ultimate generalisation of Euler-Fermat theorem

Let $a^b+u=m$ where $a,b,u,m$ are positive integers. Then

$$a^{b+k\phi (m)}+u\equiv \mathrm{\hspace{0.33em}0}(modm),$$

by the result in “Euler’s generalisation of Fermat’s theorem – a further generalisation”. Proceedings of Hawaii Intl. conference on maths & statistics 2004 (ISSN 1550–3747). Here, $k$ is a positive integer. Next,

$$a^{b^{1+k\phi (\phi (m))}}+u\equiv \mathrm{\hspace{0.33em}0}(modm).$$

(This is a corollary of “Euler’s generalisation of Fermat’s theorem – a further generalisation”.)We can proceed in a like manner till we reach

$$a^{b^{c^{{\mathrm{⋮}}^{t^{1+k\phi (\phi (\phi (\mathrm{\dots }\phi (2)\mathrm{\dots })))}}}}}.$$

At this stage onwards the function generates only multiples of $m$ and no prime number is generated. This is the ultimate generalisation of Fermat’s theorem. Please note that each step of multiple exponentiation in the above is a corollary of the theorem referred to.