Congruence relations

[Lee Yiyuan]

Hi, i am widely unfamiliar with modular arithmetic and i wanted to know (i tried hard at googling but still seemed to fail) :
if n = a (mod b), when does n = b (mod a) ?

[science_man_88]

Originally Posted by Lee Yiyuan

Hi, i am widely unfamiliar with modular arithmetic and i wanted to know (i tried hard at googling but still seemed to fail) :
if n = a (mod b), when does n = b (mod a) ?

I made a modular arithmetic thread before, question can you read a clock ? if so you can do modular (clock) arithmetic basics.

[Lee Yiyuan]

Originally Posted by science_man_88

I made a modular arithmetic thread before, question can you read a clock ? if so you can do modular (clock) arithmetic basics.

Yup, i know the basics like a + b = a (mod n) + b (mod n), etc. Saw it, thanks!

[science_man_88]

Originally Posted by Lee Yiyuan

Yup, i know the basics like a + b = a (mod n) + b (mod n), etc. Saw it, thanks!

N=a mod (b)=b(mod a) iff N|b-a and N|a-b .

[Lee Yiyuan]

Originally Posted by science_man_88

N=a mod (b)=b(mod a) iff N|b-a and N|a-b .

What about for |a - b| < |n| ?

[fivemack]

Ignore scienceman, he's got the wrong definition ...

N=a mod b = b mod a means (N-a) is divisible by b and (N-b) divisible by a, there's nothing involving a-b.

There is always (ok, for a,b>2) an infinite number of solutions to N=a mod b and N=b mod a; to find them, do 'chinese(Mod(a,b),Mod(b,a))' in pari.

[Lee Yiyuan]

Originally Posted by fivemack

Ignore scienceman, he's got the wrong definition ...

N=a mod b = b mod a means (N-a) is divisible by b and (N-b) divisible by a, there's nothing involving a-b.

There is always (ok, for a,b>2) an infinite number of solutions to N=a mod b and N=b mod a; to find them, do 'chinese(Mod(a,b),Mod(b,a))' in pari.

Thank you.
However pari outputs "inconsistent variables in chinese, b!=a

Edit: please ignore my previous comment, i am still struggling with grasping PARI

[axn]

You can do it the hard way, but essentially, the solution boils down to N=kab+a+b.

start with, n = ax+b = by+a

rearranging, a(x-1) = b(y-1)

ratio, (y-1)/(x-1) = a/b = (ka)/(kb)

general solution, y = ka+1, x = kb+1

substituing back, n = kab + a + b