Help‽ Solving algebraic equation in 2 variables‽

[Turtle]

I have this equation that has stumped me and I need some help/suggestions for solving it.

The equation is: (b/2)*((a-2)*b)-a+4)=68,921

With a and b restricted to >=3, I want to solve for a & b and in particular, integer solutions. I appreciate any help. Takk

[CraigD]

Originally Posted by Turtle

The equation is: (b/2)*((a-2)*b)-a+4)=68,921

The equation is not well formed – it has more “)”s than “(“s.

In the words of countless stereotypical space-opera robots, “does not compute! does not compute!”

[Turtle]

Originally Posted by CraigD

The equation is not well formed – it has more “)”s than “(“s.

In the words of countless stereotypical space-opera robots, “does not compute! does not compute!”

Weelforming now....

((b/2)*((a-2)*b)-a+4)=68,921

Ken we make it quadratic? I am reticent to cast my bias, as that is likely wut hast me hornswaggled. Takk.

[CraigD]

(b/2)*((a -2)*b) -a +4 =68921
Can be rearranged into
b*((a -2)*b) -2a +8 =137842
and
(a -2)*b^2 -(2a +137834) =0
Allowing b to be solved in terms of a via the quadratic equation as
b = +-sqrt((a-2)(2a +137834)) / (a -2)
which expands to
b = +-sqrt(2a^2 +137830a -275668) / (a -2)

Next step, I’d substitute
c = 2a^2 +137830a -275668
rearrange to
2a^2 +137830a +(c -275668) = 0
And solve for a in terms of c.

That’s as far as I’ve gotten before bedtime. Noticing that
(b/2)*((a -2)*b) -a +4
increases monotonically with a and b, I had a computer find all
(b/2)*((a -2)*b) -a +4 <= 68921
for the first 50,000,000 integer values of a, and found no
(b/2)*((a -2)*b) -a +4 = 68921,
which makes suspect it has no integer solution. I’d like an algebraic proof. I’ll continue work on one later, or until someone beats me to it.

[Turtle]

Originally Posted by CraigD

That’s as far as I’ve gotten before bedtime. Noticing that
(b/2)*((a -2)*b) -a +4
increases monotonically with a and b, I had a computer find all
(b/2)*((a -2)*b) -a +4 <= 68921
for the first 50,000,000 integer values of a, and found no
(b/2)*((a -2)*b) -a +4 = 68921,
which makes suspect it has no integer solution. I’d like an algebraic proof. I’ll continue work on one later, or until someone beats me to it.

Wonderful!! I am making a copy of your work for study. Your suspicion I support on grounds of the computer run. An interesting result so far, particularly as I chose the constant somewhat at random. I have to brush up my quadratic equation; I've been using it as a doorstop.

[salivating]So your program...MUMPS? [/salivating] We may have more use of it as I mean to get at a general case of the equation, which may provide the path to that proof you mentioned.

Well, don't let me keep you up.

Others' comments? Takk Alles

[Turtle]

Originally Posted by Turtle

Wonderful!! I am making a copy of your work for study.

OK, the first two lines you gave Craig.
(b/2)*((a -2)*b) -a +4 =68921
Can be rearranged into
b*((a -2)*b) -2a +8 =137842

This rearrangment comes by way of multiplying both sides by 2, correct? If so, I concur with 137842 on the right side, but oughtn't the multiplication by 2 on the left simply cancel with the denominator 2 of (b/2) and so not change the rest of the expression? I get the rearrangement
b*((a-2)*b-a+4)=137842

Following your method then I get (a-2)*b^2-ab+4b=137842.

I'll stop there on the assumption I've made another error; we can pick up with it tomorrow. Thanks again.

PS I had to read up on the definition of 'monotonic' as you used it. Do you mean as the variable values rise, then the output value also rises in an orderly fashion?

[Jay-qu]

my solutions for b are +-

[Turtle]

Originally Posted by Jay-qu

my solutions for b are +-

I'll take you at your word, as I can't see your work. But, on your word, of the value of b, what is a? How many solutions are there? Are any integers?

I am not just being difficult or joky, as some tell me I have a reputation for. The method is as important as the result in this case. By all means please continue and thank you for your help.

[Jay-qu]

assuming that is correct, I plugged it back into the original equation to get an expression for a



I do not look forward to using the cubic formular on it.. but I wade through the algebra and see what I come up with, post results in a little while

[Turtle]

Addendum & Further Musings:

I just realized by JayQ's response that I may have given the impression I know the solution. I do not know the solution. Ergo the interrobanged 'Help‽' in the title meant to imply 'I really need it, can-you/will-you give it!?"

I forgot the musings part as I drifted into...uh....


Oh, I wanted to say that that substitution Craig did by adding variable c has something of the right sense. I haven't quite grabbed it yet, but there's the details of what is substitued for to work out.

Oh, also that I am a turtle for this algebra, which is to say I have to see every step else I don't get it. So thank you Craig for shewing yur werk. I take it you have worked with me before?

Nothing further....