Question

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

Upper Level Math
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.
$$\displaystyle{x}^{{2}}+{2}{x}={u}^{{2}}$$
$$\displaystyle{x}^{{2}}-{2}{x}={v}^{{2}}$$

2021-03-08
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.