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

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

This problem is designated to

1. Testing our patience to the utmost limit, and

2. At the same time probing our problem solving skills, and last but not least,

3. If we have the persistent and perseverance to accomplish goal in fully tackling for this problem.

And the solution below is provided by a mathematics genius from Uruguay.

Okay, at first glance, this is a polynomial with not only quadratics in both $a$ and $b$, it has also the term $a^2b,\,\,ab^2$.

But, we can still force it to become a quadratic in $a$, if we rewrite it as follows:

Wait a minute, the goal to relate the given equation to a quadratic in $a$ (We could also try out the other option by having a quadratic in $b$.) is we want to gain a perspective from its discriminant, hopefully, the discriminant is a beautiful expression that we could use it as a stepping stone in our approach of solving this problem for integers $a$ and $b$.

That is to say:

4. The discriminant must be greater than or equal to zero,

5. The discriminant must be a perfect square.

Thus, our focus now is on finding the discriminant:

$ab(a+b-10)+21a-3a^2+16b-2b^2=60$

$a^2b+ab^2-10ab+21a-3a^2+16b-2b^2=60$

$a^2(b-3)+a(b^2-10b+21)+16b-2b^2-60=0$

The discriminant of the quadratic equation above

$=(b^2-10b+21)^2-4(b-3)(16b-2b^2-60)$

$=((b-3)(b-7))^2-4(b-3)(16b-2b^2-60)$

$=(b-3)^2(b-7)^2-4(b-3)(16b-2b^2-60)$

$=(b-3)((b-3)(b-7)^2-4(16b-2b^2-60)$

$=(b-3)(b^3-9b^2+27b+93)$

[MATH]\color{black}=(b-3)(\color{yellow}\bbox[5px,purple]{b^3-3b^2(3)+3b(3)^2-3^3}\color{black}+93+3^3)[/MATH]

[MATH]\color{black}=(b-3)(\color{yellow}\bbox[5px,purple]{(b-3)^3}\color{black}+93+27)[/MATH]

[MATH]\color{black}=(b-3)(\color{yellow}\bbox[5px,purple]{(b-3)^3}\color{black}+120)[/MATH]

$=(b-3)^4+120(b-3)$

It's not hard to deduce by now that

[MATH]\color{yellow}\bbox[5px,green]{(b-3)^4\lt (b-3)^4+120(b-3)}[/MATH] for $b-3\gt 3$.

We rule out the possibility where $b=3$ since 

$3a(a+3-10)+21a-3a^2+16(3)-2(3)^2-60$

$=3a^2+9a-30a+21a-3a^2+48-18-60$

$=\cancel{3a^2}+\cancel{9a-30a+21a}-\cancel{3a^2}-30$

$\ne 0$

Here comes the most challenging part of the problem...

If we can show that $b\lt \text{some positive small but not large number}$, then we could try each case out to determine all the possible value of positive integer for $b$.

E.g. if we have showed that $b\lt 5$ and that $b$ must be a positive integer, then we only have to consider the cases where $b=1,\,2,\,3,\,4$.

The problem is, how do we do that?

Okay, if we can set up the following inequality

[MATH]\color{black}\bbox[5px,orange]{(b-3)^4\lt (b-3)^4+120(b-3)\lt ((b-3)^2+1)^2}[/MATH] and determine the range of $b$ which it does not satisfy the set inequality, then we've succeeded half way.

The reason why we want to look for the range of $b$ that it falls out of the safety interval that satisfy the above inequality is due to the fact that the discriminant ($(b-3)^4+120(b-3)$) is a perfect square. Therefore, any $b$ that satisfies the above inequality means $(b-3)^4+120(b-3)$ isn't a perfect square.