1. ## Modulo Arithmetic

Consider the following:

-x+2 = 3 (mod 5)
4x = 1 (mod 5)
As the gcd(4,5) = 1 therefore there's one unique solution, namely x = 4 (mod 5)

Also...

-x + 2 = 3 (mod 8)
7x = 1 (mod 8)

Again there is a unique solution, namely x = 1 (mod 8)

Can anyone explain where these solutions come from? I understand how to get to 4x = 1 (mod 5) and 7x = 1 (mod 8) but I dont get how we get the 'unique solutions' at the end? I'd appreciate a clear explanation please. Thank you

2. Originally Posted by mitch_nufc
Consider the following:

-x+2 = 3 (mod 5)
4x = 1 (mod 5)
As the gcd(4,5) = 1 therefore there's one unique solution, namely x = 4 (mod 5)
4*4= 16= 1 (mod 5) so any number of the form 5j+ 4 satisfies 4x= 1 (mod 5). -x+ 2= 3 (mod 5) leads to -x= 1 (mod 5) and so x= -1= 4 (mod 5). Any number of the form 5j+ 4 also satisfies this. Of course, to call 4 a "unique" solution you have to call two numbers that are equal "mod 5" the same: 4, 9, 14, 19, etc. all satify those two equations but are the same "mod 5".

Also...

-x + 2 = 3 (mod 8)
7x = 1 (mod 8)

Again there is a unique solution, namely x = 1 (mod 8)
No. 7(1)= 7 which is NOT equal to 1 (mod 8). -1+ 2= 1 which is NOT equal to 3 (mod 8). 7*7= 49= 1+ 6(8)= 1 (mod 8) so the any number of the form 7+ 8n satisfies 7x= 1 (mod 8). -x+ 2= 3 (mod 8) leads to -x= 1 (mod 8) so x= -1= 7 (mod 8).

Can anyone explain where these solutions come from? I understand how to get to 4x = 1 (mod 5) and 7x = 1 (mod 8) but I dont get how we get the 'unique solutions' at the end? I'd appreciate a clear explanation please. Thank you
Are you asking why the solutions are unique? That is not unusual. Typically linear equations have only one solution. What is unusual is that there exist in each case, a number that satisfies both equations. And that is true because the two equations reduce to the same thing.

3. so are you telling me the solution for the last part is wrong? "x = 1 (mod 8)". its not a solution i got my self its from my lecturers solution manual

4. Originally Posted by mitch_nufc
so are you telling me the solution for the last part is wrong? "x = 1 (mod 8)". its not a solution i got my self its from my lecturers solution manual
Yes, that's what HallsofIvy is telling you. And he's correct.

Take a value of x that satisfies the 'solution' x = 1 mod(8). x = 9, say. Does x = 9 satisfy the equation 7x = 1 mod(8) ....? I don't think so.

Now take a value of x that satisfies the solution x = 7 mod(8). x = 7, say. Does x = 7 satisfy the equation 7x = 1 mod(8) ....?