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1 + 4 + 7 + 10 = 22;... the Nth Pentagonal Number Is Therefore

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1 + 4 + 7 + 10 = 22;... the Nth Pentagonal Number Is Therefore CHAPTER 6 1. The pentagonal numbers are 1; 1 + 4 = 5; 1 + 4 + 7 = 12; 1 + 4 + 7 + 10 = 22;... The nth pentagonal number is therefore n 1 2 − n(n 1) 3n n (3i +1) = n + 3 − = − . i 2 ! 2 X=0 Similarly, since the hexagonal numbers are 1; 1+5 = 6; 1+5+9 = 15; 1+5+9+13 = 28;... it follows that the nth hexagonal number is n 1 − n(n 1) (4i +1) = n + 4 − = 2n2 n. i 2 ! − X=0 2. The pyramidal numbers with triangular base are 1; 1+3 = 4; 1+3+6 = 10; 1+3+6+10 = 20; ... Therefore the nth pyramidal number with triangular base is n n k(k + 1) 1 1 n(n + 1)(2n + 1) n(n + 1) = (k2 + k)= + k 2 2 k 2 " 6 2 # X=1 X=1 . n(n + 1) 2n + 1 1 n(n + 1)(n + 2) = + = 2 6 2 6 The pyramidal numbers with square base are 1; 1+4 = 5; 1+4+9 = 14; 1+4+9+16= 30;... Thus the nth pyramidal number with square base is given by the sum of the squares from 1 to n, namely, n(n + 1)(2n + 1) . 6 3. In a harmonic proportion, c : a =(c b):(b a). It follows that ac ab = bc ac or that b(a + c) = 2ac. Thus the sum of− the extremes− multiplied by the mean− equals− twice the product of the extremes. 4. Since 6:3=(5 3) : (6 5), the numbers 3, 5, 6 are in subcontrary proportion. Other examples of triples− in subcontrary− proportion include 4, 10, 12 and 5, 17, 20. 5. In the example 4, 10, 12 of three numbers in subcontrary proportion, it is not true that the product of the greater and mean terms is twice the product of the mean and smaller. For in this case, the first product equals 120 while the second equals 80. 6. In a ”fifth proportion”, if a<b<c, then b : a = (b a):(c b). Thus 2, 4, 5 are in fifth proportion, because 4 : 2 = (4 2) : (5 4). Two− other triples− in fifth proportion are 3, 6, 7 and 3, 9, 11. − − 1 1 1 1 7. Let x = Diophantus’ age at death. Then x = 6 x + 12 x + 7 x +5+ 2 x + 4. It follows that 9x = 756 and x = 84. 8. To solve x + y = 20; xy = 96, we set x = 10+ z, y = 10 z. Then 100 z2 = 96; z2 = 4; and z = 2. Thus x = 12 and y = 8. − − 31 32 Chapter 6 9. To find two squares whose difference is 60, set x2 = smaller square and x2 +60 = larger square. Then x2 +60= (x + 3)2, where 3 is arbitrarily chosen. This equation reduces 17 289 1 1 to 6x = 51 and therefore x = 2 . The two squares are therefore 4 = 72 4 and 132 4 . In the general case, the two squares are x2 and x2 + b =(x + a)2, where a2 <b. It follows b a2 that x = −2a and the two squares are found to be 2 2 b a2 b + a2 − and . 2a ! 2a ! 10. Let x2 be the least square and (x + m)2 = x2 + 2mx + m2 be the middle square. The difference is 2mx + m2. Therefore the largest square is x2 + 2mx + m2 + n(2mx + m2) = x2 + (2m + 2mn)x + m2 + nm2 = (x + b)2 = x2 + 2bx + b2. Provided that m2(1 + n) <b2 <m2(1 + n)2, the solution is b2 m2 nm2 x = − − . 2m + 2mn 2b − 11. To solve x 6 = u2, x 7 = v2, we subtract and get u2 v2 = 1. It follows that − 1 − 1 − 5 3 (u + v)(u v) = 2 2 . If we set u + v = 2 and u v = 2 , we get u = 4 , v = 4 , and 121 − · − x = 16 . 12. To solve x y = 10, x3 y3 = 2170, set x = z + 5 and y = z 5. It follows that (z + 5)3 (−z 5)3 = 2170.− This equation reduces to 30z2 = 1920− or z2 = 64 or z = 8. Thus x =− 13− and y = 3. In the general case, if x y = a and x3 y3 = b, we set a a − − x = z + 2 and y = z 2 If we substitute for x and y in the second equation, we get as − 3 2 4b a equation in z which reduces to z = 12−a . It follows that this latter expression must be a square. 13. To solve x + y = 20, x3 + y3 = 140(x y)2, set x = 10+ z, y = 10 z. Then (10 + z)3 + (10 z)3 = 140(2z)2. This equation− reduces to 2000 + 60z2 −= 560z2 or 500z2 = 2000 or−z2 = 4 or z = 2. Thus the solution is x = 12, y = 8. In general, if a a x + y = a, x3 + y3 = b(x y)2, we set x = + z, y = z. On substituting into the 2 2 3 −a 3 a 2 2 − a 2 2 second equation, we get 2( 2 ) + 6 2 z = b(2z) , which reduces to 4 + 3az = 4bz or a3 a3 = (4b 3a)z2. Thus z2 = . This equation has a rational solution provided 4 − 4(4b 3a) that the right side is a square, and− that condition is equivalent to Diophantus’ condition that a3(b 3 a) is a square. − 4 14. Simply divide the given square a2 into two squares. This is possible by II–8. 15. We want to solve x + y = (x3 + y)3. We set x = 2z and y = 27z3 2z (so that x + y = (3z)3). Then (x3 + y)3 = (35z3 2z)3 = (3z)3. It follows that 35−z2 = 5. This is impossible for rational z. But now note− that 35 = 27 + 8 = 33 + 23 and 5 = 3 + 2. In order that the equation in z be solvable in rationals, we need two numbers a and 3 3 a +b b (to replace the 3 and 2) so that a+b is a square. So let a + b = 2 (where 2 is arbitrary). Then b = 2 a and a3 + b3 must equal 2 times a square. This implies that a3 + (2 a)3 = 8 12−a + 6a2 = 2(square) or that 4 6a + 3a2 is a square. So set 4 6a +− 3a2 = (2 −4a)2 and solve for a. We get a = 10 −and therefore b = 16 . Since it is − − 13 13 The Final Chapters of Greek Mathematics 33 only the ratio of a and b which is important, we can choose a = 5, b = 8 and therefore put x = 5z, y = 512z3 5z and repeat the initial calculation. We then get 637z3 = 13z 2 1 1 − 5 267 and z = 49 , so z = 7 . Then x = 7 ; y = 343 is the desired solution. 16. In this case, set DA = 3, AB = 1, BE = 9. The problem then becomes to divide 13 into two squares such that one of them lies between 3 and 4. Since 13 = 22 + 32, it is convenient to take this square equal to (2 x)2. Then 13 (2 x)2 = (3+ mx)2. 4 6m − − − If we solve for x, we get x = 2 . Since x must be less than 2 √3, we choose to m−+1 − solve the inequality x < 1 . This becomes m2 + 24m 15 > 0. A convenient rational 4 − solution to this inequality is m = 5 . Then x = 16 , 2 x = 162 , and 3 + mx = 277 . We 8 89 − 89 89 have ( 162 )2 +( 277 )2 = 13. It then follows that the parts of unity are (162 )2 3 = 2481 89 89 89 − 7921 and ( 277 )2 9 = 5440 . To show the necessity, note that if x + y = 1, x + a = c2, and 89 − 7921 y + b = d2, then x + y + a + b = c2 + d2, or a + b +1= c2 + d2. 17. Suppose the right triangle has legs a, b, hypotenuse c, and angle bisector d. Let r be the length of that part of leg a from the right angle to the point where the bisector intersects the leg. To make the right triangle with the angle bisector as hypotenuse a rational triangle, we can set d = 5x and r = 3x. It follows that b = 4x. If we then let a = 3, we have from Elements VI–3 that c : (a r) = b : 4 or that c : (3 3x) = 4x : 3x. Thus c = 4 4x and the reason why a was− chosen to be 3 is evident.− By the Pythagorean theorem,− we have (4 4x)2 = 32 + (4x)2 or 16 32x + 16x2 = 16x2 + 9. Thus 32x = 7 7 − − and x = 32 . To get integral answers, we can multiply through by 32. Thus the original triangle is (96, 28, 100) and the bisector equals 35. 18. The diagram for Elements VI–28 is Fig. 2.28. Let us assume that the proposed rectangle has been constructed with base AS and area equal to c and that the defect is a square. If we set AB = b, and BS = x, then AS = b x and x(b x)= c. Since the maximum − b − b 2 of the ”function” f(x)= x(b x) occurs when x = 2 , and since this maximum is ( 2 ) , − b 2 it follows that c cannot exceed the value ( 2 ) . This means that the area c of the given rectilinear figure must not be greater than the area of the square on half the given line of length b. 19. Assume that the theorem is true. Then AB2 +BC2 = 3AC2.
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