CHAP, xvi] CONGRUENT NUMBERS. 463 



relatively prime. The dependence of the work of Paciuolo upon that of 

 Leonardo was pointed out in detail by B. Boncompagni 23 and by G. Libri. 24 

 F. Ghaligai 25 also borrowed [Libri, 24 III, 145] from Leonardo; he gave 

 5 2 +24 = 7 2 , 5 2 24=1, stating that 24 is the least congruent number. 

 To find another, start with 1 and 3; add and double the sum, getting 8; 

 multiply by 3 1, getting 16; multiply by 1X3, getting 48; its double 96 

 is a congruent number; in fact, l+3 2 = 10 and 10 2 -96 = 2 2 , 10 2 +96 = 14 2 . 



F. Feliciano 26 gave the same congruent numbers and rule as had 

 Paciuolo. 20 He gave x 2 = 6| as the solution of z 2 d=6= D. 



P. Forcadel de Beziers 27 employed right triangles with one leg less by 

 unity than the hypotenuse h, citing h 5, 13, 25, 41, 61. Their squares are 

 " congrus " numbers, the corresponding " congruens " being 24, 120, 336, 

 720, 1320 [double the products of the two legs]. He gave, 28 for n = 1, 2, 3, 

 4, 5, the congrus (4n 2 +l) 2 and corresponding congruens 8n(4n 2 1). 



N. Tartaglia 29 quoted two rules of Leonardo, as given by Luca Paciuolo, 

 for forming congruent numbers, one rule by use of two consecutive numbers, 

 the other by use of 30 (a 2 +6 2 ) 2 4a&(a 2 -6 2 ) = D. 



G. Gosselin 30 treated (f. 75 verso) the problem: Given a square 100, 

 to find the congruent number. Separate the double 20 of the side into two 

 parts 2L and 20 2L whose product equals the product of two other 

 numbers of difference 20, say L, 20+L. Thus L = 4 and 8X12 = 4X24 

 is the required congruent number 96. Conversely, given a congruent 

 number, to find the square (f. 77, verso). '' This is the problem which 

 Luca, Pisano, Tartaglia, Cardan and Forcadelus found so difficult, in investi- 

 gating which they consumed not a little oil; nevertheless they did not 

 succeed and it remained unsolved up to the present; let us now explain 

 that difficult thing." Given the congruent number 96, to find the square Q 

 such that 96+Q is the required square. Hence the sum 192+Q of the latter 

 and 96 must be a square. Thus the difference of two squares is 96 = 4 

 .24 = 6-16 = 8-12. But (8 + 12) = 10 is excluded since 100 + 192+Q, while 

 i(4+24) = 14and 14 2 = 192+Q gives Q = 4, yielding the answer 96+Q = 100. 



Beha-Eddin 32 (1547-1622) listed, among the seven problems remaining 

 unsolved from former times, as Prob. 2 that to make z 2 +10 and z 2 10 

 both squares. As noted by Nesselmann, it is impossible. 



23 Annali di Sc. Mat. e Fis., 6, 1855, 135-154. 



24 Hist. Sc. Math, en Italic, ed. 2, Halle, 1865, II, 39; III, 137-140, 265-271. 



25 Summa de Arithmetica, Florence, 1521, f. 60; Practica d'arithmetica, Florence, 1552, 



1548, f. 61, left. 



26 Libro di Arithmetica & Geometria speculatiua & praticale : Francesco Feliciano . . . Inti- 



tulato Scala Grimaldelli, Venice, 1526, etc., Verona, 1563, etc., ff. 3-5 (unnumbered 

 pages 7, 8). 



27 L'arithmeticqve, I, 1556, Paris, ff. 8, 9. 



28 The related right triangle has the sides 4n, 4n 2 1, 4n 2 +l. 



29 La Seconda Parte del General Trattato di numeri et misure, Venice, 1556, ff. 143-6. 



30 The final factor, given as a+b, was corrected by the translator, G. Gosselin, 1578, 91. 



31 De Arte magna, seu de occulta parte num., Paris, 1577. 



32 Essenz der Rechenkunst von Mohammed Beha-eddin ben Alhossain aus Amul, arabisch u. 



deutsch von G. H. F. Nesselmann, Berlin, 1843, p. 55. French transl. by Aristide Marre: 

 Khelasat al Hisab, ou Essence du Calcul de Beha-eddin Mohammed ben al-Hosain al- 

 Aamouli, Nouv. Ann. Math., 5, 1846, 313; ed. 2, corrected and with new notes, Rome, 

 1864. 



