Question

if a|(b-1) and a|(c-1) then a|(bc-1)

Answer 1

What is the '|', if I may ask?

Answer 2

divides

Answer 3

like 5|10 because 5(2)=10

Answer 4

5|6 is not a true statement because there is no such integer a so that we have 5a=6

Answer 5

Oh, yeah! :) So you have to work out the proof, I suppose?

Answer 6

yep

So you get to play with the following:
ak=b-1 where k is an integer
ai=c-1 where i is an integer
You use the above to show:
ap=bc-1 where p is an integer.

Answer 7

you get to play with what you are given to show the then part

have you tried doing any playing with those expressions?

Answer 8

I used the first two equations that you used and i substituted them into the formula a|(ab-1) to obtain a|(a^2ki + ak+ai +1 - 1 but I don't think that is the correct approach

Answer 9

it looks like a great approach

Answer 10

look at (b-1)(c-1)

Answer 11

and that one a is suppose to be a c but yeah

Answer 12

so if i end up with a|a(aki + k + i) then that is correct?

Answer 13

well on sec
this is the way I would say
bc-1=(a(aki+k+i))

so this means a|(bc-1)
since aki+j+i is an integer

Answer 14

that makes a lot more sense than what i put

Answer 15

but your approach was good

Answer 16

(b-1)(c-1)=bc-c-b+1=bc-1-(c-1)-(b-1)

so

(b-1)(c-1)+(c-1)+(b-1)=bc-1

Answer 17

a divides (b-1)(c-1)+(c-1)+(b-1)


therefore a divides bc-1

Answer 18

cute stuff zarkson

Answer 19

thank you both for the help. Appreciate it and you both helped it make a lot more sense

Answer 20

another question. if a = 7(mod 8) and b = 3(mod 8) then a * b = 1(mod 8). I feel like that is a false statement

Answer 21

well what is the remainder when dividing 21 by 8?

Answer 22

not 1 that is why i thought it was false

Answer 23

i think you are right :)

Answer 24

i knew i had a little knowledge of this math stuff haha thanks

Answer 25

np