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Josalynn

Answered question

2021-03-07

Find a square number such that when twice its root is added to it or subtracted from it, one obtained other square numbers. In other words, solve a problem of the type.

x^{2} + 2 x = u^{2}

x^{2} - 2 x = v^{2}

Answer & Explanation

comentezq

Skilled2021-03-08Added 106 answers

Step 1
To solve the given problem.

x^{2} + 2 x = u^{2}

...(i)

x^{2} - 2 x = v^{2}

...(ii)
Step 2
Let us assume three squares such that

a^{2}, b^{2} and c^{2}

be three consecutive terms of an arithmetic progression with a common difference d.
so

a^{2} = b^{2} - d and c^{2} = b^{2} + d

...(iii)
put the value of

x = \frac{2 b^{2}}{d} \in

...(i)

x^{2} + 2 x = {(\frac{2 b^{2}}{d})}^{2} + 2 (\frac{2 b^{2}}{d})

= \frac{4 b^{4}}{d^{2}} + \frac{4 b^{2}}{d}

= \frac{4 b^{2} (b^{2} + d)}{d^{2}}

= \frac{4 b^{2} c^{2}}{d^{2}}

from(iii)

= {(\frac{2 b c}{d})}^{2}

Step 3
Similarly Put the value of x in (ii)

x^{2} - 2 x = {(\frac{2 b^{2}}{d})}^{2} - 2 (\frac{2 b^{2}}{d})

= \frac{4 b^{4}}{d^{2}} - \frac{4 b^{2}}{d}

= \frac{4 b^{2} (b^{2} - d)}{d^{2}}

= \frac{4 b^{2} a^{2}}{d^{2}}

from(iii)

= {(\frac{2 b a}{d})}^{2}

Hence proved.

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