by a message from Chris Whitrow:

I just stumbled across your site and noticed your page on triangular and
square numbers. I couldn't help dropping this note to point out the curious fact that there is also an infinite set of numbers which are simultaneously both triangular and square. There is a recurrence relation for generating them:

(1)

S(n+1) = 4·S(n)·[8·S(n) + 1]

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

(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 = 35^{2}.

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 T_{n} = n(n + 1)/2 denote the nth triangular number. The solution follows from the observation that if T_{n} is a perfect square, so is T_{4n(n+1)}. Indeed, let n(n + 1)/2 = k^{2}. Then 4n(n + 1) = 8k^{2}. Continue:

T_{4n(n+1)} = T_{8k2}

= 8k^{2}(8k^{2} + 1)/2 = 4k^{2}(8k^{2} + 1)

= 4k^{2}[4n(n + 1) + 1] = 4k^{2}[4n^{2} + 4n + 1]

= 4k^{2}(2n + 1)^{2}.

All such numbers can be found by solving a Pell equation: 8x^{2} + 1 = y^{2} which plays a prominent role in number theory.

Numbers Simultaneously Square and Triangular

8x^{2} + 1 =

y^{2}

Triangular number

Side of triangle

Side of square

8·1^{2}+1 =

3^{2}

1

1

1

8·6^{2}+1 =

17^{2}

36

8

6

8·35^{2}+1 =

99^{2}

1225

49

35

8·204^{2}+1 =

577^{2}

41616

288

204

8·1189^{2}+1 =

3363^{2}

1413721

1681

1189

8·6930^{2}+1 =

19601^{2}

48024900

9800

6930

8·40391^{2}+1 =

114243^{2}

1631432881

57121

40391

All such numbers can be obtained by squaring the members of the sequence generated from the recurrence relation