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




Re: Solitary numbers / Friendly numbers (help needed)
Posted:
Sep 15, 2004 9:11 AM


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



