Question concerning simple proof

Question

I am having trouble trying to prove the following statement:

Let $a,b \in \mathbb{Z}$ . If $ab = a+b$, then $a = b$.

(In truth, the statement was a prove/disprove, but I assumed it to be true).

I initially tried to prove the contrapostive, letting $b = a+k$ and aiming to show that $ab \ne a+b$, and then tried to prove it directly using basic algebra rules/manipulation, or at least through a simple contradiction (e.g., get to a point where one side is an integer, and the other isn't). But I have yet to make progress on either, so I would appreciate a nudge in the right direction.

Thank you kindly!

Note: sorry for the vague tags, I wasn't sure which to put for this particular question.

Answer 1

Hint: assume $a,b \ne 0\,$, then $a(b-1)=b \implies a \mid b$. By symmetry, also $b \mid a$. Therefore $a=b$.

Answer 2

Hint 1: the thing you are trying to prove doesn't happen often (if it is true, $a^2=2a$ just in 2 cases), so not by induction;

Hint 2: it works if both are $0$. $ab=a+b \rightarrow a(b-1)=b$ assume one of them, say $b \neq 0$ and check for divisibility: $b$ is prime w.r.t $b+1$ so...?

Answer 3

HINT: $\;ab=a+b\iff (a-1)(b-1)=1.$