Conjecture: Integer Solutions for a^3b-1 and ab^3+1

[Albert1]

$a,b\in N$
$\dfrac {a^3b-1}{a+1},\,\, and \,\, \,\,\dfrac {ab^3+1}{b-1}$ also $\in N$
find all pairs of $(a,b)$

 

[Albert1]

$a,b\in N$
$\dfrac {a^3b-1}{a+1},\,\, and \,\, \,\,\dfrac {ab^3+1}{b-1}$ also $\in N$
find all pairs of $(a,b)$

hint:

Spoiler

$\because \dfrac{a^3b-1}{a+1}\in N,\,\,\therefore \,\,a+1\mid a^3b-1---(1)$

$\because\dfrac{ab^3+1}{b-1}\in N,\,\,\therefore \,\,b-1\mid ab^3+1---(2)$

 

[Albert1]

hint:

Spoiler

$\because \dfrac{a^3b-1}{a+1}\in N,\,\,\therefore \,\,a+1\mid a^3b-1---(1)$

$\because\dfrac{ab^3+1}{b-1}\in N,\,\,\therefore \,\,b-1\mid ab^3+1---(2)$

sol of others:

Spoiler

from (1):$a+1\mid a^3b-1- b(a+1)$
or :$a+1\mid ab(a^2-1)-(b+1)$
we get :$a+1\mid b+1-----(3)$
from (2):$b-1\mid ab^3+1- a(b-1)$
or :$b-1\mid ab(b^2-1)+(a+1)$
we get :$(b-1)\mid (a+1)\,\,\,from (3)\,\, \,\,this \,\,implies \,(b-1)\mid (b+1)-----(4)$
$b-1\mid b+1-(b-1)$
$b-1\mid 2---(5)$
$\therefore b=2,3---(6)$
put $(6) to (3)$
we have :$(a,b)=(2,2),(1,3),(3,3)$