ganeshie8
  • ganeshie8
find the integer ordered pairs (a, b) that satisfy a^3+19b = 22
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Said another way find all \(b\) such that\[\sqrt[3]{22-19b}\]is an integer
lochana
  • lochana
\[a,b \in \mathbb{Z} ?\]
ganeshie8
  • ganeshie8
yes for both...

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More answers

ikram002p
  • ikram002p
it has no integer solution i need to prove why i suppose xD
ikram002p
  • ikram002p
we need to show that a^3 can't be of the form 19b+16 maybe errrrm :-\ (just conjecturing)
ganeshie8
  • ganeshie8
do you mean a^3 cannot be of form 19b + \(\color{red}{22}\) ?
ikram002p
  • ikram002p
well 19b-22 so like -22 mod 19 :3
Kainui
  • Kainui
I thought it was positive integers only and thought it was simple :(
ganeshie8
  • ganeshie8
Haha I see.... its trivial if we allow only positive integers
ikram002p
  • ikram002p
@Kainui if its only positive proving it would be like a^3+19b>22 for a,b>2
ikram002p
  • ikram002p
?
Kainui
  • Kainui
Hahaha yeah it would be, but it's not that simple xD
Kainui
  • Kainui
Has anyone dared to try to show this yet? \[a^3 \cancel \equiv 3 \mod 19\]
imqwerty
  • imqwerty
we can agree to the fact that one of them (a or b) will be positive and the other will be negative \[a^3+19b=22\]\[a^3-2^3=14-19b\]\[(a-2)[a^2+4+2a]=14-19b\]\[(a-2)[(a+2)^2-2a]=14-19b\] now if b is negative then the right hand side is positive a has to be positive and a has to be >2 (a-2) becomes positive [(a+2)^2 - 2a] is negative so LHS=negative but the RHS is positive so no solution Now we have to prove it for b positive and a negative
Kainui
  • Kainui
Very nice @imqwerty that's really clever I like it
ganeshie8
  • ganeshie8
that is a very good try :) looks below conclusion does not hold : `[(a+2)^2 - 2a] is negative `
imqwerty
  • imqwerty
oops i made a mistake there ima try again
ikram002p
  • ikram002p
@Kainui i dared to think but not to try lol xD
ganeshie8
  • ganeshie8
i don't have an elementary solution for this... but honestly, after seeing your solution on yesterday's dan's problem, i believe that you can provide one ... good luck !
Kainui
  • Kainui
http://www.wolframalpha.com/input/?i=n%5E3%3D3%20mod%2019&t=crmtb01 :P
ikram002p
  • ikram002p
well i think kai already figured it out ur blonde as* :P
ikram002p
  • ikram002p
http://www.wolframalpha.com/input/?i=n%5E3%3D16+mod+19
Kainui
  • Kainui
Yeah but this method sucks, cause it's just brute force + luck that it has no solutions. I am more interested in what qwerty is working on
ikram002p
  • ikram002p
=)
Kainui
  • Kainui
I like the thinking of a and b having opposite signs and then immediately turning it into a difference of cubes sorta like building up some structure outta nothing like magic haha
ikram002p
  • ikram002p
try that kai as both same sign ovc don't work.
ganeshie8
  • ganeshie8
One way to prove it is by considering the 19 cases as kai was saying : \[a^3 \pmod{19}\] plugin a = 0,1,2,...,18 and show that none of them reduce to \(3\). but 19 cases might make you dizzy...
Kainui
  • Kainui
Here, I proved it myself instead of just plugging into wolfram alpha just to feel like I actually did something haha. https://repl.it/B0Js
ganeshie8
  • ganeshie8
you're making the java compiler dizzy... ;p
imqwerty
  • imqwerty
lol If you got here with no exceptions, then you proved it! xD
Kainui
  • Kainui
I think Java is better at compiling some math problems than my brain is haha.
Kainui
  • Kainui
If we had programming a long time ago, I wonder if people would still care about 'closed form solutions' lol. Even \(\sqrt{2}\) or \(\ln 2\) are in some ways really just algorithms in disguise even though they're also numbers haha. But anyways I digress... I still wanna get dizzy solving this some other way lol.
lochana
  • lochana
Just to support @imqwerty and @kainui , I found out that it doesn't have solutions for a and b; -1000

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