Question

Let p be an odd prime number and Tp be the following set of 2*2 matrices.
Tp=
{ A= [a11 =a ,a12=b,a21=c,a22=a] }
a,b,c€{0,1,...,p-1}

The number of a in Tp such that trace of A is not divisible by p but det A is divisible by p is

Answer 1

The number of A in Tp is there in the second last line instead of the number of a in Tp.

Answer 2

@ganeshie8

Answer 3

whats the trace of A ?

Answer 4

2a

Answer 5

\(a,b,c\in \{0,1,2,\ldots,p-1\}\)

We don't want \(2a\) to be divisible by \(p\). How many integers in the given set satisfy this requirement ?

Answer 6

All of them since p is an odd prime number.

Answer 7

2a won't b divisible by an odd prime number for any value attained by a.

Answer 8

Except for one value that is 0.

Answer 9

Is a is 0 then it is divisible by p.

Answer 10

If*

Answer 11

Good. So we see that there are \(p-1\) choices for \(a\) that satisfy the first condition.
Lets look at second condition.

Answer 12

whats the determinant of A ?

Answer 13

a^2-bc

Answer 14

Yes, lets prove a little lemma first.

Answer 15

Consider the complete set of residues mod \(p\)
\[\{0,1,2,\ldots,p-1\}\]

If \(\gcd(b,p)=1\), then below set also forms a complete set of residues mod \(p\)
\[\{0*a,1*a,2*a,\ldots,(p-1)*a\}\]

Answer 16

Residues mod p?
Does it mean that all the values that a,b,c can attain?

Answer 17

for example,
let \(p=5\).

The complete set of residues mod \(5\) are
\[\{0,1,2,3,4\}\].

Let \(a=3\), then below set also forms the complete residue set mod \(5\)
\[\{0*3,1*3,2*3,3*3,4*4\}\]

Answer 18

residue is same as remainder

Answer 19

simplify the remainders in the last set and convince yourself you indeed get the set \(\{0,1,2,3,4\}\)

Answer 20

Okay!

Answer 21

Are you there?
@ganeshie Sir