((my) mod n ) congruent to n-1

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smslca

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If given a 'n' value and m = floor ( squareroot )
then is there any way to find the value of 'y' , such that

((m*y) mod n) is congruent to (n-1)

---------- Post added at 07:02 ---------- Previous post was at 05:38 ----------

with the help of a friend
i figured out that, if m is the divisor of n, it wont be possible to get a solution .
But what about the other values?
 

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