Help‽ Solving algebraic equation in 2 variables‽

[moo]

The two equations Craig posted are equivalent. The two that Moo posted aren't.

 

N - (a + 4) is like N - a -4

 

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. B)

 

[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]

However, the fact remains that the original equation

((b/2)*((a-2)*:)-a+4)=68921

Isn’t equivalent to

((b/2)*((((a-2)*B)-a)+4))=68921.

Agreed, but the first thing I asked was for clarification on the unbracketed "-a+4" portion. And the formula I used as a result does work with whole numbers. B)

 

moo

[CraigD]

Agreed, but the first thing I asked was for clarification on the unbracketed "-a+4" portion. And the formula I used as a result does work with whole numbers.

You are correct, and I applaud your skill with using numeric methods. B)

 

It’s interesting to note that a slight changes in the right side constant of the original equation, such as

((b/2)*((a-2)*:)-a+4)=68920

can, but doesn’t always causes it to have whole number solutions (a=4056, b=6).

 

So, what about the question about a general case, of either or both equations: “for what values of c are there integer solutions a and b?”

[moo]

I applaud your skill with using numeric methods.

Thanks, but I am nowhere near you and these other guys. :)

 

So, what about the question about a general case, of either or both equations: “for what values of c are there integer solutions a and b?”

No idea. I'm really not very good with algebra, especially concerning formulas of more than one unknown. Most of my math requirements are programming related and fairly straightforward (hence the hack to run down these answers). :)

 

moo

[Turtle]

Morning Everybody! Just woke, so I'm still groggy. Awake enough to see good progress.

 

I think I mis-wrote the equation originally, per the bracketing issue moo et all brought up. So indeed it should have read:

((b/2)*((((a-2)*:)-a)+4))=68921

 

So the solutions moo gave are correct it seems. :) I am after the non-trivial solutions, that is a & b >=3.

 

I did program this at some time in the past to spit out the list, but obviously I missed the extra parentheses; likely because I knew what result to expect so just worked it correct.

 

I will explain further this afternoon as I have to go for a flu shot and I'm still too groggy to address all the issues brought up, and they are all germane.

 

Great stuff everyone; most beneficial. Turtle Happy! :)

[CraigD]

I think I mis-wrote the equation originally, per the bracketing issue moo et all brought up. So indeed it should have read:

((b/2)*((((a-2)*:)-a)+4))=68921

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

 

While I was writing this post, I had my computer find solutions for values of c from 68900 to 69001 (when I hit the “break” key), with the following results:

((b/2)*((((a-2)*:)-a)+4))=68900
((b/2)*((((a-2)*:)-a)+4))=68901 a=56 b=51
((b/2)*((((a-2)*:eek_big:-a)+4))=68902 a=11485 b=4
((b/2)*((((a-2)*:eek_big:-a)+4))=68903
((b/2)*((((a-2)*:eek_big:-a)+4))=68904 a=22969 b=3
((b/2)*((((a-2)*:eek_big:-a)+4))=68905 a=1533 b=10
((b/2)*((((a-2)*:eek_big:-a)+4))=68906
((b/2)*((((a-2)*:eek_big:-a)+4))=68907 a=22970 b=3
((b/2)*((((a-2)*:eek_big:-a)+4))=68908 a=3283 b=7
((b/2)*((((a-2)*:eek_big:-a)+4))=68909
((b/2)*((((a-2)*:eek_big:-a)+4))=68910 a=22971 b=3
((b/2)*((((a-2)*:eek_big:-a)+4))=68911
((b/2)*((((a-2)*:hihi:-a)+4))=68912
((b/2)*((((a-2)*B)-a)+4))=68913 a=1916 b=9
((b/2)*((((a-2)*B)-a)+4))=68914 a=11487 b=4
((b/2)*((((a-2)*B)-a)+4))=68915 a=6893 b=5
((b/2)*((((a-2)*B)-a)+4))=68916 a=1046 b=12
((b/2)*((((a-2)*B)-a)+4))=68917
((b/2)*((((a-2)*B)-a)+4))=68918
((b/2)*((((a-2)*B)-a)+4))=68919 a=22974 b=3
((b/2)*((((a-2)*B)-a)+4))=68920 a=11488 b=4
((b/2)*((((a-2)*B)-a)+4))=68921 a=86 b=41
((b/2)*((((a-2)*B)-a)+4))=68922 a=82 b=42
((b/2)*((((a-2)*B)-a)+4))=68923
((b/2)*((((a-2)*B)-a)+4))=68924
((b/2)*((((a-2)*B)-a)+4))=68925 a=6894 b=5
((b/2)*((((a-2)*B)-a)+4))=68926 a=214 b=26
((b/2)*((((a-2)*B)-a)+4))=68927
((b/2)*((((a-2)*B)-a)+4))=68928 a=22977 b=3
((b/2)*((((a-2)*B)-a)+4))=68929 a=3284 b=7
((b/2)*((((a-2)*B)-a)+4))=68930
((b/2)*((((a-2)*B)-a)+4))=68931 a=4597 b=6
((b/2)*((((a-2)*B)-a)+4))=68932 a=100 b=38
((b/2)*((((a-2)*B)-a)+4))=68933
((b/2)*((((a-2)*B)-a)+4))=68934 a=22979 b=3
((b/2)*((((a-2)*B)-a)+4))=68935 a=6895 b=5
((b/2)*((((a-2)*B)-a)+4))=68936
((b/2)*((((a-2)*B)-a)+4))=68937 a=22980 b=3
((b/2)*((((a-2)*B)-a)+4))=68938 a=11491 b=4
((b/2)*((((a-2)*B)-a)+4))=68939
((b/2)*((((a-2)*B)-a)+4))=68940 a=22981 b=3
((b/2)*((((a-2)*B)-a)+4))=68941
((b/2)*((((a-2)*B)-a)+4))=68942
((b/2)*((((a-2)*B)-a)+4))=68943 a=22982 b=3
((b/2)*((((a-2)*B)-a)+4))=68944 a=2464 b=8
((b/2)*((((a-2)*B)-a)+4))=68945 a=6896 b=5
((b/2)*((((a-2)*B)-a)+4))=68946 a=4598 b=6
((b/2)*((((a-2)*B)-a)+4))=68947
((b/2)*((((a-2)*B)-a)+4))=68948
((b/2)*((((a-2)*B)-a)+4))=68949 a=1917 b=9
((b/2)*((((a-2)*B)-a)+4))=68950 a=1534 b=10
((b/2)*((((a-2)*B)-a)+4))=68951
((b/2)*((((a-2)*B)-a)+4))=68952 a=95 b=39
((b/2)*((((a-2)*B)-a)+4))=68953 a=52 b=53
((b/2)*((((a-2)*B)-a)+4))=68954
((b/2)*((((a-2)*B)-a)+4))=68955 a=6897 b=5
((b/2)*((((a-2)*B)-a)+4))=68956 a=11494 b=4
((b/2)*((((a-2)*B)-a)+4))=68957
((b/2)*((((a-2)*B)-a)+4))=68958 a=22987 b=3
((b/2)*((((a-2)*B)-a)+4))=68959
((b/2)*((((a-2)*B)-a)+4))=68960
((b/2)*((((a-2)*B)-a)+4))=68961 a=4599 b=6
((b/2)*((((a-2)*B)-a)+4))=68962 a=11495 b=4
((b/2)*((((a-2)*B)-a)+4))=68963
((b/2)*((((a-2)*B)-a)+4))=68964 a=22989 b=3
((b/2)*((((a-2)*B)-a)+4))=68965 a=886 b=13
((b/2)*((((a-2)*B)-a)+4))=68966
((b/2)*((((a-2)*B)-a)+4))=68967 a=22990 b=3
((b/2)*((((a-2)*B)-a)+4))=68968 a=11496 b=4
((b/2)*((((a-2)*B)-a)+4))=68969 a=509 b=17
((b/2)*((((a-2)*B)-a)+4))=68970 a=22991 b=3
((b/2)*((((a-2)*B)-a)+4))=68971 a=3286 b=7
((b/2)*((((a-2)*B)-a)+4))=68972 a=2465 b=8
((b/2)*((((a-2)*B)-a)+4))=68973 a=22992 b=3
((b/2)*((((a-2)*B)-a)+4))=68974 a=11497 b=4
((b/2)*((((a-2)*B)-a)+4))=68975 a=6899 b=5
((b/2)*((((a-2)*B)-a)+4))=68976 a=141 b=32
((b/2)*((((a-2)*B)-a)+4))=68977
((b/2)*((((a-2)*B)-a)+4))=68978
((b/2)*((((a-2)*B)-a)+4))=68979 a=22994 b=3
((b/2)*((((a-2)*B)-a)+4))=68980 a=11498 b=4
((b/2)*((((a-2)*B)-a)+4))=68981 a=1256 b=11
((b/2)*((((a-2)*B)-a)+4))=68982 a=1047 b=12
((b/2)*((((a-2)*B)-a)+4))=68983
((b/2)*((((a-2)*B)-a)+4))=68984
((b/2)*((((a-2)*B)-a)+4))=68985 a=1918 b=9
((b/2)*((((a-2)*B)-a)+4))=68986 a=11499 b=4
((b/2)*((((a-2)*B)-a)+4))=68987
((b/2)*((((a-2)*B)-a)+4))=68988 a=22997 b=3
((b/2)*((((a-2)*B)-a)+4))=68989
((b/2)*((((a-2)*B)-a)+4))=68990 a=365 b=20
((b/2)*((((a-2)*B)-a)+4))=68991 a=4601 b=6
((b/2)*((((a-2)*B)-a)+4))=68992 a=20 b=88
((b/2)*((((a-2)*B)-a)+4))=68993
((b/2)*((((a-2)*B)-a)+4))=68994 a=22999 b=3
((b/2)*((((a-2)*B)-a)+4))=68995 a=1535 b=10
((b/2)*((((a-2)*B)-a)+4))=68996
((b/2)*((((a-2)*B)-a)+4))=68997 a=23000 b=3
((b/2)*((((a-2)*B)-a)+4))=68998 a=11501 b=4
((b/2)*((((a-2)*B)-a)+4))=68999
((b/2)*((((a-2)*B)-a)+4))=69000 a=659 b=15
((b/2)*((((a-2)*B)-a)+4))=69001

Here is the MUMPS program that produced it

f v=68900:1 w !,"((b/2)*((((a-2)*B)-a)+4))=",v s h0=$h,c=0 f a=3:1 q:$h-h0*86400+$p($h,",",2)-$p(h0,",",2)>30!(c=v)  f b=3:1 s c=((b/2)*((((a-2)*B)-a)+4)) w:c=v " a=",a," b=",b q:c'<v

The program abandons looking for a solution after 30 seconds, so shouldn’t be considered completely authoritative.

 

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 :)

[Turtle]

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

Agreed, with the caveat that the correct expression is imperitive. In answer to Dave's question, "I'm just curious - why this particular equation? Is it just a random equation that you came up with, or does it have a real-world application?", the answer is no, it is not random, and yes it has application.

 

 

While I was writing this post, I had my computer find solutions for values of c from 68900 to 69001 (when I hit the “break” key), with the following results:[]The program abandons looking for a solution after 30 seconds, so shouldn’t be considered completely authoritative.

 

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 :)

 

Affirmative on the general correctness of the list. A smaller correct list is what has prompted this question, so bear with me as we work through the algebra. I want to make sure I understand your c. As near as I can tell from your list however, the values with no integer solutions are the buggers I'm after.

 

Breaking to clear cobwebs...

[Turtle]

Affirmative on the general correctness of the list. A smaller correct list is what has prompted this question, so bear with me as we work through the algebra. I want to make sure I understand your c. As near as I can tell from your list however, the values with no integer solutions are the buggers I'm after.

 

Breaking to clear cobwebs...

 

OK. Some things I know about ((b/2)*((((a-2)*;)-a)+4))=c

(Positive integer values only)

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

►If b=1 then c=1 for all integers a.

►All integers c have trivial solutions when a=c and b=2, and when b=c and a=2

►Some integers c have more than one pair of non-trivial integer solutions. For example, when c=15 there is the solution pair a=3 & b=5, as well as the pair a=6 & b=3.

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

►Break time! Smoke 'em if ya got 'em. :) er....:)

[Jay-qu]

Working on a general algebraic solution for the new equation (thanks turtle :)) in terms of c, churn through some stuff and see what comes out!

 

:)

[Turtle]

Working on a general algebraic solution for the new equation (thanks turtle :D) in terms of c, churn through some stuff and see what comes out!

 

:)

At least I knew to stop and ask directions when I saw I was lost. One has to also consider I may misinterpret the directions as well. ;)

I was thinking how different the discussion would be if we were all in a room together with blackboards. :eek: :smart: :smart: :smart: :smart: :) :smart:

Well, steady as she goes. Carry on. :thumbs_up

[Jay-qu]

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:

[Jay-qu]

solving the quadratic

[math]0=(a-1)b^2+(2-a)b-c[/math]

pending

:steering:

[Jay-qu]

heres what I found

[math]0=(a-1)b^2+(2-a)b-c[/math]

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

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

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

 

the quadratic inside the square root can be factorised, but its ugly..

 

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

 

I am going to keep playing around looking for a more elegant solution ;)

[Turtle]

heres what I found

...I am going to keep playing around looking for a more elegant solution :)

Acknowledged. I think I'll just idle along too and give Craig et al a chance to look over our work and say what they saw. :)

[moo]

In answer to Dave's question, "I'm just curious - why this particular equation? Is it just a random equation that you came up with, or does it have a real-world application?", the answer is no, it is not random, and yes it has application.

Do we get to find out what that "real-world application" is? :)

 

moo

[Turtle]

Do we get to find out what that "real-world application" is? ;)

 

moo

 

I'm glad you put that in quotes. B) Not only did I not want to 'spoil' the math with extant bias, I didn't want to have a harangue over what 'real' or 'application' means.:lol:

 

For a fuller exposition, the last dozen or so pages of the katabataks thread covers the expression in depth, as well as my need for assistance. :doh:

[Jay-qu]

I subbed that solution back into the original equation and played around with it for a while.. nothing really cool has fallen out.. yet B)

 

If I get a chance I will Latex it later today, it could take some time!