In the prior post on this topic, I showed that S(n)=6S(n-1) - S(n-2) and that
Q(n) = 6Q(n-1)-Q(n-2) + 2. It can further be shown that for S(1) = Sqrt(a),
S(2) = Sqrt(b) that S(3) = 6S(2) - S(1) = Sqrt(34b - a + 2*[(S(1)^2) -6S(1)*S(2)
+ (S(2)^2)]). The term within the brackets, S(1)^2 -6S(1)*S(2) + S(2)^2, is a
constant, K, for all a,b, as adjacent terms in the series S(n) = 6S(n-1)-S(n-2)

--- In Triangular_and_Fibonacci_Numbers@yahoogroups.com, "ramsey2879"
<ramseykk2@...> wrote:
>
> Generalized Proof that an infinite number of square triangular
> numbers exist.
>
> I prove below that in general that if n^2 – T(q) = K, there are an
> infinite number of distinct pairs n` and q` such that (n`)^2 –T(q) =
> K. First a little history of the problem.
> It is widely recognized that in general, any odd square minus 1,
> divided by 8 is a triangular number, i.e. (2n +1)^2 – 1 = 8* T(n).
> Thus to find square triangular numbers can be reduced to solving the
> Pell equation 8n^2 + 1 = m^2. Solutions along this line are given
> at http://www.cut-the-knot.org/do_you_know/triSquare.shtml ,
> http://mathpages.com/home/kmath159.htm , and
> http://mathforum.org/library/drmath/view/55927.html . It has been
> shown in these sites that if S_n is the series of numbers that are
> squared and Q_n is the corresponding series such that (S_n)^2 = T
> (Q_n) then S_n = 6*S_(n-1) – S_(n-2) and Q_n = 6*Q_(n-1) – Q_(n-2) +2.
> However, these solutions are quite involved and are not capable of
> being generalized to other cases where (S_n)^2 – T(Q_n) <> 0. A
> related web site http://www.cut-the-
> knot.org/do_you_know/triSquare2.shtml takes advantage of this
> solution and a relationship between the two series S_n) and Q_n to
> give two recursive formulas for finding other values of Q_n and S_n
> when a single value of S_n and a corresponding value of Q_n are
> known. This too is restricted to the case K = 0. The purpose of this
> paper is to prove the infinity of solutions more general expression,
> including K <> 0, i.e. if there is one solution to n^2 – T(q) = K for
> the integers n and q, then there are an infinite number of such
> solutions as well as to show how the recursive formulas also work for
> complex as well as real numbers, and to show other related results.
>
> First I developed two algebraic identities that I got by studying the
> differences between squares and triangular numbers and noting the
> patterns. They are:
>
> 1) (a-b)^2 = T(2a-b) +T(b) – 2*T(a)
> 2) 2*T(a) – T(b) = 2*T(a_1) – T(b_1) where a_1 = 3a – 2b and b_1
> = 3b – 4a –1.
>
Where T(b) = 2*T(a), (a-b)^2 are the square triangular numbers.
> By manipulation of the above identities, generalized expressions for
> the recursive series S_n and Q_n can be found. The resulting initial
> conditions are, S_0 = a +b +1, S_1 = -a+b, Q_0 = 2*a+b+1, and Q_1
> = 2*a -b (where a and b can be any number including a complex number
> or a radical). Regardless of the values for a and b, the resulting
> values of S_n and Q_n solve the equation (S_n)^2 = T_(Q_n) – K(a,b)
> for all n where K is a constant that is equal to 2*T_a – T_b. This
> is true even for complex numbers "a" and "b".
>
> Proof
> The reader may easily verify identities 1) and 2) by expansion and
> collection of like terms. Identity 2 gives the recursive relation as
> follows:
> a_0 = a
> b_0 = b
>
> a_1 = 3a-2b
> b_1 = 3b-4a-1
>
> a_2 = 3*(a_1) – 2*(b_1)
> = 3*(3a-2b) – 2*(3b-4a-1)
> = 17a – 12b +2
> = 6*a_1 – a_0 + 2
>
> b_2 = 3*(b_1) – 4*(a_1) – 1
> = 3*(3b-4a-1) –4*(3a-2b) –1
> = 17b – 24a – 4
> = 6*b_1 – b_0 + 2
>
> Since the same relationships above are calculated regardless of the
> initial choices of a and b, a_n = 6*a_(n-1) – a_(n-2) +2 and b_n
> = 6*b_(n-1) – b_(n-2) +2 for all n. The corresponding series of a_n
> and b_n can be worked both backward and forward infinitely.
> Although, a single series set exists for K = 0, 1, 2 and 5, this is
> not so for the case of K = 6, 20, etc. For instance for the case of K
> = 6 there are two sets of distinct series for instance, with a_0 = 2,
> b_0 = 0 and the other starting with a_0 = b_0 = 3. Certain values of
> K, e.g. 4, 7, 8, 13, 16 etc., have no solution to K = 2*T(a) – T(b)
> in integers.
>
> For identity 1, we let the value of T(b) – 2*T(a) = -K. By identity
> 2, T(b_n) – 2 * T(a_n) = -K for all n. Since identity 1 works for
> all a and b we can make the substitution b_n` = -(b_n+1)
> without changing the value of K since T(b) = T(-b-1). Thus
>
> (a_n + b_n +1)^2 = T(2*a_n + b_n + 1) - K for all n
>
> S_n = a_n + b_n +1
> S_0 = a+b+1
> S_1 = -a + b
> S_2 = -7a + 5b –1
> S_n = 6*S_(n-1) – S_(n-2) for all n > 1
>
> Q_n = 2a_n + b_n +1
> Q_0 = 2a + b +1
> Q_1 = 2a – b
> Q_2 = 10a-7b +1
> Q_n = 6*Q_(n-1) – Q_(n-2) +2 for all n > 1
>
> Again, this works for all a and b, both real or complex. If we know
> S_0 and S_1, the solution of the resulting simultaneous equations
> gives:
> a = (S_0 – S_1 –1)/2 and b = (S_0 + S_1-1)/2. For instance the first
> two square triangular numbers, not counting 0, are 1 and 36. Thus
> S_0 = +/- 1 and S_1 = +/- 6, which gives two basic solutions for Q_n
> and K. That is K can be 0 or 9 in this case since 1 = T_1-0 and 36 =
> T_8-0; and since 1 = T_4 – 9 and 36 = T_9 – 9. The other two
> solutions simply give S_n = - S_n and T_n = T_(-n-1) etc.
>