Abstract
Let G be a simple graph and let $\={G}$ denotes its complement. We say that G is integral if its spectrum consists entirely of integers. If $\overline{aK_{a}\;{\bigcup}\;{\beta}K_{b}}$ is integral we show that it belongs to the class of integral graphs $[\frac{kt}{\tau}\;{x_0}\;+\;\frac{mt}{\tau}\;z}\;K_{(t+{\ell}n)+{\ell}m}\;\bigcup\;[\frac{kt}{\tau}\;{y_0}\;+\;\frac{(t\;+\;{\ell}n)k\;+\;{\ell}m}{\tau}\;z]n\;K_{em)$, where (i) t, k, $\ell$, m, $n\;\in\;\mathbb{N}$ such that (m, n) = 1, (n,t) = 1 and ($\ell,\;t$) = 1 ; (ii) $\tau\;=\;((t\;+\;{\ell}n)k\;+\;{\ell}m,\;mt)$ such that $\tau\;$\mid$kt$; (iii) ($x_0,\;y_0$) is a particular solution of the linear Diophantine equation $((t\;+\;{\ell}n)k\;+\;{\ell}m)x\;-\;(mt)y\;=\;\tau\;and\;(iv)\;z\;{\geq}\;{z_0}$ where $z_{0}$ is the least integer such that $(\frac{kt}{\tau}\;{x_0}\;+\;\frac{mt}{\tau}\;{z_0})\;\geq\;1\;and\;(\frac{kt}{\tau}\;{y_0}\;+\;\frac{(t+{\ell}n)k+{\ell}m}{\tau}\;{z_0})\;\geq\;1$.