Help‽ Solving algebraic equation in 2 variables‽

[Qfwfq]

http://hypography.com/forums/143416-post18.html

 

Turtle said: "Take the original equation, drop the constant, let a=3 and b=4 and then solve. Whichever bracketing gives the answer 10 is correct."

 

That's how I rolled. :warped:

 

[EDIT] Btw if you were referring to my post #17, the purpose was to point out that different bracketing would NOT give the same results. That's why I asked for clarification. :)

 

moo

Moo, the purpose of my post was to say that, while the two version you gave are not equivalent, those posted by Craig ARE.

 

I was very brief Friday because I had no more time, with more of it I could have shown that only a finite range of values needs to be checked. I don't understand why people are agreeing that ((b/2)*((((a-2)*;)-a)+4)) isn't the same as (b/2)*(((a-2)*:)-a+4), those extra parentheses don't change anything. I can try to be of help if there are no doubts about the algebra.

[moo]

I don't understand why people are agreeing that ((b/2)*((((a-2)*:warped:-a)+4)) isn't the same as (b/2)*(((a-2)*;)-a+4), those extra parentheses don't change anything.

[EDIT] Actually they simply clarify the order of calculation for the "-a+4" part.

 

moo

[Qfwfq]

And those two orders give the same results.

 

►Inputting integer values for a and b always gives integer values for c.

Have you tried it with even b and odd a?

►Some integers c have no integer solutions for both a and b. (These are the buggers I want!)

That changes things somewhat, I can show that for a given c such as the one you initially posted a finite search is conclusive, I'll give a thought about arbitrary c when I can. It should be possible to check all c up to a given max without excessive difficulty.

:D

[moo]

And those two orders give the same results.

Tell those who were getting different results until the parenthesis were added. I only did it the one way myself. :D

 

moo

[Qfwfq]

It should be possible to check all c up to a given max without excessive difficulty.

:D

Yeah, sure, the expression increases monotonically with a (providing b > 1) and with b (providing a > 2) so it's enough to iterate over both, terminating the inner loop when the given max is passed and the outer one when the inner one terminates straight away. Further, for even values of b it's enough to consider even values of a so I would make the outer loop in b and alternate the inner one accordingly.

:cup:

 

P. S. This would of course be like a sieve, eliminating c values as they come up so as to leave any ones that have no solution.

[moo]

Qfwfq, one problem here is that the equation you compared is not the original. ;)

 

Original... ((b/2)*((a-2)*:D-a+4) <- from Turtle's post #3

Yours...... (b/2)*(((a-2)*:cup:-a+4)

 

moo

[Turtle]

:doh:

Yeah, sure, the expression increases monotonically with a (providing b > 1) and with b (providing a > 2) so it's enough to iterate over both, terminating the inner loop when the given max is passed and the outer one when the inner one terminates straight away. Further, for even values of b it's enough to consider even values of a so I would make the outer loop in b and alternate the inner one accordingly.

:cup:

 

P. S. This would of course be like a sieve, eliminating c values as they come up so as to leave any ones that have no solution.

 

This is more or less to my original intent, that is whether or not we can find values of c with no integer solutions in a & b by means other than sieving or trial-and-error. It was in iterating for all integer values of a & b to make a list of c's that I noticed that some integer values of c were missing.

For a & b >3, here is my short list of integers c with no integer solutions in a & b.

{7:8:11:13:14:17:19:20:23:26:29:31:32:37:38:41:43: 44:47:50:53:56...}

 

That would be so cool :D

 

ok, through some re-arranging I was able to form a quadratic.

 

[math]\frac{b}{2}((((a-2)b-a)+4)=c[/math]

[math]\frac{b}{2}(ab-2b-a+4)=c[/math]

[math]c=\frac{ab^2}{2}-b^2-\frac{ab}{2}+2b[/math]

[math]c=(a-1)b^2+(2-a)b[/math]

and there is your quadratic :)

 

:steering:

I found some old notes in which I tried to work this out myself, and I have the same for the first 2 lines you give. For the next line however I wrote:

(ab^2-2b^2-ab+4b)/2

It's still early for me & I need to convince myself we are not in disagreement, but rather it's a matter of a different form.

Mooving along iteratively,

Turtle

[Jay-qu]

now that I look at that again it looks wrong :naughty:

[Jay-qu]

More like:

 

[math]\frac{b}{2}((((a-2)b-a)+4)=c[/math]

[math]\frac{b}{2}(ab-2b-a+4)=c[/math]

[math]c=\frac{ab^2}{2}-b^2-\frac{ab}{2}+2b[/math]

[math]c=(\frac{a}{2} -1)b^2+(2-\frac{a}{2})b[/math]

 

which then gives

 

[math]b=\frac{-(2-\frac{a}{2})+-\sqrt{(2-\frac{a}{2})^2-4(\frac{a}{2}-1)(-c)}}{2(\frac{a}{2}-1)}[/math]

[Qfwfq]

Qfwfq, one problem here is that the equation you compared is not the original. ;)

 

Original... ((b/2)*((a-2)*b)-a+4) <- from Turtle's post #3

Yours...... (b/2)*(((a-2)*b)-a+4)

 

moo

Well, blow me down! :doh: But, could you and Craig not have said so more clearly and sooner? I would have wasted a bit less time yesterday.

 

The clearest thing is to knock out the superfluous parentheses instead of adding even more of them...

 

((b/2)*((a-2)*b)-a+4) = (b/2)*((a-2)*b)-a+4

 

or even more clearly:

 

[math]\frac{(a-2)b^2}{2}-a-4[/math]

W LaTeX!!!!

/forums/images/smilies/mad_2.gif

 

<Qfwfq wearily gropes for Friday's scribblings and... yup, there it is, just like the above!>

[moo]

But, could you and Craig not have said so more clearly and sooner? I would have wasted a bit less time yesterday.

Lol sorry, it's a fluke that I noticed it at all given a severe lack of sleep at the time. :doh:

 

moo

[Turtle]

Regardless of which equation is used, the question “for what values of c are there integer solutions a and b?” remains interesting.

...

I’d love to see a computationally brief rule deciding if a solution exists for a give c, but suspect that there is a deep/hard problem at the root of this simple question (I’ve a specific one in mind). I’d love to see someone prove that, and describe the hard problem. I’ll be working on it myself :)

 

My bad :doh: ;I never gave the origin of the expression and deep question directly. :) ;)

 

The general question is what integer values have no integer solutions for the expression (b/2)*((a-2)*B)-a+4 when a & b>= 3. That expression is the generalized expression for figurate (otherwise called polygonal numbers) where a is the number of sides and b is the ordinal in the list of a-sided numbers. For example, if the question is what is the 4th triangular number (3-sided), then substituting 3 for a and 4 for b and solving the equation we get 10; 10 is the 4th triangular number.

 

So, my general deep question is what are, and how do we find, the integers that are not figurate? (note all integers c are trivially figurate when a=c and b=2)

 

That's all I got to add. :hihi: From post #58:

For a & b >3, here is my short list of integers c with no integer solutions in a & b.

{7:8:11:13:14:17:19:20:23:26:29:31:32:37:38:41:43: 44:47:50:53:56...}

:D ..............:eek: