264 Proceedings of Royal Society of Edinburgh. [sess. 
Note on Pairs of Consecutive Integers the Sum of whose 
Squares is an Integral Square. By Thomas Muir, 
LL.D. 
(Read January 21, 1901.) 
(1) The solution of the problem of finding such pairs of integers 
is not a thing of yesterday, as may be seen by consulting Hutton’s 
translation of Ozanam’s Recreations , i. pp. 46-8 (1814). It may 
he enunciated thus : — 
If p r /q r be the r th convergent to J 2, then p r p r+1 and 2q r q r+1 
are consecutive integers , and the sum of their squares is equal to 
(q 2 r + q 2 r+l) 2 0 »'(q 2 r«) 2 (i) 
(2) By introducing the idea of a continuant, — which enables us 
to leave out any direct reference to ^2, —we have the alternative 
form of enunciation : — 
If the continuant (2, 2, 2, ... ) be denoted by a, b, c when the 
number of 2’s is r - 1 , r , r + 1 respectively , then (a + b) (b + c) and 
2bc are consecutive integers , and the sum of their squares is equal to 
the square of b 2 + c 2 . (2) 
(3) Neither of these enunciations, however, indicates which of 
the two consecutive integers is the less ; * and the merit of Mr 
Christie’s enunciation {Math. Gazette , i. p. 394) arises from the 
fact that he has hit upon a general expression for the less of the 
two. The most striking way of putting his result is as follows : — 
The solution of the equation x 2 + (x + l) 2 = y 2 in integers is 
x = 2 0 + 2 l + 2 2 + ... + 2 2r _! ) 
V = Z‘2r > , 
where 2 r is the simple continuant of the r th order whose diagonal 
elements are all 2’s. . . . . . . . (3) 
(4) By way of proof of (2) we note that two properties of con* 
tinuants give ac — b 2 = ±1 and c = 25 + a ; that consequently 
The first is less when r is even, and the second when r is odd. 
