Mock AIME 1 2010 Problems/Problem 1
Since ,
is congruent to
mod
. Taking the square root of both sides, we obtain
is congruent to
mod
. Solving,
is congruent to either
or
mod
. The maximum of
is
, of which the square root is between
and
. So,
must be less than
. The series of solutions for
is
and
. However,
does not work because
, a value not possible for
. Thus, there are
different solutions for a total of
solutions.