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 ,
$$\frac{\sigma(p)}{p}&gt;1$$
$$ \implies3 \cdot 2^{k-1}&gt;2^{k+1}-1$$
Which can be showed impossible for $k&gt;0$.

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 $\frac{\sigma(q)}{q}&gt;1$ can be shown not possible.

Thus no friendly pair of $10$ is possible. Thus $10$ is solitary.
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.
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