266 Proceedings of Royal Society of Edinburgh. [sess. 
Similarly, by adding 2 4s+2 + 2 4s+3 , we shall obtain 
2 0 + 2 4 + 2 2 + . . . + 2^+3 = (2 2s+2 + 2 2 s+ 1 )(2 2h _ 1 + 2 2s ) - 1 , 
which we know otherwise (§ 3, footnote) equals 2*2 2s+2 *2 2s+1 . 
It is thus clear that if the proposition hold for any particular 
case where r is even, it must hold for the next two cases, and 
therefore for the next two, and so on ; and as its validity for the 
case r = 2 is readily verified, the proposition may be considered to 
be established. 
(7) When we have got one instance of an integer whose square, 
together with the square of the next higher integer, gives an 
integral square, there is a very simple means of getting the next 
instance. The theorem is : — 
If & be an integer such that a 2 + (a+ l) 2 = z 2 , where z is integral , 
then 3a + 1 + 2z is the next integer of this kind. 
To establish this we have to show that 
3(2 0 + 2i + 2 2 + . . . +2 2r _ 1 ) + 1 + 2*2 2r = 2 0 + 2 1 + 2 2 + . . . +2 2r+1 , 
that is, that 
2(2 0 + 2 1 + 2 2 + . . . +2 2r _i) + 1 + 2 2r = 2 2r+1 . 
How, if we know one case of this to be true, we can immediately 
prove the next case ; for, suppose that 
2(2 0 + 2! + 2 2 + . . . + 2 2m _ 4 ) + 1 + 2 2m = 2 2m+1 ; 
then by adding 2 2m + 2 2m+1 we obtain 
2(2 0 + 2 1 + 2 2 + . . . +2 2m ) + 1 + 2 2m+1 = 2*2 2m+1 + 2 m = 2 2m+2 . 
It remains only to show that it is true when m=l, and this is 
self-evident. 
(8) From the foregoing we have 
2 0 + 2 1 + 2 2 + . . . +2 2r _! = J(2 2r+1 - 2 2r - 1) , 
= J(2 2r + 2 2r-1 - 1) ; 
and we are thus led to the theorem : — 
The solution of the equation x 2 + (x + l) 2 = y 2 in integers is 
