It is given that \(\displaystyle{x}_{{1}},{x}_{{2}},{y}_{{1}},{y}_{{2}}\) are all integers.

\(\displaystyle{a},{x}_{{1}},{y}_{{1}},{b}\) is part of an arithmetic sequences.

Thus, \(\displaystyle{x}_{{1}}−{a}={y}_{{1}}−{x}_{{1}}\) the common difference is same in arithmetic sequences.

\(\displaystyle{x}_{{1}}−{a}={y}_{{1}}−{x}_{{1}}\)

\(\displaystyle{2}{x}_{{1}}={y}_{{1}}+{a}\)

\(\displaystyle{a}={2}{x}_{{1}}−{y}_{{1}}\)

So, \(\displaystyle{a}\) is integer as \(\displaystyle{x}_{{1}}\) and \(\displaystyle{y}_{{1}}\) are integers. (1)

\(\displaystyle{y}_{{1}}−{x}_{{1}}={b}−{y}_{{1}}\) the common difference is same in arithmetic sequences.

\(\displaystyle{y}_{{1}}−{x}_{{1}}={b}−{y}_{{1}}\)

\(\displaystyle{2}{y}_{{1}}={x}_{{1}}+{b}\)

\(\displaystyle{b}={2}{y}_{{1}}−{x}_{{1}}\)

So, \(\displaystyle{b}\) is integer as \(\displaystyle{x}_{{1}}\) and \(\displaystyle{y}_{{1}}\) are integers. (2)

It is given that \(\displaystyle{a},{x}_{{2}},{y}_{{2}},{b}\) is part of a geometric sequence. Hence, \(\displaystyle{\frac{{{x}_{{2}}}}{{{a}}}}={\frac{{{y}_{{2}}}}{{{x}_{{2}}}}}\) the common ratio is same in geometric sequence.

\(\displaystyle{x}_{{2}}{x}_{{2}}={a}{y}_{{2}}\)

\(\displaystyle{a}={\frac{{{{x}_{{2}}^{{{2}}}}}}{{{y}_{{2}}}}}\)

\(\displaystyle{x}_{{2}}=\sqrt{{{a}{y}_{{2}}}}\)

Similarly, \(y_2=\sqrt{bx_2}\)

Also, \(\displaystyle{\frac{{{x}_{{2}}}}{{{a}}}}={\frac{{{b}}}{{{y}_{{2}}}}}\) the common ratio is same in geometric sequence.

\(\displaystyle{a}{b}={x}_{{2}}{y}_{{2}}\)

\(\displaystyle{a}{b}\) is an integer as \(\displaystyle{x}_{{2}}\) and \(\displaystyle{y}_{{2}}\) are integerand the product of two integer is an integer.

Thus, from equations it is proved that there is a pair of integers \(\displaystyle{\left({a},{b}\right)}\).