Tricky Totient

Find the total number of positive integers n n such that ϕ ( n 2 + 1 ) = 6 n \phi(n^2+1)=6n .

Details and assumptions

The Euler's totient function ϕ ( m ) \phi(m) is the number of integers from 1 1 to m 1 m-1 that are coprime to m . m.


The answer is 2.

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1 solution

Raviteja Meesala
May 20, 2014

For every 10 natural numbers we have a minimum of 2 prime numbers .. So if the minimum number of prime numbers exceed the number co primes expected then we dont need to check those numbers.. The limit comes down to 30.. now n^2 + 1 >=6n .. implies n>=6 ; By inspection we get 2 possible answers 6 and 8...

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