Let us define \(Z(n)\) , such that for the input of natural \(n\) , \(Z(n)\) gives the immediate prime number next to \(n\) .

Then \[ a^{Z(n)} \equiv a \mod n \]

=> \(a\) and \(n\) belong to the set of natural numbers where \(n\) is not a prime number and greater than one ..

## Comments

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TopNewestThe claim is not true.

For example, with \( n = 32, Z(n) = 37 \), we have \( 3^{37} \equiv 19 \pmod{32} \).

I believe what happened was that you were testing small cases where the prime factors of \(n\) were small and distinct, which combined with euler's theorem led to this being true in several cases. As such, I went with a prime power, and then used one that was large enough. – Calvin Lin Staff · 9 months, 1 week ago

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– Chinmay Sangawadekar · 9 months, 1 week ago

Can we find range if solutions for this ?Log in to reply

Awesome result! Is it original?

I'll try proving it. – Harsh Shrivastava · 9 months, 1 week ago

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– Chinmay Sangawadekar · 9 months, 1 week ago

Yeah it is , but it is disproved for large values :( .Log in to reply