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Topic: Solitary numbers / Friendly numbers (help needed)
Replies: 3   Last Post: Sep 17, 2004 4:30 PM

 Messages: [ Previous | Next ]
 David Einstein Posts: 389 Registered: 12/6/04
Re: Solitary numbers / Friendly numbers (help needed)
Posted: Sep 15, 2004 9:11 AM

Geert van der Wulp wrote:
&gt; Lately I have taken up an interest in solitary numbers: numbers which
&gt; do not have "friends". The number n has a friend m iff:
&gt;
&gt; sigma(n)/n = sigma(m)/m
&gt;
&gt; Now I have read and verified the claims that a number n has NO friends
&gt; (so it is solitary) if:
&gt; 1) n is prime,
&gt; 2) n is a prime power,
&gt; 3) if the greatest common divisor of n and sigma(n) is 1.
&gt;
&gt; Now another source claimed that for certain numbers it is not
&gt; difficult to prove that they are solitary, even though they fall
&gt; outside the above mentioned rules. This is supposedly the case for the
&gt; number 18.
&gt;
&gt; Now I have tried to prove this claim, but I did not succeed:
&gt;
&gt; Assume first that we have found a number x for which
&gt;
&gt; sigma(x)/x = sigma(18)/18 = (1+2+3+6+9+18)/18 = 13/6
&gt;
&gt; This implies that 6 is a divisor of x and 13 is a divisor of sigma(x).
&gt;
&gt; How to proceed?
Ok.

We know that 2*3=6 divides x, so we know that x=2^a * 3^b * n where n is
relatively prime to 6. Now it is easy to see that if sigma(x)/x = 13/6
and x != 18 then a=b=1. (Can you see why?) Now if x=6*n with n
relatively prime to 6 then the numerator of sigma(x)/x reduced to lowest
terms will be even, and the denominator will be relatively prime to 6,
in particular it will not equal 13/6.

Date Subject Author
9/15/04 Geert van der Wulp
9/15/04 Michael JÃ¸rgensen
9/15/04 David Einstein
9/17/04 Keith A. Lewis