CHAP. VI] SUM OF TWO SQUARES. 245 



the series being continued as long as x ri 2 is positive; if x is a square, 

 s(0) is replaced by x/2. The proof is by use of the series for 



Q = (1 - ff)(l - <? 2 )(1 - <? 3 ) = (1 + ?)(! + <? 2 ) (l-2g+2g-2g+ - - .). 



Hence if x is not a square and no x n 2 is a square, sfc) is a multiple of 4. 

 Thus s(x) is a multiple of 4 when a; is not a square or a El. If also z is a 

 prune, x is of the form 4m 1, since s(x) = x + 1. Hence every prime 

 not a El is of the form 4m 1, so that every prime 4m + 1 is a El. 



S. Re'alis 105 proved that every prime 4n + 1 is the quotient of x 2 + y 2 

 by the common factor of x z and ?/ 2 , where 



x = 2 + /5 2 - 7 2 , y = (7 - ) 2 + (7 - $ 2 - 7 2 - 

 For the latter values and 



u = a 2 + (a - 7) 2 ~ (a - 0) 2 , w = /3 2 + (/S - 7)' - (0 - or) 2 , 



we have 2 + y 2 = w 2 + v 2 , identically, and they furnish all the solutions. 



E. Lucas 106 proved that every prime 4k + 1 is a E] by use of " satins " 

 n a formed of the points (x, y) with x = 0, 1, , n such that y is the residue 

 of ax modulo n where a is prune to n and a < n. Since each parallel to 

 the ?/-axis contains one and but one point of the satin, ax = 1 (mod ri) 

 has a unique solution. If / 2 + 1 = is solvable, y = fx gives fy=f 2 x^ x, 

 and the satin n/ is unaltered by a rotation through a right angle and is a 

 square satin. If n is a prime p = 4k + 1, we can separate 2, 3, , p 2 

 into (p 5)/4 sets of four numbers like a, a, p a, p a, where aa. = 1 

 (mod p), and one set p, p f, such that /(p / ) = 1, whence/ 2 +1 = 

 is solvable. Th\is p divides a sum of two squares. Since the satin is 

 formed of squares having p as a side, p. is a sum of two squares. 



T. Harmuth 107 proved that every prime p = 4n + 1 divides a sum of 

 two relatively prune squares. Let g be an odd primitive root of p and 

 set f- = 2 (mod p). Then g ze + 2 2 = (mod p),e = \ + (p - l)/4. 



S. Gunther 108 proved (1) by use of lattice (gitter) points. No three 

 lattice points are vertices of a regular triangle. The geometrical proof by 

 Lucas shows that 



x 2 + ?/ 2 = u? + v z = 2(ux + vy) 



have no rational solutions. If a 2 is a E] , a is a E] . 



For the knight's path problem in chess, we have (pp. 14-16) the system 

 of equations 



(Xi - x i+l }* + (y, - 7/ i+ 2 = 5 (i = 1, 2, . . ., n 2 - 1), 



and, if the path is closed, also 



+ (ym - yi) 2 = 5. 



105 Nouv. Ann. Math., (2), 18, 1879, 500-4. 



106 L'Ingegnere Civile, Turin, 1880; French transl., Aseoc. franj}., 40, 1911, 72-87. Cf. 



A. Aubry, 1'enseignement math., 13, 1911, 200; Sphinx-Oedipe, num6ro special, Jan., 

 1912, 10-13. 



107 Archiv Math. Phys., 66, 1881, 327-8. 



108 Zeitschrift Math. Naturw. Unterricht, 13, 1882, 94-98, 102. 



