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

Skilled2021-03-08Added 106 answers

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

$a}^{2},{b}^{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{c}^{2$ be three consecutive terms of an arithmetic progression with a common difference d.

so

${a}^{2}={b}^{2}-d{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{c}^{2}={b}^{2}+d$ ...(iii)

put the value of$x=\frac{2{b}^{2}}{d}\in$ ...(i)

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

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

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

$=\frac{4{b}^{2}{c}^{2}}{{d}^{2}}$ from(iii)

$={\left(\frac{2bc}{d}\right)}^{2}$

Step 3

Similarly Put the value of x in (ii)

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

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

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

$=\frac{4{b}^{2}{a}^{2}}{{d}^{2}}$ from(iii)

$={\left(\frac{2ba}{d}\right)}^{2}$

Hence proved.

To solve the given problem.

Step 2

Let us assume three squares such that

so

put the value of

Step 3

Similarly Put the value of x in (ii)

Hence proved.

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