472 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xvi 



and meets the curve at a new point M i whose coordinates are rational and 

 easily found. Thus if we employ Leonardo's solution x = 41/12, y = 49/12, 

 z = 31/12 when a = 5, b = 5, we obtain in succession an infinitude of rational 

 solutions. Or we may find the points with rational coordinates, on the 

 hyperbola y 2 x~ = a, by setting x+y = u, y x = a/u, and identifying their 

 abscissas with those of the analogous points x=(v* b}f(2v), y= (v 2 +6)/(2y) 

 on z 2 x z = b, obtaining the condition uv(u v)=av bu. The tangent at 

 a rational point (u, v] meets the cubic at a new rational point. Finally, 

 x 2 -\-a = y 2 , x' 1 a = z- have the solutions y, z = x (cos 0sin 6); x 2 = a/sin 26; 

 hence the rational solutions are given by those rational values of tan 6/2 for 

 which sin 26 is a rational square; there are none if a = 1 or 2. 



For congruent numbers of order n, see papers 200-1, 210, 222 of Ch. 

 XXIII, and 320 of Ch. XXII. 



CONCORDANT FORMS: x~+my z , x*+ny* BOTH SQUARES. 



RELATED PROBLEMS. 



Diophantus, II, 15, required x, m, n such that x 2 +m and x 2 +n are 

 squares, given the sum m+n. He took m = 4o:+4, n = 6;r-f9, ?7i+n = 20, 

 whence z = 7/10. In II, 16, (x+2) 2 m and (z+2) 2 n are squares if 

 ra = 4#+4, n = 2x+3; for m+n = 20, x = 13/6. The same problems occur 

 in III, 23, 24. 



Diophantus, II, 17, required x, m, n such that x^-\-m and x 2 -\-n are 

 squares, given the ratio m/n. He took x = 3, m/n = 3, n= (?/+3) 2 9. The 

 condition is 3 2 +3{(?/+3) 2 -9} =3y 2 +lSy+9= D, say (2y-3) 2 , whence 

 2, =30. 



Certain Arab writers 1 - 2 of the tenth century treated the special problem 

 to make x--\-k and x 2 k both squares, taking k as given, unlike Diophantus. 



Rafael Bombelli 75 divided 40 into two parts (30 and 10) such that if 

 each be subtracted from the same square (30|) the remainders are squares. 



L. Euler 7 treated the problem, equivalent to Diophantus II, 17: If a 

 and b are given integers, find z, p, q, r, s such that 



(1) p 2 +a2# 2 = r 2 , p 2 +fo? 2 = s 2 . 



Eliminating z, we get p 2 = (6r 2 as 2 )/(6 a). Since the latter is a square for 

 r = S) S et r = s+(b-a)t. Then p 2 = s 2 + 2bst+b(b-a)P. Set p = s+tx/y. 

 Equate to t the numerator of the resulting fraction for t/s. Thus 



Simplifications arise if we set x = v-\-l>y] then 



p = v z aby 2 , q-z = 4:vy( 



Thus we take v and y arbitrary, and choose q z to be the greatest square 

 factor of the final expression. It is shown ( 230) that p z +q z and p 2 +3g 2 

 are not both squares. Cf. Euler. 33 



76 L' Algebra Opera, Bologna, 1579, 461. 



Algebra, St. Petersburg, 2, 1770, 225-230; French tranel., Lyon, 2, 1774, pp. 286-302; 

 Opera omnia, (1), I, 456-464. Cf. Euler 27 of Ch. XV. 



