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?
 A: $a+b+ab=2020 \iff (a+1)(b+1)=2021=47 \cdot 43$
A: Your proof is not valid, assuming I understood you correctly.

This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$

The problem is you started out by assuming $a + b = 88$.  You just assumed what you wanted to prove!
Here is the question again:

Can you prove this:
if $a,b$ are positive integers,
and if $a + b + ab = 2020$,
then $a+b=88$?

Notice that the fact that $a, b$ are positive integers is very necessary here.  Otherwise, you could pick any rational $a,b$ with $(a + 1)(b + 1) = 2021$ (in this case $a$ can be any rational number) and you would have a solution.  In general $a + b \ne 88$ if $a,b$ are rational.
Thus your proof should be suspicious: you haven't used any properties about integers, as far as I can see. Your proof would also conclude that $a + b = 88$ if $a,b$ are rational, and this is not a true result!  Therefore, your proof cannot be valid.
The correct proof is as leticia gives:  write $(a + 1)(b+1) = 2021 = 47 \cdot 43$, and use prime factorization -- a property of positive integers -- to derive your result.
A: Key Idea $\ $ Completing a square generalizes to two-variables $\,x,y\,$ to completing a product
$$\begin{eqnarray} && xy + bx + cy\\
&=\,& x(y\!+\!b) + cy \\
&=\,& x(y\!+\!b) + c(y\!+\!b)-cb \\
&=\,& \qquad\, (x\!+\!c)(y\!+\!b) - cb
\end{eqnarray}\qquad\qquad\quad$$
${\rm So}\ \ b=1=c\ \ {\rm yields}\ \ n = xy\!+\!x\!+\!y = (x\!+\!1)(y\!+\!1)-1\!\iff\! n\!+\!1 = (x\!+\!1)(y\!+\!1)\ $ which has solutions $\,x,y = j\!-\!1,k\!-\!1\,$ for each factorization $\,n\!+\!1 = jk.\,$
Remark $\ $ In the same way we can show that if $\,a\ne 0\,$ then
$$\ axy\! +\!bx\!+\!cy = n\!\! \overset{\ \times\,a}\iff\!  (ax\!+\!c)(ay\!+\!b) = an\!+\!cb$$
Alternatively one can  mechanically derive the above from the first special case $(a = 1)$ using the $\rm AC$-method as described here.
A: 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)
