in the funii aw- + Inf + cz- + da- 1'7 



(6-4) /i= 3 (mod 4). 



There is only a finite number of exceptions. Take 

 iV^=4M8/A + 7). 

 If X^l, take ti=l. Then 



N — nil- = 1 or 5 (mod 8). 

 If X = 0, take n = 2. Then 



N - nu? = 3 (mod 8). 

 In either case the proof is completed as before. 



In order to determine precisely which are the exceptional 

 numbers, we must consider more particularly the numbei'S between 

 n and 4» for which X = 0. For these \i must be 1, and 



N -nu-= (mod 4). 

 But the numbers which are multiples of 4 and which cannot be 

 expressed in the form .x- + y ' + z' are the numbers 



4''(8i. + 7), (/c = l, 2, 3, ..., v^O,l, 2, 3, ...)• 

 The exceptions required are therefore those of the numbers 



n + ¥{^v + 1) (6-41) 



which lie between n and 4» and are of the form 



8;i + 7 (6-42). 



Now in order that (6'41) may be of the form (6'42), k must be 

 1 if 11 is of the form 8A- 4- 3 and k may have any of the values 

 2, 3, 4, ... if n is of the form 8A;+7. Thus the only numbers 

 which cannot be expressed in the form (5'2), in this case, are those 

 of the form 4^ (8/i + 7) less than n and those of the form 



?i + 4''(8i/+7), (y-0, 1, 2, 3, ...), 



lying between n and 4?i, where k=\ if n is of the form 8A; + 3, 

 and K>\ if ?r is of the form 8A; + 7. 



(6-5) n = 1 (mod 8). 



In this case we have to prove that 



(i) if n ^ 33, there is an infinity of integers which cannot be 



expressed in the form (5"2) ; 

 (ii) if n is 1, 9, 17, or 25, there is only a finite number of 

 exceptions. 

 In order to prove (i) suppose that iV = 7 . 4^. Then obviously 

 u cannot be zero. But if u is not zero n^ is always of the form 

 4''(8t/+l). Hence 



N - nu^ = 7 . 4^ - ?i . 4" (8v + 1). 



Since n ^33, X must be greater than or equal to k + 2, to ensure 

 that the right-hand side shall not be negative. Hence 



N - jui^ = ^^ (Sk + 7), 



VOL. XIX. PT. I. 2 



