Potential Flaw in The Second Solution for the problem $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

In this blog post, I will discuss why the last step in my attempt to look for the positive integers $a$ and $b$ doesn't sound so right.

Let's see from the point where we have gotten up to

$(a-2)(b-3)(a+b-5)=30$

$(a-2)(b-3)(a-2+b-3)=30$

Note that not only we have to consider the cases where:

1. $a-2\lt 0$ and $b-3\lt 0$,

2. $a-2\gt 0$ and $b-3\gt 0$,

We have to also consider the cases where

3. $a-2\lt 0$, and $b-3\gt 0$,

4. $a-2\gt 0$ and $b-3\lt 0$.

Then and only then we can safely list out all the possible solutions that we have found.

In my previous post (my last step), I only took care of the first two cases, and stupidly ignored the last two cases:

For the third case, we see that:

$a-2\lt 0$ and we note that $a\ne 2,\,b\ne 3$ or the LHS of the expression would become zero

and see where that leads us.

When $a-2\lt 0$,

$(\text{-ve})(b-3)(a-2+b-3)=30$

We can therefore set $b-3\gt 0$ to get:

$(-\text{ve})(+\text{ve})(-\text{ve}+\text{ve})=30$

In order to get the product of positive $30$, the last factor must yield a negative, i.e.

$a-2+b-3\lt 0$

$a\lt 5-b$

$0\lt a\lt 5-b$

This yields $b\lt 5$.

Combining both inequalities $b-3\gt 0$ and $b\lt 5$, this leads to $b=4$ and substituting this back into the equation
$(a-2)(b-3)(a+b-5)=30$ gives $a=7$.

Note that $b-3\lt 0$ has already been addressed in the first case.

For fourth case, we have:

We have to also take into consideration the possibility where $a-2\gt 0$:

When $a-2\gt 0$,

$(+\text{ve})(b-3)(a-2+b-3)=30$

We then let $b-3\lt 0$, so we have:

$(+\text{ve})(-\text{ve})(+\text{ve}+(-\text{ve}))=30$

In order to get the product of positive $30$, the last factor must yield a negative, i.e.

$a-2+b-3\lt 0$

$a\lt 5-b$

$0\lt a\lt 5-b$

This yields $b\lt 5$.

Combining both inequalities $b-3\lt 0$ and $b\lt 5$, this leads to $b=1,\,2$ of which there are no solutions for these two values of $b$.

For $a-2\gt 0$ and $b-3\gt 0$, that is just the last case, the one I covered in the last blog post. I will hence stop it from here, you can follow this link to check out the full solution.

All in all, we have to think and cover for ALL cases if we want to fully solving any given mathematical problem for best outcome.