What exactly is a mutually incongruent solution of a congruence? Can you give an example?

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What exactly is a mutually incongruent solution of a congruence? Can you give an example?

Mathematics
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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

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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
But how does the number 7 or any number congruent to 7 mod 11 capture the idea of "mutually incongruent" solution? if 7 not congruent m (mod 11), what numbers play in the role of m according to the definition of mutually incongruent?
there are no other integer solutions, it is like asking for all the mutually perpendicular vectors that are solutions, if there is only one vector, you give only that.
every other solution is congruent to 7, so there are no other incongruent solutions
huhm... ok, let me give another example where it has more than 1 mutually incongruent solution. 9x ≡ 12 (mod 15). Since gcd(9,15) = 3, there will be 3 mutually 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.....???
let's get back to 9x ≡ 12 (mod 15). There will be exactly 3 mutually incongruent solutions. And all solutions are x = 3 + 5k
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}
notice that any integer greater than or equal to 15 is congruent to some integer in the set \(\{0,1,2,\ldots,14\}\)
so we don't need to bother about anything greater than 14
and we don't need to bother about anything less than 0
the set of residues \(\{0, 1, 2, \ldots, n-1\}\) is our most favorite one, when somebody asks you for incongruent solutions, they expect you to give them from this set
So {3,8,15} are the set of mutually incongruent because 3 + 5k not congruent 8 + 5s not congruent 15 + 5t, for all integer k,s,t Is this what the incongruent solutions really mean?
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
I think I understand it better now although I still need to do some examples to clear up any confusion. @ganeshie8 thank you :)
basically we're "lifting" the solutions from mod5 to mod15, so \(x\equiv 3\pmod{5}\) gives 3 incongruent solutions in mod 15 : \(\{3,8,13\}\)
yeah, that's what I noticed earlier when I said 3 + 15 = 18 8 + 15 = 23 13 + 15 = 28
so instead of expressing all solutions in the form of 3 + 5k, we can express all solutions in terms of mutually incongruent soltuions. I.e {x | x = 3 + 15k or x = 8 + 15k or x = 13 + 15k }
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
if you're not explicitly asked for the form of solutions, leaving it as \(x\equiv 3\pmod{5}\) looks neat
awesome. Thanks

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