Step 1

To solve the given problem.

\(\displaystyle{x}^{{2}}+{2}{x}={u}^{{2}}\)...(i)

\(\displaystyle{x}^{{2}}-{2}{x}={v}^{{2}}\)...(ii)

Step 2

Let us assume three squares such that

\(\displaystyle{a}^{{2}},{b}^{{2}}{\quad\text{and}\quad}{c}^{{2}}\) be three consecutive terms of an arithmetic progression with a common difference d.

so

\(\displaystyle{a}^{{2}}={b}^{{2}}-{d}{\quad\text{and}\quad}{c}^{{2}}={b}^{{2}}+{d}\)...(iii)

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

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

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

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

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

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

Step 3

Similarly Put the value of x in (ii)

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

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

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

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

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

Hence proved.

To solve the given problem.

\(\displaystyle{x}^{{2}}+{2}{x}={u}^{{2}}\)...(i)

\(\displaystyle{x}^{{2}}-{2}{x}={v}^{{2}}\)...(ii)

Step 2

Let us assume three squares such that

\(\displaystyle{a}^{{2}},{b}^{{2}}{\quad\text{and}\quad}{c}^{{2}}\) be three consecutive terms of an arithmetic progression with a common difference d.

so

\(\displaystyle{a}^{{2}}={b}^{{2}}-{d}{\quad\text{and}\quad}{c}^{{2}}={b}^{{2}}+{d}\)...(iii)

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

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

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

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

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

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

Step 3

Similarly Put the value of x in (ii)

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

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

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

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

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

Hence proved.