Question

how many order (a,b) such that :
(a^3 + 1)/(ab - 1) is an integer
with a, b are integer number

Answer 1

@rational

Answer 2

oh honey

Answer 3

do cube root factoring first

Answer 4

a^3 + 1 = (a + 1)(a^2 - a + 1)

Answer 5

okay

Answer 6

(a + 1)(a^2 - a + 1)
-----------------
(ab-1)

Answer 7

multiply by conjugate see where that gets us

Answer 8

or well lets consider the bounds here

Answer 9

conyugate of ab - 1 ?

Answer 10

ya nvm lets do conjugate

Answer 11

can u tell me some pretext to this problem like are u working on mod functions or remainder theorem or anything?

Answer 12

(a + 1)(a^2 - a + 1)
----------------- = K , where K is an integer
(ab-1)

Answer 13

this is just theory number question from my book but idont know what the start to do it

Answer 14

maybe this is obvious but
b=a^2+2/a , is one form of solution right

Answer 15

and b=2/a

Answer 16

how you get that ?

Answer 17

by look at ab-1

Answer 18

since the top has (a+1) and its (a^3+1)

Answer 19

also from the factor (a^2-a+1)
we get that
ab-1=a^2-a+1
b=(a^2-a+2)/a
b=a-1+2/a

Answer 20

getting the int solutions for these will give us some solutions but i dont think these wud be all of the solutions

Answer 21

from b=2/a
(1,2) is the only solution u can get from here

Answer 22

also
(2,1) should work as well
since that also makes the denominator 1

Answer 23

anything divided by 1 will be an integer so we are fine there

Answer 24

any int* div by 1

Answer 25

wait, how to get b = 2/a ?

Answer 26

that makes the denominator 1

Answer 27

ab-1 =1
b=2/a

Answer 28

another way to get 2/a is that

Answer 29

oh, okay :D

Answer 30

(a^3 + 1) look at its factors
(a^3 + 1)=(a + 1)(a^2 - a + 1)

so if
ab-1 = a+1 or
ab-1=a^2-a+1

Answer 31

we know it will divide with the result as an integer if this holds true

Answer 32

so u get
ab-1=a+1
b=(a+2)/a
=1+2/a

again this leads us back to the fact that
b=1+2/a
so integer is coming from the fact that a can be 1 and 2 here

Answer 33

all other integers of a will give us a remainder

Answer 34

theresore
we have a new solution here

Answer 35

(1,2)
(2,1)
(2,3)
(2,2)

Answer 36

lets see if theres anymore from the other equations

Answer 37

for b = 1 + 2/a than a = {+-1 , +-2} ?

Answer 38

sorry that shud be (1,3) not (2,3)

Answer 39

oh yes negative too good point

Answer 40

oh and there is a restriction which is
ab=/=1

Answer 41

(1,1) is basically the only way that will happen and its not a solution anyway

Answer 42

and -1,-1

Answer 43

also from the factor
(a^2-a+1)
we get that
ab-1=a^2-a+1
b=(a^2-a+2)/a
b=a-1+2/a

Answer 44

from case b = 1 + 2/a
i get (a,b) = (1,2),(-1,-1),(2,2),(2,0)

Answer 45

tell me the extra solutions from that equation now

Answer 46

ok so we throw out (-1,-1)

Answer 47

like we agreed

Answer 48

u are missing solution there

Answer 49

(1,3)

Answer 50

nvm just that,

Answer 51

b=a-1+2/a
a=1
b=2
a=-1
b=-2-2 = -4

a=2 and -2
b=2, -4

Answer 52

1,2
-1,-4
2,2
-2,-4

Answer 53

ok, for the solution i can cover it. let's ta2e conclusion as all cases satisfy :
* ab -1 = a + 1
** ab - 1 = a^2 - a + 1
only 2 cases right ?

Answer 54

*take

Answer 55

there more

Answer 56

i got my cases from this
(a^3+1) =ab-1
so b=a^2+2/a
-------------------
(a+1)=ab-1
b=1+2/a
------------------

ab - 1 = a^2 - a + 1
b=a-1+2/a
-------------------
and
ab-1=1
b=2/a

Answer 57

get solutions from those 4

Answer 58

not sure if there anything im missing

Answer 59

ok, nove . i will cover all cases to get all solutions. thank you very much for your explain.

Answer 60

*dan815
how about ab - 1 = -1 ? is this make extra case ?

Answer 61

@dan815

Answer 62

oh yes! it definately does

Answer 63

integer number divided by -1 = integer also

Answer 64

thank you :D

Answer 65

well actually it might be covered though

Answer 66

did we see that 0,0 is a solution anywhere

Answer 67

yes from the case ab - 1 = -1 make ab = 0
(a,b) = (0,0)

Answer 68

oh true wow there an infinite case u found!

Answer 69

(N,0), where N is any integer

Answer 70

(0,N) , Where N is any integer another infinite case solution

Answer 71

wait, ab = 0
solution a = 0 or b = 0

Answer 72

and a=0 and b= anything
a=anythign and b=0

Answer 73

so, it would be infinity order for (a,b) ?

Answer 74

yep!

Answer 75

wah... :D