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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– Otto Bretscher · 1 year ago

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

@Otto Bretscher – Chinmay Sangawadekar · 1 year ago

Not actually , as \(\phi (a)\) can't be less than one right ?Log in to reply

– Otto Bretscher · 1 year ago

\(\phi(a)\equiv -1\pmod{a}\)Log in to reply

@Otto Bretscher – Chinmay Sangawadekar · 1 year ago

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

– Otto Bretscher · 1 year ago

You know, I'm a guy with a short attention span. I soon get tired of a certain "type" of problem (like those congruences) and move on to something else. I'm sure you can come up with great examples.Log in to reply

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– Otto Bretscher · 1 year ago

who knows ;)Log in to reply

@Otto Bretscher – Chinmay Sangawadekar · 1 year ago

Experienced Mathematician like you ;)Log in to reply