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Asked Jul 27, 2019
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Question attached.  I have found solutions to this probem on the internet, but I don't understand the explanations.  Can you provide a "layman's terms" explanation?  Thanks.  

Prove that for any given positive integer N there exist at most finitely many integers
denotes the Euler totient function. Conclude in
n with(n)
N, where
particular that (n) tends to infinity as n tends to infinity.
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Prove that for any given positive integer N there exist at most finitely many integers denotes the Euler totient function. Conclude in n with(n) N, where particular that (n) tends to infinity as n tends to infinity.

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Expert Answer

Step 1

Let N be a given positive integer and let p be the least prime number greater than N + 1.

Let n be an integer such that

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(n) N

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Step 2

If q > p is a prime divisor of n, then n = (qk) m for some k > 1and m with q not diving m.

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(n)(')(m) (q -1)> (p-) N

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Step 3

It is contraction to our assumption.

Thus, there is no prime divisor of n is greater than N + 1.

In p...

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P1 P2 P Consider the primes are m. for some 0 a, where 1sim a Pm Now it can be write as n = P^" - P2 That is Ф(m) -о(p*) -о(p:*). o(p.*) - (2**) (P-1)-(р.*)(P.-1). . (р, з)(р. - _ )(Pm-1) iśm a-l >(p1P-1)N for sufficiently large a where 1

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