CHAP. VI] SUM OF TWO SQUARES. 233 



that 4C 2 + 1 R 2 must be a square. The problem thus reduces to finding 

 two squares with the sum 4C 2 + 1> treated in the first case. 



The methods employed to express A as a US or ED are no better than the 

 similar one of subtracting from A in turn squares, or sums of two squares, 

 and ascertaining if the remainder is a El. 



Kausler 36 extended his table of pronic numbers to Q = 1000, and gave 

 their halves and quarters, and applied them as in the former paper. Given 

 A = a 2 -f- b 2 , to solve x 2 + ?/ 2 = A, set x = a + 2ma, y = 2na b. Then 

 a = (nb ma)/(m? + n 2 ) is to be integral. Let m, n be relatively prime. 

 Then b = an -f flm, where = (am + a)/n is an integer. The latter gives 

 n = pa + /ax, m = qa + /* where p/q is a convergent to a/0. Then the 

 former gives a relation between a, 0, p, <?, /z which is not solved. 



C. F. Gauss 37 applied the theory of binary quadratic forms to prove that 

 every prime 4n + 1 is a CH in a single way. In a foot-note he considered 

 M = 2^Sa a b ft - , where a, b, are distinct prunes of the form 4n + 1, 

 and S is the product of all the prime factors 4n + 3 of M. If S is not a 

 square, M is not a GO . It is stated that, if $ is a square, there are 



decompositions of M into a sum of two squares, when one of the exponents 

 a, /3, is odd; but A; + ^ if a, /?, are all even. Here the squares and 

 not their roots are counted. 



A. M. Legendre 38 had already given the last result. 



Legendre 39 developed Vp into a continued fraction with the 



quotients a a /? n v /3 a 2a - -, 



la m m fo f 



convergents XT ' * ' _... -, 



01 no n g Q g 



where / 2 pg* = 1. Then by use of the convergents corresponding 

 to 



f m(n/no) , 2 



- = - ; , " - , / = mn + m n , g = n 2 + n\. 

 g n(n/n Q ) + n Q 



Substituting these values into / 2 pg~ = (mn Q m n) 2 , we get 



m 2 pn z = (ml pnl). 



But if (Vp + 7 )/Do and (Vp + -^)/^ are the complete quotients corre- 

 sponding to mo/no, mfn, then 



m 2 pn 2 = (mn Q m ri)D, ml pnl = (mn m ri)D . 

 Hence A> = D, so that DD Q + P = p gives p = D 2 + P. _ 



M Nova Acta Acad. Petrop., 14, ad annos 1797-8 (1805), 232-267. 

 " Disquisitiones Arith., 1801, Art. 182; Werke, I, 1863, 159-163. 

 '"ThSorie des nombres, 1798, p. 293; ed. 3, 1830, I, 314 (transl. by Maser, I, 309). 

 Th6orie des nombres, ed. 2, 1808, 59-60; ed. 3, 1830, I, 70-1. (Maser, I, 71-73). Cf. 

 Dirichlet, 88 83, long footnote. Cf. Euler 72 (end), of Ch. XII. 



