284 HISTORY OF THE THEORY OF NUMBERS. [CHAP, vm 



impossible unless K is a 02 . For K = a 2 + 6 2 + c 2 , the equation becomes 



Hh(an + r) = Ff(q + bn) + Gg(p + en), 

 H = an - r p = q - bn G = P ~ cn 



h f g 



It is stated that H = F = G, and that the linear equation in n, r, derived 

 by eliminating p, q requires n = / 2 + <7 2 + ^ 2 , whence 



P = (/ 2 - 2 + #)c + 20/Ki - 2/76, 



<? = (- / 2 + 9* + W + 2Aa 

 r = (/2 + 02 _ p) a _j_ 2 fhb 



[For these, n 2 (a 2 + 6 2 + c 2 ) = p 2 + q* + r 2 , identicaUy in /, g, h, a, b, c.] 

 There is a similar treatment of the corresponding problem for 4 or 5 squares. 

 If Malfatti had proved his statement that K must be a sum of the like 

 number of squares, he could have deduced Bachet's theorem from Euler's 8 

 result that every integer is a sum of four rational squares. 



P. Barlow 20 gave a " simplification of Legendre's 15 proof." To show that 

 any prime A divides a sum of SI, he proved at length that x 2 + w 2 1 = mA 

 is solvable [evidently by x = 1, w = 0!] and stated that a like proof shows 

 that y 2 + z 2 + 1 = nA is solvable. The proof probably meant for the 

 latter is as follows. If p = y* (mod A), either (p + 1) is a quadratic 

 residue (= 2 2 ) and the result follows, or it is a non-residue and hence p + 1 

 a residue, since 1 is a non-residue (otherwise our equation holds for 

 y 0). But p, p + 1, p + 2, are not all residues. The proof is thus 

 only a slight modification of that by Euler. 10 



A. Cauchy's proof in 1813 of Fermat's theorem on 3 triangular numbers, 

 4 squares, 5 pentagons, etc., was considered in Ch. I. It is in place to 

 mention here the theorems on sums of squares upon which his proof rests, 

 especially since special cases were cited above from the correspondence 

 of Euler and Goldbach. If 



(9) k = t z + v? + y 2 + w*, s = t + u + v + w, 

 then 



(10) 4k - s 2 = (t + u - v - wY + (t - u + v - w) 2 + (t - u - v + w) z . 



But if 4 a is the highest power of 4 dividing a, then a is a 02 if and only if 

 a/4 a is not of the form 8n + 7. If & is even, the three sums in (10) are 

 even, so that k s 2 /4 is a S] . By (9), k = s (mod 2). Cauchy proved that, 

 if k is even, sufficient conditions for (9) are that s be even and between 

 V3AT ^1 and VSfe, and k-- s 2 /4 =}= 4(8n + 7). With the exception of 



s > "vSA l, these were seen above to be necessary conditions. For k 

 odd, sufficient conditions for (9) are that s be odd and between ^3k 2 1 

 and V4AJ; there exists such an s for any k. As to the former case, he proved 

 that for any k there exists an integer between ^3k and Vi/b and congruent 

 to k modulo 2 except when k = 1, 5, 9, 11, 17, 19, 29, 41, 2, 6, 8, 14, 22, 

 24, 34. 



80 New Series of Math. Repository (ed., Leybourn), 2, 1809, II, 70; Theory of Numbers, 

 London, 1811, 212. 



