by . Now

take

we have no solution and there are

(take

) and there are and consider the

numbers . For all of these numbers,

such numbers. Since all square residues have solutions

square residues modulo including zero, this means that for the number

we have a all nonsquare residues

, the equation must have no solution. However, for

is not a square residue

but

solution which is a contradiction.

For the

number so

for is not a square residue

but

has a solution

since

obtain a contradiction so the second lemma is also proved.

Now,

since is a square residue. Hence we again can take every value

modulo

and

if and we must

have and

and also we must have

is a nice pair. So, can take only three

values. We will consider each case separately:

If

If

means

If

means

If we count these possibilities,

.

4.解（dinoboy）

First, remark that it suffices for

to be injective modulo

For modulo

simply note that we require

For

modulo

or we require . for

some

that and .

if i am not wrong, we

get then we must

have

then we must

have which

which

then we must

have which

means