By Jacques Sesiano
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Extra info for An Introduction to the History of Algebra
1 1 be added to 6 + 2 will be 400 , and it makes a square, with side 51 20 . It is then necessary, in dividing 13 into two squares, to construct the side of each as close as possible to 51 20 ; and I look for that which, when diminished by 3 and augmented by 2, gives this, namely 51 20 . I then set two squares, one with (side) 11x + 2, the other with (side) 3 − 9x. The sum of their squares 5 . Thus, the side of one of is 202x2 + 13 − 10x, equal to 13. Then x is 101 257 258 the squares will be 101 , and that of the other 101 .
It (still ) remains that their sum must be 13. But this sum is 5x2 + 13 − 8x. This equals 13, and x is 85 . (Let us return) to the initial hypotheses. I had set the side of the ﬁrst to 1 be x + 2, it will be 18 5 , and the side of the second to be 2x − 3, it will be 5 . 1 325 As for the squares, they will be 324 25 and 25 , respectively. Their sum is 25 , making up the proposed 13. Consider, generally, u2 + v 2 = k = k12 + k22 , with k, k1 , and k2 given. By setting u = x + k1 and v = mx − k2 , we obtain x2 (m2 + 1) − 2x(mk2 − k1 ) + (k12 + k22 ) = k, and thus x= Each rational m with m > k1 k2 2(mk2 − k1 ) .
26 2. ALGEBRA IN ANCIENT GREECE One of them is29 ⎧ ⎪ ⎪u + v + w + t = 9900 ⎪ ⎨v − u = 1 u 7 ⎪ w − (u + v) = 300 ⎪ ⎪ ⎩ t − (u + v + w) = 300. This is solved as follows. We will express each of the four desired quantities in terms of an auxiliary unknown x. Set u = 7x. By the second equation, then v = 8x, so u+v = 15x. By the third equation, w = u+v+300 = 15x+300, so u+v+w = 30x+300. By the fourth equation, t = u+v+w+300 = 30x+600. Then, by adding these last two equalities and using the ﬁrst equation, we obtain u + v + w + t = 60x + 900 = 9900, so that x = 150, and u = 1050, v = 1200, w = 2550, and t = 5100.
An Introduction to the History of Algebra by Jacques Sesiano