Question

all pairs (a,b) of positive integer satisfying :
(a^2 + b)|(a^2*b+a) and (b^2 - a)|(ab^2 + b)

Answer 1

i guess to solve one by one for (a^2 + b)|(a^2*b+a) and (b^2 - a)|(ab^2 + b) then find intersect, right ?

Answer 2

is it not that the condition is \(a^2b+a\) is divided by \(a^2+b\)??

Answer 3

dan interpreted the problem in another way, right?

Answer 4

i dunno man im just moving stuff aroudn

Answer 5

u cud do that too i guess, and do some sythetic division

Answer 6

yeah, in other words a^2 + b is divisible by a^2 * b + a and .... @Loser66

Answer 7

ya u are right

Answer 8

a * b = c * d
a b c d are all integers so

Answer 9

but it doesnt mean they have to be divisible

Answer 10

ah , if i use the long division get
a = b^2 for the first one

Answer 11

like
(p1*p2) * (p3*p4) = (p3*p2*p1) * (p4)

this could happen and p1*p2 is not divisible by p4

Answer 12

is b = a+1 the ansur ?

Answer 13

how ? btw it is not options question, so i dont know

Answer 14

it seem to work, check once

Answer 15

I'll put the solution shortly..

Answer 16

omg i didnt see that Division sign thing

Answer 17

give me a little hint

Answer 18

lmao

Answer 19

thats hard to see on my computer lol

Answer 20

can you double check if it is really working or not

i don't have a rigorous proof yet, still working..

Answer 21

so we are basically solving a system of modular equatiosn right

Answer 22

yes, b = a + 1 is satisfying both for (a^2 + b)|(a^2*b+a) and (b^2 - a)|(ab^2 + b)