Famous Theorems of Mathematics/Number Theory/Totient Function

This page provides proofs for identities involving the totient function and the Möbius function .

Sum of integers relatively prime to and less than or equal to n

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The proof of the identity

 

uses the fact that

 

because if   and   then   and if   and   then  

This means that for   we may group the k that are relatively prime to n into pairs

 .

The case   does not occur because   is not an integer when n is odd, and when n is even, we have   because we assumed that   There are

 

such pairs, and the constituents of each pair sum to

 

hence

 

The case   is verified by direct substitution and may be included in the formula.

Proofs of totient identities involving the floor function

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The proof of the identity

 

is by mathematical induction on  . The base case is   and we see that the claim holds:

 

For the induction step we need to prove that

 

The key observation is that

 

so that the sum is

 

Now the fact that

 

is a basic totient identity. To see that it holds, let   be the prime factorization of n+1. Then

 

by definition of   This concludes the proof.

An alternate proof proceeds by substituting   directly into the left side of the identity, giving  

Now we ask how often the term   occurs in the double sum. The answer is that it occurs for every multiple   of  , but there are precisely   such multiples, which means that the sum is

 

as claimed.


The trick where zero values of   are filtered out may also be used to prove the identity

 

The base case is   and we have

 

and it holds. The induction step requires us to show that

 

Next observe that

 

This gives the following for the sum

 

Treating the two inner terms separately, we get

 

The first of these two is precisely   as we saw earlier, and the second is zero, by a basic property of the Möbius function (using the same factorization of   as above, we have  .) This concludes the proof.

This result may also be proved by inclusion-exclusion. Rewrite the identity as

 

Now we see that the left side counts the number of lattice points (a, b) in [1,n] × [1,n] where a and b are relatively prime to each other. Using the sets   where p is a prime less than or equal to n to denote the set of points where both coordinates are divisible by p we have

 

This formula counts the number of pairs where a and b are not relatively prime to each other. The cardinalities are as follows:

 

and the signs are  , hence the number of points with relatively prime coordinates is

 

but this is precisely   and we have the claim.

Average order of the totient

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We will use the last formula of the preceding section to prove the following result:

 

Using   we have the upper bound

 

and the lower bound

 

which is

 

Working with the last two terms and using the asymptotic expansion of the nth harmonic number we have

 

and

 

Now we check the order of the terms in the upper and lower bound. The term   is   by comparison with  , where   is the Riemann zeta function. The next largest term is the logarithmic term from the lower bound.

So far we have shown that

 

It remains to evaluate   asymptotically, which we have seen converges. The Euler product for the Riemann zeta function is

 

Now it follows immediately from the definition of the Möbius function that

 

This means that

 

where the integral   was used to estimate   But   and we have established the claim.

Average order of φ(n)/n

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The material of the preceding section, together with the identity

 

also yields a proof that

 

Reasoning as before, we have the upper bound

 

and the lower bound

 

Now apply the estimates from the preceding section to obtain the result.

Inequalities

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We first show that

 

The latter holds because when n is a power of a prime p, we have

 

which gets arbitrarily close to 1 for p large enough (and we can take p as large as we please since there are infinitely many primes).

To see the former, let nk be the product of the first k primes, call them  . Let

 

Then

 

a harmonic number. Hence, by the well-known bound  , we have

 

Since the logarithm is unbounded, taking k arbitrarily large ensures that rk achieves values arbitrarily close to zero.

We use the same factorization of n as in the first section to prove that

 .

Note that

 

which is

 

The upper bound follows immediately since

 

We come arbitrarily close to this bound when n is prime. For the lower bound, note that

 

where the product is over all primes. We have already seen this product, as in

 

so that

 

and we have the claim. The values of n that come closest to this bound are products of the first k primes.

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