CHAP, xvi] CONGRUENT NUMBERS. 40 1 



all of which are even. By (1), s(a 6) <t(a-\-fy, whence m<q. Now 

 v = mq is the sum of m consecutive odd numbers 



(3) g-( m -l), , q-3, g-1, g+1, g+3, , g+(w-l). 



Similarly, u = np is the sum of n consecutive odd numbers equidistant 

 by twos from p. Thus the numbers of terms in the sums for v and u have 

 the ratio m :n = s :t = a :b. There are (q ni)/2 odd numbers <q m; 

 their sum Zi is (q m) 2 /4. The sum 2 2 of the odd numbers <q-\-m is 

 (<7+w) 2 /4. Between g ?ft and q-\-m lie the ?w consecutive odd numbers (3), 

 so that their sum is v. But 



m+n=(s+)(a 6) = (a+6)(s i)=p q, q+m = p n. 



Thus the n odd numbers between p n and p+n, whose sum is u, are the 

 n odd numbers which follow q-\-m. Finally, the sum 3 of the odd numbers 

 <p+n is (p+n) 2 /4. Hence Zi+v=z 2 , z 2 +u=z s , while z b z 2 , z 3 are squares; 

 further, 



v = mq = st(a 6) (a+6) = wp = u. 



Thus the proposed problem is solved by taking 8 



y = v = u, xl=z it xl = Zi, xl = z 3 . 



Next, if the inequality sign in (1) is reversed, we have only to inter- 

 change m and q in the definitions (2), which were used only to obtain 

 q-\-m = p n, v u. As the latter hold also now, the preceding discussion 

 holds for the present case also. The case a :b = a+b : a 6 is shown to 

 be impossible in integers. 9 



Leonardo 10 gave several numerical examples. For a = 5, 6 = 3, then 

 2/ = 240, x l = 7, z 2 = 17, Z 3 = 23. For a = 3 or 2, 6 = 1, then ?/ = 24, &i = l, 

 z 2 = 5, #3 = 7. For a = 5, 6 = 2, then ?/ = 840, x l = l, z 2 = 29, a? 3 = 41. For 

 a = 7, 6 = 5, then y = 840, x l = 23, x 2 = 37, z 3 = 47. Note 11 that 24 is the least 

 congruent number for which the three squares xl are integers; but with 

 fractions, we can find smaller as shown later. 



For, Leonardo 12 proved that if a and 6 are relatively prime and if a+6 

 is even then a6(a+6)(a 6) is divisible by 24 and stated 13 that a similar 

 proof holds if a and 6 are not relatively prime. He proved also that, if 

 one of a and 6 is even and the other odd, 2a-26(a+6)(a 6) is divisible by 24. 

 Thus he was able to state 14 that any congruent number is a multiple of 24. 



The product 15 of 24 by any square /i 2 is a congruent number and the 

 corresponding squares are the products of those for 24 by /i 2 . We also get 

 congruent numbers by multiplying 24 by a sum of squares 1 2 + 2 2 +3 2 H 



8 B. Boncompagni, Annali di Sc. Mat. e Fis., 6, 1855, 135, quoted Leonardo's solution to be 



y = 4a6(a 2 -6 2 ), z 2 = a 2 +6 2 , Xi, x 3 =2a6(6 2 -a 2 ). But this corresponds only to the 

 case s = 2a, t=2b. 



9 Tre Scritti, 96; Scritti, II, 271. Genocchi, 7 pp. 292-3. 



10 Tre Scritti, 88-92; Scritti, II, 268-70. Genocchi, 7 pp. 278-9. 



11 Tre Scritti, 90-93; Scritti, II, 269-270. Genocchi, 7 pp. 280-1. 



12 Si duo numeri, Tre Scritti, 80; Scritti, II, 264. 



13 Tre Scritti, 82; Scritti, II, 265. 



14 Tre Scritti, 92; Scritti, 11, 270. Genocchi, 7 pp. 273-4. 



enim 24, Tre Scritti, 93; Scritti, II, 270. Genocchi, 7 p. 283, p. 254. 



