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Can you prove this:

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

This is the attempt given:

$\frac{2020-88}{a b}=1$


Substituting for b using the 2nd 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?

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I do not see in your proof how you get to $\frac{2020 - 88}{ab} = 1$. Can you explain this step?? Haven't you just assumed $a + b = 88$ to get to this step? – 6005 May 10 '14 at 17:53
Yes, I have, that is the problem. – bobbym May 10 '14 at 17:55
Are you trying to prove $a + b = 88$, or is that a "given" from which you are trying to derive the values $a, b$. I ask because you state the proposition as an "if, then" statement. To prove it, you can only assume the "if" condition, in order to prove the then conclusion. If that's what you're trying to do, you cannot assume (or use) what it is you are to prove. – amWhy May 10 '14 at 17:56
I do not know, the question is exactly as asked, what do you think? – bobbym May 10 '14 at 17:57
What you have proved is the other way around: if $a+b+ab=2020$ and $a+b=88$, then $a,b\in\mathbb N$. – user26486 May 11 '14 at 16:17
up vote 19 down vote accepted

Your proof is not valid, assuming I understood you correctly.

This is the attempt given:

$\frac{2020-88}{a b}=1$


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.

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Very nice answer. +1 – Prism May 10 '14 at 20:50

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

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That was one other proof provided. – bobbym May 10 '14 at 18:00
@bobbym See my answer for how to view this as a generalization of completing squares to completing products, a general method which works for any Diophantine equation of this form. – Bill Dubuque May 10 '14 at 20:56
@Bill Dubuque Looking at it now. – bobbym May 11 '14 at 2:38

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.

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Nice explanation. – rugi Feb 29 at 5:48

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)

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