Sequential sums $1+2+\cdots+N$ that are squares While playing with sums $S_n = 1+\cdots+n$ of integers,
I have just come across  some "mathematical magic"
I have no explanation and no proof for.
Maybe you can give me some comments on this:
I had the computer calculating which Sn are squares,
and it came up with the following list:  
Table
row $N$        sum($1+\cdots+N$)   M (square root of sum)
r=1     N=1     sum=1       M=1
r=2 N=8         sum=36      M=6
r=3 N=49        sum=1225        M=35
r=4 N=288       sum=41616       M=204
r=5 N=1681  sum=1413721     M=1189
r=6     N=9800  sum=48024900    M=6930
Of course we have $1+\cdots+N = \frac{N(N+1)}{2}$,
but this gives no indication for which N the sum $1+\cdots+N$  is a square.
Can you guess how in this table we can calculate the entries in row 2 from the entries in row 1?
Or the entries in row 3 from the entries in row 2? 
Or the entries in row 4 from the entries in row 3? 
Or the entries in row 5 from the entries in row 4? 
I looked at the above table and made some strange observations: 


*

*The value of the next M can be easily calculated from the previous entries:
 Take the M from the previous row, multiply by 6 and subtract the M from two rows higher up.
         $M(r) = 6*M(r-1)–M(r-2)$
 How is this possible?
The S(r)  we  calculate as $S(r) = M(r)^2$. Note that we do not know whether this newly constructed
 number $S_r$ is in fact of the type $1+\cdots+k$ for some $k$. 

*The value of  the next N can be calculated as 
          N(r) = Floor($M(r)*\sqrt 2$), 
where Floor means “rounding down  to the next lower integer“.
Somewhat surprising, $S(r)$ is the sum $1+\cdots+N(r)$  !

*It looks as if outside the entries in the above table there are no other cases.
With other words, the method  $M(r) = 6*M(r-1)–M(r-2)$
seems to generate ALL solutions n where the sum $1+\cdots+n$ is a square. 
Problems:
Is there a proof for any of the three observations? 
Do observations 1 and 2 really work for the infinite number of rows in this table?
Is there an infinite number of rows in the first place?
Puzzled, 
Karl
 A: I see. Nobody answered this the way I would have... Taking $u = 2 n+1,$ we are solving $$ u^2 - 8 m^2 = 1. $$ A beginning solution is $(3,1).$  Given a solution $(u,m),$ we get a new one $$ (3 u + 8 m, u + 3 m).  $$ Then $n = (u-1)/2$ for each pair.
So, with $n^2 + n = 2 m^2$ and $u = 2 n + 1,$ we get triples
$$ (n,u,m)   $$
$$ (1,3,1)   $$
$$ (8,17,6)   $$
$$ (49,99,35)   $$
$$ (288,577,204)   $$
$$ (1681,3363,1189)   $$
$$ (9800,19601,6930)   $$
$$ (57121,114243,40391)   $$
$$ (332928,665857,235416)   $$
$$ (1940449,3880899,1372105),   $$
With my letters, each is a similar sequence, let us use $r$ for "row,"
$$  m_1 = 1, m_2 = 6, \; \; m_{r+2} = 6m_{r+1} - m_r,  $$
$$  u_1 = 3, u_2 = 17, \; \; u_{r+2} = 6u_{r+1} - u_r,  $$
$$  n_1 = 1, n_2 = 8, \; \; n_{r+2} = 6n_{r+1} - n_r + 2.  $$
A: This problem shows up often when working with Pythagorean triangles with consecutive sides. (3,4,5), (20,21,29) etc.
I will answer half the question and the other half is similar. 
$$\frac{N (N+1)}{2}$$ will be perfect square only when
$N$ is square and $(N+1)/2$ is a square or $N/2$ is a square and $N+1$ is a square (this is true since $N$ and $N+1$ are co-prime).
Now consider the first case.  $N$ is necessarily odd; so
$$ N = (2k+1)^2 = 4k^2+4k+1$$
and 
$$ \frac{N+1}{2} = 2k^2 + 2k + 1 = k^2 + (k+1)^2$$
Hence we need
$$ k^2 + (k+1)^2 = m^2$$
The solution is just Pythagorean triangle with consecutive sides. Refer to any elementary number theory book for finding these.
