Bad proof that if $a + b + ab = 2020$ then $a+b=88$ Can you prove this:

Let $a,b \in \mathbb{N}$. If $a + b + ab = 2020$ then $a+b=88$.

This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$
Substituting for $b$ using the $2$nd equation.
$2020-88 = a (88-a)$
That is a quadratic that is easily solved and gets $a = 42$ and $a = 46$.
So we have $a = 42$ and $b = 46$ or $a = 46$ and $b = 42$.
Was the question answered? A valid proof?

You demonstrated the that both can co-exist and have a solution:

$\exists! (a, b) \in ℕ$ such as $a+b=88$ and $a+b+ab=2020$

Now if you can prove the uniqueness of the second part

$\exists! (a, b) \in ℕ$ such as $a+b+ab=2020$

That would be sufficient. (however, proving the uniqueness of the solution is quite likely to result in actually finding that solution, rendering the first part of the reasoning useless)
