What's wrong in this proof of $10$ is a solitary number? Friendly numbers are two or more natural numbers with a common abundancy, the ratio between the sum of divisors of a number and the number itself. Two numbers with the same abundancy form a friendly pair.
A number that is not part of any friendly pair is called solitary. I have found a very elementary proof of 10 being a solitary number which is an open problem. So I would like to find a mistake in my proof. Let $\sigma(n)$ be the sum of divisor function.
For $10$ to have a friendly pair we have to find another solution to $$\sigma(5x)=9x$$
 $$6 \ \sigma(x) = 9x \ \text{(Assuming (5,x)=1)}$$
 $$2\ \sigma(x)=3x$$
Now Let $x=2^kp$ where $p \in \mathbb{N}$ ,$(p,z)=1$
$$2(2^{k+1}-1) \sigma(p)=3 \cdot 2^kp$$
$$\frac{\sigma(p)}{p}=\frac{3 \cdot 2^{k-1}}{2^{k+1}-1}$$
Now as any number has atleast two prime factors one and itself hence ,
$$\frac{\sigma(p)}{p}>1$$
$$ \implies3 \cdot 2^{k-1}>2^{k+1}-1$$
Which can be showed impossible for $k>0$.
Now  let us assume that $x$ is in the form $x=5^t2^kq$ 
$$\frac{2(5^{t+1}-1)(2^{k+1}-1) \sigma(q)}{4}=3 \cdot 5^t \cdot 2^k \cdot q$$
$$(5^{t+1}-1)(2^{k+1}-1) \sigma(q)=3 \cdot 5^t \cdot 2^{k+1} \cdot q$$
$$\frac{\sigma(q)}{q}=\frac{3 \cdot 5^t \cdot 2^{k+1} }{(5^{t+1}-1)(2^{k+1}-1)}$$
Which by proceeding as above ie. by the fact $\frac{\sigma(q)}{q}>1$ can be shown not possible.
Thus no friendly pair of $10$ is possible. Thus $10$ is solitary.
 A: Here is the main problem. First of all, since:
$$\frac{\sigma(10)}{10} = \frac{\sigma(2)\sigma(5)}{10} = \frac{(1+2)(1+5)}{10} = \frac{18}{10} = \frac95$$
a hypothetical friend $x\in\mathbb{N}$ of $10$ must satisfy:
$$\frac{\sigma(x)}{x}=\frac95$$
You search for $2\sigma(x)=3x$ which is wrong. In your very first equation the "5" is inside the $\sigma$ function when it should be outside. Therefore "5" becomes "6" in your proof attempt which is wrong. The irrelevant equation $\frac{\sigma(x)}{x}=\frac32$ is solved by $x=2$ only, by the way. The fact that $2$ is solitary is not new.
A: I believe that the problem lies in the last part of your proof. The sum of factors of q over q is definitely more than 1. Remember that the sum of factors include q as a factor as well. Therefore, the sum of factors of q over q should be more than 1 and a friend is still possible.
A: Notice that the first part is straightforward and correct, even if it is missing some helpful details. This part demonstrates simply that 10 cannot divide its friend(s).
In the second part, your conclusion that the result holds is based on the assumption that $k$ (the exponent for 2) is nonzero. Allowing $k=0$, the RHS of the offending equation becomes $$\frac{6\cdot5^{t}}{5^{t+1}-1}$$ which is definitely greater than 1.
A second error in the second part is in the setup. Specifically, you used the $$2\sigma(x)=3x$$ setup in a way that was unjustified, since it hinged on $(5,x)=1$. The proper setup for the second part would be $$\frac{(2^{k+1}-1)(5^{t+1}-1)\sigma(q)}{4\cdot2^k5^tq}=\frac{9}{5}$$ since this is what happens when $5|x$.
A: I, too, after seeing that ten has not been proved as a solitary number jumped straight to your proof. However, the error is in the first step where you claim $\sigma(5x) = 6\sigma(x)$. Disproving this is merely a game of examples where it does not work and that is when x is a multiple of $5$. For example, $10$; $\sigma(50) = 93$, while $6\sigma(10) = 108$.
A: according to my proof 0 is not solitary
a solitary number is a number which does not have any friends.
Solitary Number include all primes, power of prime(i can give you the proof) and number which have g.c.d.(n,Sigma(n))=1
So here n = 10
And Sigma(n=10)=18[can be found easily
so g.c.d.(10,Sigma(10))=2
therefore 10 is not solitary number 
in case of any defect you can mail me the defect in the proof
thanks
