Thursday, July 23, 2015

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

I want to apology for posting the unfinished blog post on last Saturday (solve-in-positive-integers-equation.html), and I am truly sorry for that.

Here is the other part of the solution that I unwittingly left out.

[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.

Please bear in mind that $(b-3)^4+120(b-3)$ must be a perfect square!

We know there is no way $(b-3)^4+120(b-3)$ could be a perfect square for any $b$ positive integer values that satisfy the above inequality. Therefore, we need to find out that range of values of $b$ that satisfies the above inequality and then and only then we look for each case for the values of $b$ outside that range, since $(b-3)^4+120(b-3)$ must be a perfect square.

Solve for the LHS inequality, we have:

$(b-3)^4\lt (b-3)^4+120(b-3)$

$\cancel{(b-3)^4}\lt \cancel{(b-3)^4}+120(b-3)$

$3\lt b$

Solve for the RHS inequality, we got:

$(b-3)^4+120(b-3)\lt ((b-3)^2+1)^2$

$(b-3)^4+120(b-3)\lt (b-3)^4+2(b-3)^2+1$

$\cancel{(b-3)^4}+120(b-3)\lt \cancel{(b-3)^4}+2(b-3)^2+1$

$0\lt 2(b-3)^2-120(b-3)+1$

$2b^2-132b+379\gt 0$

Since $b$ is integer, so we obtain $b\le 3$ or $b\ge 63$.

Combining both results from the LHS and RHS ranges we got $3 \le b\gt 63$.

Thus, we need to consider the cases for which $3\le b\le 62$ so to determine all the positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

Wow...that seems like there are a total of $63$ (including $0$) cases to look for...but then we know if we continue and checking out all cases from $b=0,\,1,\,2,\cdots,62$, we will reach to the correct answers for certain.

$(3-3)^4+120(3-3)=0$ gives $a=\text{complex solutions}$

$(4-3)^4+120(4-3)=121=11^2$ gives $a=-4,\,7$

$(5-3)^4+120(5-3)=121=16^2$ gives $a=-3,\,5$

$(6-3)^4+120(6-3)=21^2$ gives $a=-3,\,4$

$(7-3)^4+120(7-3)\ne\text{perfect square}$

$(8-3)^4+120(8-3)=35^2$ gives $a=-4,\,3$

$(9-3)^4+120(9-3)\ne\text{perfect square}$

$(10-3)^4+120(10-3)\ne\text{perfect square}$

$(11-3)^4+120(11-3)\ne\text{perfect square}$

$(12-3)^4+120(12-3)\ne\text{perfect square}$

$\,\,\,\,\,\,\,\vdots$

$(62-3)^4+120(62-3)\ne\text{perfect square}$

Therefore, the answers to this problem are:

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