relationship between totatives and divisors


Theorem 1.

Let n be a positive integer and define the sets In, Dn, and Tn as follows:

  • •

    In={m∈ℤ:1≤m≤n}

  • •

    Dn={d∈In:d>1 and d|n}

  • •

    Tn={t∈In:t is a totativeMathworldPlanetmath of n}

Then Dn∪Tn=In if and only if n=1, n=4, or n is prime.

Proof.

If n=1, then Dn=∅ and Tn={1}. Thus, Dn∪Tn=In.

If n=4, then Dn={2,4} and Tn={1,3}. Thus, Dn∪Tn=In.

If n is prime, then Dn={n} and Tn=In∖{n}. Thus, Dn∪Tn=In.

Sufficiency:

This will be proven by considering its contrapositive.

Suppose first that n is a power of 2. Then n≥8. Thus, 6∈In. On the other hand, 6 is neither a totative of n (since gcd⁡(6,n)=2) nor a divisorMathworldPlanetmathPlanetmath of n (since n is a power of 2). Hence, Dn∪Tn≠In.

Now suppose that n is even and is not a power of 2. Let k be a positive integer such that 2k exactly divides n. Since n is not a power of 2, it must be the case that n=2k⁢r for some odd integer r≥3. Thus, n=2k⁢r>2k+1. Therefore, 2k+1∈In. On the other hand, 2k+1 is neither a totative of n (since n is even) nor a divisor of n (since 2k exactly divides n). Hence, Dn∪Tn≠In.

Finally, suppose that n is odd. Let p be the smallest prime divisorPlanetmathPlanetmath of n. Since n is not prime, it must be the case that n=p⁢s for some odd integer s≥3. Thus, n=p⁢s>2⁢p. Therefore, 2⁢p∈In. On the other hand, 2⁢p is neither a totative of n (since gcd⁡(2⁢p,n)=p) nor a divisor of n (since n is odd). Hence, Dn∪Tn≠In. ∎

Title relationship between totatives and divisors
Canonical name RelationshipBetweenTotativesAndDivisors
Date of creation 2013-03-22 17:09:15
Last modified on 2013-03-22 17:09:15
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 15
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 11A25