I have added the surface of my two squares: 28;20.[The text uses sexagesimal numbers. In our usual decimal system, \(28;20 = 20\cdot 1 + 28 \cdot 60 = 1700\).]
(The side of) one square is a quarter (the side of) the (other) square.
You put down 4 and 1.
You multiply 4 by 4: 16.
You multiply 1 by 1: 1.
You add 1 and 16: 17.
The reciprocal of 17 cannot be solved.
What must I put to 17 [i.e. What must I multiply 17 by] which will give me 28;20 [1700]? 1;40 [100].
This is the square of 10.
You raise 10 by 4 and 40 is (the side of) one square.
You raise 10 by I and 10 is (the side of) the second square.
A modern symbolic solution: Let x and y be the side lengths of the squares. We know that
\[x^2 + y^2 = 1700 \] and (without loss of generality assume \(y < x\) ) \[ x = 4y \]
Substitute \(x = 4y\) into \(x^2 + y^2 = 1700\) and get
\begin{eqnarray*}
16y^2 + y^2 & =& 1700 \\
17y^2 &=& 1700\\
y^2 &=& 100\\
y &=& \pm 10
\end{eqnarray*}
Since the problem is talking about physical squares, we discard the negative solution and get \(y = 10\), \(x=4y=40\).
You can find problems exactly like this in any present day basic algebra textbook, right down to the use of nice integers which tempt natural arithmeticians to not show their work and thus receive only partial credit on graded homework and tests.
My number theory textbook says that the mathematician whose book Kitab al-jabr introduced Europe to algebra, Abu Ja'far Mohammed Ibn Musa Al-Khwarizmi, was from the town of Khiva (Kowarzizm) in modern day Uzbekistan. I tried to do a short research report on Khiva my first year of college, and could (at the time) put my hands on almost zero substantive material. The most anyone could seem to say about the town was that it (a) wasn't quite as notorious a slave-trading center as its neighbor town, Merv, and (b) has a lovely, though relatively recent, half-finished minaret:
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| image from Wikipedia |
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