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\(1\) and \(6\) are not mutually incongruent in mod \(5\)
because \(1\equiv 6\pmod{5}\)

\(1\) and \(3\) are mutually incongruent in mod \(5\)
because \(1\not\equiv 3\pmod{5}\)

incongruent is just an opposite of congruent

ok, so two numbers m and n are mutually incongruent mod c means
m not congruent n (mod c)

Yep!

For this example: Find a complete set of mutually incongruent solutions of 7x≡5 (mod 11)

x = 7, so all solutions are x = 7 + 11k, k an integer

the answer should be just \(\{7\}\)

why {7} ? and not, say {18} ??

sure, you can put any single integer that is congruent to 7

but the numbers 0 to n-1 are more convenient and look better

every other solution is congruent to 7, so there are no other incongruent solutions

looks good

no wait, how do you know there exists a solution ?

because gcd(9,15) = 3 | 12

I see x = 3 is one of the solution so all solution is
x = 3 + k * 15/gcd(9,15)
x = 3 + 5k

"If a solution exists", then there will be exactly 3 incongruent solutions

consider below congruence :
\[9x\equiv 2\pmod{15}\]
how many solutions ?

none, gcd(9,15) does not divide 2

right?

Yep!

so.....???

what are the mutually incongruent solutions?

let k = 0,1,2

the incongruent solutions are \(\{3,8,13\}\)
they are mutually incongruent in mod \(15\)

how do you know to let k = 0,1,2??

they are the most convenient ones

letting k = 222, 223,224 is technically fine
but you're not keeping things simple here

ok, let 0 <= <= 10, then
x = 3,8,13, 18, 23, 28, 33, 38, 43, 48, 53

3 and 18 are not incongruent

right. I was trying to understand how to find the incongruent solutions by listing a few solutions

what i notice is
3 + 15 = 18
8 + 15 = 23
13 + 15 = 28

it's somehow like {3,8,15}, {18,23,28}

so we don't need to bother about anything greater than 14

and we don't need to bother about anything less than 0

mod 15 btw

{3, 8, 13} right

opps yeah. Typo :D

3 + 5k not congruent 8 + 5s not congruent 13 + 5t (mod 15), for all integer k,s,t

3 + 15k not congruent 8 + 15s not congruent 13 + 15t, for all integer k,s,t
in mod 15

notice i have changed 5k to 15k

3 + 5k refers to mod 5
3 + 15k refers to mod 15

yeah, that's what I noticed earlier when I said
3 + 15 = 18
8 + 15 = 23
13 + 15 = 28

you may simply say, the solutions are \(x\equiv 3\pmod{5}\)

you could also say \(x\equiv 3,8,13\pmod{15}\)

right right. That's exactly what I was trying to say :D

awesome. Thanks