Question

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

Answer 1

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

Answer 2

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

Answer 3

yes for both...

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

?

Answer 12

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

Answer 13

Has anyone dared to try to show this yet?

\[a^3 \cancel \equiv 3 \mod 19\]

Answer 14

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

Answer 15

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

Answer 16

that is a very good try :)

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

Answer 17

oops i made a mistake there
ima try again

Answer 18

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

Answer 19

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 !

Answer 20

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

:P

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

=)

Answer 25

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

Answer 26

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

Answer 27

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...

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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.

Answer 33

Just to support @imqwerty and @kainui , I found out that it doesn't have solutions for a and b; -1000<a<1000 and -1000<b<1000