Jacques Sesiano's An Introduction to the History of Algebra PDF

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By Jacques Sesiano

ISBN-10: 0821844733

ISBN-13: 9780821844731

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As the square of the middle value 2 2 must be 1, we multiply this solution by 27 ; then 1 = 97 , 1 = 57 , 1 , and finally x = u = 49, v = 98. whence r = 49 Example 7. 9, where we were asked to represent a number as the sum of two squares, one such decomposition was assumed to be already known. In addition, finding any other representation sufficed. 9 imposes a restriction: the two squares to be found must lie within specific limits and, in fact, must be as close as possible to one another. Moreover, the problem leads to decomposing an expression depending on the given number into a sum of two squares.

2 a2 1−γ 1 • If a1 > 0 > a2 , then 1− γ2 − • If 0 > a1 > a2 , then > γ2 > 1 γ2 γ2 = 1 > 1, and 1, and a1 . |a2 | 1 > |a1 | − 1 . 19 to determine three squares u2 , v 2 , w2 such that the quotient of the differences equals this known value. In our case, one of the squares is given (γ 2 ), so we then need only to multiply the answer obtained by an appropriate factor to obtain the desired solution. γ2 42 2. ALGEBRA IN ANCIENT GREECE Example 5. 42 We wish to find two numbers, one cubic and the other square, such that both the sum and the difference of the cube of the cube and the square of the square are square numbers.

We will first solve this question as above, limiting ourselves to using only identities, and then consider the actual calculations. Let P + A = k, with k a given integer. Then (u + v + w) + A = k. Since, by identity (2 ), (u + v)2 − w2 = 4A, we deduce that A= (*) u+v+w 2 u+v−w 2 . It follows that P +A=u+v+w + = u+v+w 2 u+v−w u+v+w 2 2 u+v−w +2 2 = k. Write u+v+w =p 2 u+v−w = q − 2, 2 so that k is the product of p and q. By adding and subtracting these two expressions (thus by applying the usual half-sum and half-difference identities), we obtain the two relations (α) (β) u+v =p+q−2 w = p − (q − 2).

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An Introduction to the History of Algebra by Jacques Sesiano

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