Josalynn

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.
${x}^{2}+2x={u}^{2}$
${x}^{2}-2x={v}^{2}$

comentezq

Step 1
To solve the given problem.
${x}^{2}+2x={u}^{2}$...(i)
${x}^{2}-2x={v}^{2}$...(ii)
Step 2
Let us assume three squares such that
${a}^{2},{b}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{c}^{2}$ be three consecutive terms of an arithmetic progression with a common difference d.
so
${a}^{2}={b}^{2}-d\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{c}^{2}={b}^{2}+d$...(iii)
put the value of $x=\frac{2{b}^{2}}{d}\in$...(i)
${x}^{2}+2x={\left(\frac{2{b}^{2}}{d}\right)}^{2}+2\left(\frac{2{b}^{2}}{d}\right)$
$=\frac{4{b}^{4}}{{d}^{2}}+\frac{4{b}^{2}}{d}$
$=\frac{4{b}^{2}\left({b}^{2}+d\right)}{{d}^{2}}$
$=\frac{4{b}^{2}{c}^{2}}{{d}^{2}}$ from(iii)
$={\left(\frac{2bc}{d}\right)}^{2}$
Step 3
Similarly Put the value of x in (ii)
${x}^{2}-2x={\left(\frac{2{b}^{2}}{d}\right)}^{2}-2\left(\frac{2{b}^{2}}{d}\right)$
$=\frac{4{b}^{4}}{{d}^{2}}-\frac{4{b}^{2}}{d}$
$=\frac{4{b}^{2}\left({b}^{2}-d\right)}{{d}^{2}}$
$=\frac{4{b}^{2}{a}^{2}}{{d}^{2}}$ from(iii)
$={\left(\frac{2ba}{d}\right)}^{2}$
Hence proved.

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