Putnam 1968/B5 solution

29th Putnam 1968

Problem B5

Let F be the field with p elements. Let S be the set of 2 x 2 matrices over F with trace 1 and determinant 0. Find |S|.

Solution

Answer: p² + p.

Let the matrix be


 a  b

 c  d

If we take any a not 0 or 1, then d is fixed as 1 - a and is also not 0. Hence bc is fixed as a(1 - a) and is non-zero. So we can take any b not 0 and d is then fixed. Altogether that gives us (p - 2)(p - 1) possibilities.

If we take a as 0 or 1, the d is fixed as 1 - a and bc must be 0. That gives 2p-1 possibilities for bc, or 4p-2 possibilities in all.

So in total we have (p - 2)(p - 1) + 4p - 2 = p² + p.

29th Putnam 1968