## Monday, July 20, 2015

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

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

Second solution:

This second solution is much neater than the first, but in solving mathematical problems, solving it is what matters, isn't it? :D

So, I really shouldn't compare one to the other, so I retract back what I said...

We can rewrite the given equation in such a way that:

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

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

[MATH]\color{yellow}\bbox[5px,purple]{a^2(b-3)+b^2(a-2)-10ab+21a+16b=60}[/MATH]

It's hard not to be tempted to try out the cases when $a=2$ and $b=3$ to see what we got...hey, aren't we all taught to prompt into action whenever we saw something like that? :P

When $a=2$, we have:

[MATH]\color{yellow}\bbox[5px,purple]{a^2(b-3)+b^2(a-2)-10ab+21a+16b=60}[/MATH]

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

$4b-12+0-20b+42+16b=60$

$30=60$

When $b=3$, we have:

[MATH]\color{yellow}\bbox[5px,purple]{a^2(b-3)+b^2(a-2)-10ab+21a+16b=60}[/MATH]

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

$0+9a-18-30a+21a+48=60$

$30=60$

In other words, $a=2$ and $b=3$ are roots to the equation $y=ab(a+b-10)+21a-3a^2+16b-2b^2-30$.

That is, $a-2$ and $b-3$ are two factors of $y=ab(a+b-10)+21a-3a^2+16b-2b^2-30$.

Note that $(a-2)(b-3)=ab-3a-2b+6$, so now we can perform the polynomial long division to factor $y$ completely:

[MATH]\begin{array}{r}a+b-5\hspace{160px}\\ab-3a-2b+6\enclose{longdiv}{a^2b+ab^2-10ab-3a^2+21a-2b^2+16b} \\ -\underline{\left(a^2b+0ab^2-2ab-3a^2+6a\right)} \hspace{88px} \\ ab^2-8ab+15a-2b^2+16b-30 \hspace{30px} \\ -\underline{\left(ab^2-3ab+0a-2b^2+6b\right)} \hspace{70px} \\ -5ab+15a+10b-30 \hspace{70px} \\ -\underline{\left(-5ab+15a+10b-30 \right)} \hspace{60px} \\  0 \hspace{70px} \end{array}[/MATH]

Therefore, to summarize, what we have gotten now is the following:

$ab(a+b-10)+21a-3a^2+16b-2b^2-30=(a-2)(b-3)(a+b-5)$

So, in order to solve for the positive integers for $ab(a+b-10)+21a-3a^2+16b-2b^2=60$, it is equivalent to solve for the positive integers for

$ab(a+b-10)+21a-3a^2+16b-2b^2-30=30$

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

Note that we could cross the ranges for both $a$ and $b$ where $a\le 2$ and $b-3\le 0$, that is a really useful info, since we are dealing with the product of three positive terms that is equal to $30$:

$(a-2)(b-3)(a-2+b-3)=1\cdot 30=1\cdot 2\cdot 15=1\cdot 2\cdot 3\cdot 5$

Out of the possibilities above, it's obvious that

\begin{align*}(a-2)(b-3)(a-2+b-3)&=1\cdot 2\cdot 3\cdot 5\\&=1\cdot 5\cdot 6\\&=1\cdot 5\cdot (1+5)\\&\stackrel{\text{or}}{=}5\cdot 1\cdot (5+1)\end{align*}

which results in $(a,\,b)=(3,\,8),\,(7,\,4)$.

Or

\begin{align*}(a-2)(b-3)(a-2+b-3)&=2\cdot 3\cdot 1\cdot 5\\&=2\cdot 3\cdot 5\\&=2\cdot 3\cdot (2+3)\\&\stackrel{\text{or}}{=}3\cdot 2\cdot (3+2)\end{align*}

which results in $(a,\,b)=(4,\,6),\,(5,\,5)$.

Therefore, the answers to this problem are:

$(a,\,b)=(3,\,8),\,(4,\,6),\,(5,\,5),\,(7,\,4)$

Do you think I have done nothing wrong so far? I encourage you to think about it, and I will tell what doesn't sound okay in my last step to look for the positive integers of $a$ and $b$.