find the integer ordered pairs (a, b) that satisfy
a^3+19b = 22

- ganeshie8

find the integer ordered pairs (a, b) that satisfy
a^3+19b = 22

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- schrodinger

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- anonymous

Said another way find all \(b\) such that\[\sqrt[3]{22-19b}\]is an integer

- lochana

\[a,b \in \mathbb{Z} ?\]

- ganeshie8

yes for both...

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## More answers

- ikram002p

it has no integer solution i need to prove why i suppose xD

- ikram002p

we need to show that a^3 can't be of the form 19b+16 maybe
errrrm :-\ (just conjecturing)

- ganeshie8

do you mean a^3 cannot be of form 19b + \(\color{red}{22}\) ?

- ikram002p

well 19b-22 so like -22 mod 19 :3

- Kainui

I thought it was positive integers only and thought it was simple :(

- ganeshie8

Haha I see.... its trivial if we allow only positive integers

- ikram002p

@Kainui if its only positive proving it would be like
a^3+19b>22 for a,b>2

- ikram002p

?

- Kainui

Hahaha yeah it would be, but it's not that simple xD

- Kainui

Has anyone dared to try to show this yet?
\[a^3 \cancel \equiv 3 \mod 19\]

- 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

Very nice @imqwerty that's really clever I like it

- ganeshie8

that is a very good try :)
looks below conclusion does not hold :
`[(a+2)^2 - 2a] is negative `

- imqwerty

oops i made a mistake there
ima try again

- ikram002p

@Kainui i dared to think but not to try lol xD

- 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

http://www.wolframalpha.com/input/?i=n%5E3%3D3%20mod%2019&t=crmtb01
:P

- ikram002p

well i think kai already figured it out ur blonde as* :P

- ikram002p

http://www.wolframalpha.com/input/?i=n%5E3%3D16+mod+19

- 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

=)

- 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

try that kai as both same sign ovc don't work.

- 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

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

you're making the java compiler dizzy... ;p

- imqwerty

lol If you got here with no exceptions, then you proved it! xD

- Kainui

I think Java is better at compiling some math problems than my brain is haha.

- 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.

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