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

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.
