formula is thus obtained, it can be used for the purpose of finding the number of ways in which P can be a polygonal number. The portion of the geometrical argument which has to be supplied is, it is true, somewhat long, and its length and difficulty may, as Wertheim suggests, account for the copyist having failed, as it were, to see his way through it and having stopped through discouragement when he had lost his bearings. I shall now reproduce Wertheim's suggested restoration of the rest of the problem. The figure requires some extension, and I accordingly give a new one after Wertheim. -{2 (P-1)-(n − 1)(a− 2)} = 2 + (n-1) (a−2) FN=2P+n(a−2), from above GN=2(P-1)-(n − 1)(a−2) RN-FN-2AB=n(a−2) {2P+n (a− 2)} {2 (P-1)-(-1) (a-3)} = 4P(P− 1) 17 Now RN-FN-FR = FM-NM - FR = FM-GM-FR = BL−}GM-2AB=BD + }DK-}GM–2AB = BD+2AB + 2BH− } GM − 2AB = BD + 2BH-GM, RN=BD + 2BH-GM = BD+2BH−}BL+}FG = BD + 2BH-BD - DL + FG = }BD + 2BH− } DL + } FG =}BD + 2BH-−(AB+BH) + FG = }BD-AH + FG = (BD + FG-2AH). Thus the double of any polygonal number must be divisible by its side, and the quotient is the number arrived at by adding a to the product of (side 1) and (number of angles - 2). For a triangular number the quotient is n + 1, and is therefore greater than the side; and, as the quotient increases by n-1 for every increase of 1 in the number of angles (a), it is always greater than the side. I We can therefore use the above formula (17) to find the number of ways in which a given number P can be a polygonal number. Separate 2P into two factors in all possible ways, excluding 1. 2P. Take the smaller factor as the side (n). Then take the other factor, subtract 2 from it, and divide the remainder by (n − 1). If (n − 1) divides it without a remainder, the particular factors taken answer the purpose, and the quotient increased by gives the number of angles (a). If the second factor diminished by 2 is not divisible by (n-1) without a remainder, the particular division into factors is useless for the purpose. The number of ways in which P can be a polygonal is the number of pairs of factors which answer the purpose. There is always one pair of factors which will serve, namely a and Pitself. The process of finding pairs of factors is shortened by the following considerations. Thus in choosing values for the factor n we need not go beyond that shown in the right-hand expression. Example 1. In what ways is 325 a polygonal number? Here I + √(1 + 8P) = = −1+√(2601) = 50. Therefore cannot be greater than 25. Now 2.325 = 2.5.5.13, and the only possible values for n are therefore 2, 5, 10, 13, 25. The corresponding values for a are shown in the following table. 1. 39. - a = m (x − b). x x + a = m (x + b). a x = m (b−x).. x+b = m (a− x). x+b= m (x − a). (a + x)b + (b + x) a = 2 (a + b) x, ) (a + b) x + (b + x) a = 2 (a + x) b, (a> b). Determinate systems of equations of the first degree. I. 3. x-y=a, x= a) {1. 13. X1+Xq=Y1+Y1 =*+*=a} (x,>*a, Yı>Ya, &1 >%), x1 = my1, Y1 = n%2, %1 = pxg) 1. 15. x+am (y − a), y + b = n(x − b). § 1. 16. 1. 17. 1. 18. y + z + w = a, z + w + x = b, w + x + y = c, x + y + z = d. L 19. y+z+w− x = a, z + w + x−y = b, w+x+y − z = c, IV. 36. _yz=m (y + z), zx = n (x + x), xy = p(x+y). Determinate systems reducible to equations of second degree. xy + (x + y) = co − 15. (IV. 34. ys + (y + z) = a3 − 1, sx + (8 + x) = b2 − 1, − 1, zx − (3 + x) = b3 − 1, xy − (x + y) = c − 1. zx = n (x + y + z), xy = p (x + y + s). ex = b2, xy=c. Systems of equations apparently indeterminate but really reduced, by arbitrary assumptions, to determinate equations of the first degree. 1. 14. xy = m (x + y) [value of y arbitrarily assumed]. I I I m x2-y2 = m (x − y) + a [Diophantus assumes x-y=2]. - I I I -4x+--+--+ [value of y assumed) I -+--+ m n x=% %+ +y I x m |