Bad proof that if $a + b + ab = 2020$ then $a+b=88$

$a+b+ab=2020 \iff (a+1)(b+1)=2021=47 \cdot 43$

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.

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.