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

[smslca]

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

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

[smslca]

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?

[ccorn]

Originally Posted by smslca

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

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

Yes, for any m coprime to n. The key algorithm is fundamental and famous: Look up the Extended Euclidean Algorithm (XGCD). You ask for y being the negated multiplicative inverse of m mod n. The XGCD gives you the multiplicative inverse, which is n-y in your terms.