Moonlight of Mathematics, Part 1

Statement. The sum of the squaresoriginal: "vargayogaḥ." The result of $x^2 + y^2$. is 100. The differenceoriginal: "antaram." Here referring to the difference between the two quantities, $x - y$. is 2. From twice the sum of the squares (200), subtract the square of the difference (4), which leaves 196. The square root of this is the sum (14). Using the rule "the sum placed in two positions," the two quantities produced are 8 and 6.
The logic used here is: $(x + y)^2 = 2(x^2 + y^2) - (x - y)^2$. Thus, $\sqrt{2(100) - 2^2} = \sqrt{196} = 14$. Then, by the rule of concurrence: $(14+2)/2 = 8$ and $(14-2)/2 = 6$.

The square of the product should be treated as the "product" term, and the difference of the squares as the "difference" term.
From these two, by the method of concurrence, the results are the squares of the two quantities; their square roots are the quantities themselves. ॥ 4 ॥

If the difference of the squares of two quantities is one hundred and seventy-five,
and their product is three hundred, O child,
tell me those two quantities if you know them.

Statement. The product of the two quantities is 300. The difference of their squares is 175. Here, the square of the product [to be used in calculation] is 60,000Calculated as $4 \times \text{product}^2$. Note: $4 \times 300^2$ is actually 360,000; the value 60,000 in the transcription likely reflects an error in the manuscript or OCR, though the final results are correct.. The difference of the squares is 175. By the rule to be mentioned later, "From the square of the difference of the two quantities added to...", the sumHere, the sum of the squares ($x^2 + y^2$). produced is 625. Using the rule "the sum placed in two positions," the two quantities [squares] produced are 400 and 225. The square roots of these are 20 and 15; these are the actual quantities.
This follows the identity: $(x^2 + y^2)^2 = (x^2 - y^2)^2 + 4(xy)^2$. Substituting the values: $(x^2 + y^2)^2 = 175^2 + 4(300)^2 = 30,625 + 360,000 = 390,625$. The square root of 390,625 is 625. Then, through concurrence with the difference 175: $(625+175)/2 = 400$ and $(625-175)/2 = 225$.

$= \text{First unknown}^2 + \text{Second unknown}^2$. Twice this $= 2(\text{First unknown})^2 + 2(\text{Second unknown})^2$. Subtracting from this the square of the difference $(\text{First unknown} - \text{Second unknown})^2 = \text{First unknown}^2 - 2(\text{First unknown} \times \text{Second unknown}) + \text{Second unknown}^2$, the square of the sum is produced $= \text{First unknown}^2 + 2(\text{First unknown} \times \text{Second unknown}) + \text{Second unknown}^2$. The square root of this is indeed the sum. This is exactly what was stated by BhaskaraBhaskara II (c. 1114–1185), a preeminent Indian mathematician and astronomer. in the rule beginning "From twice the square of the hypotenuse..."

1. Here is the rationale. Let the two quantities be 'ya' (the first unknown) and 'ka' (the second unknown). Then, according to the problem:

Product = (ya)(ka)        } Product² = ya²ka²
                    {
Diff. of squares = ya² - ka² } (Diff. of squares)² = ya⁴ - 2ya²ka² + ka⁴

Therefore: (Diff. of squares)² + 4(Product²) = ya⁴ + 2ya²ka² + ka⁴ = ( ya² + ka² )²

Then: $\sqrt{(\text{Diff. of squares})^2 + 4(\text{Product})^2}$ = ya² + ka²
                   Diff. of squares      = ya² - ka²

From these two, by concurrence, the squares of the quantities are found, and from their square roots, the quantities themselves are determined. This is how it is proved.