Well these congruences are pretty basic and trivial but I didn't find anything about them in books or on net.

It is interesting to note the following congruences.

\[1) \phi(a)^n \equiv \phi(a) \mod a\] Here , \(a\) is arbitary prime number and \(n\) is any odd number.

\[2) \phi(a)^{n'} \equiv 1\mod a\] Here \(a\) is any arbitary prime and \(n'\) is an arbitary even number.

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TopNewestComment deleted Mar 06, 2016

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Yes, they are true... it amounts to \((-1)^n=-1\) for odd \(n\) and \((-1)^n=1\) for even \(n\).

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Not actually , as \(\phi (a)\) can't be less than one right ? @Otto Bretscher

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@Otto Bretscher

Ah , how could I miss that !! anyway can you construct any problem which could be solved by these congruences ? I had one problem in my mind , simplifying the towers of integers becomes a lot easy by these congruences...Log in to reply

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Comment deleted Mar 06, 2016

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@Otto Bretscher

Experienced Mathematician like you ;)Log in to reply