a+b=a*b, a-b=? if a and b are positive integers So this is a question off Facebook (I know). $a, b$ are positive integers.
The answer to the problem turns out to be $a-b=0$ as the only two obvious solutions are $a=b=0, 2$ (and we select two as it can't be zero).
My question is how do we prove this? I was trying
$a+b=ab$
$\dfrac{a}{a-1}=b$
But I can't proceed by substituting this in $a-b$ and solving the polynomial as I would have to first assume that $a-b=0$.
How do we prove this? How can we prove that the only value $a$ and $b$ can take is $2$ (and hence $a-b=0$)? Alternatively, how can we prove that if $a-b$ is non-zero, then $a$ and $b$ aren't integers? How can we prove that there are infinitely many solutions if the constraint is shifted to real numbers?
 A: Since $a =b(a-1),\;b|a$, and since $b=a(b-1),\;a|b$.
Therefore $a=b$, so $2a=a^2\implies a=2=b$.
A: $$a+b=ab$$
$$\frac{a+b}{ab}=1$$
This approach prevents either $a$ or $b$ being $0$.
$$\frac1a+\frac1b=1$$
The only integer values of $a,b$ that satisfy this are $a=b=2$
 ($a\gt2\implies1\lt b\lt2$).
We can see $a=b$ more directly by letting $b=ka$. Then $a(k+1)=ka^2\to a(k+1-ka)=0$. So either $a=0$ (and so $a=b$) or $a=\frac{k+1}{k}$. As $a\in\mathbb{Z}, k=1\to a=b=2$.
If we drop the constraint that $a\in\mathbb{Z}$, then for example consider $k=2$. Then $a=\frac32, b=3, a+b=ab=\frac92$.
Finally $a-b=ab-2b=b(a-2)$ which is only zero if $b=0$ or $a=2$.
A: $a+b = ab\\
a(1-b) +b = 0\\
a = \frac {b}{b-1}$
$b-1$ divides $b$  when $b = 0$ or $b-1 = 1$  i.e. everything divides $0,$ and $1$ divides everything.
$\gcd (b,b-1) = 1$  which can be demonstrated by the Euclidean algorithm.
There is no $b>2$ such that $(b-1)$ divides $b.$
A: As $a$ and $b $ are positive $a=b=0$ is not an acceptable solution.
As you note: $ab=a+b; ab-b =a;b=a/(a-1) $ if $a-1\ne 0$.  So $a-1|a $ so $a-1|a-(a-1)=1$.
So $a-1=1$ and $a=2$ and $b=2$
... unless $a-1=0$ but then we get the contradiction $1*b=1+b $
A: To solve this question we need another condition.
$$a=b$$
then solve value of a and b
$$a+b=ab$$
replace a to b cause we know $$a=b$$
$$b+b=b*b$$
$$2b=b^2$$
$$b^2=2b$$
$$b^2-2b=0$$
$$b(b-2)=0$$
$$b=(0,2)$$
as we know $$a=b$$
meaning $$a=(0,2)$$
so, $$a-b=0$$
