By Jacques Sesiano
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Extra resources for An Introduction to the History of Algebra
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).
An Introduction to the History of Algebra by Jacques Sesiano