Let p be a prime. a) Determine the number of irreducible polynomials over Zp of the form x^2+ax+b.?

Question

The second part b) Determine the number of irreducible quadratic polynomials over Zp.

Answer 1

The answer to a) is the binomial coefficient p_C_2 = p(p-1)/2.
The answer to b) is p(p-1)²/2.

Examples:
x² + x + 1 is the only irreducible polynomial over Z2;
x² + 1, x² + x + 2, x² + 2x + 2 are irreducible over Z3;
x² + 2, x² + 3, x² + x + 1, x² + x + 2, x² + 2x + 3, x² + 2x + 4,
x² + 3x + 3, x² + 3x + 4, x² + 4x + 1, x² + 4x + 2 are over Z5.

Proof: a) the number of all polynomials x² + ax + b is p² /we have p choices: 0, 1, 2, . . , p-1 for a and same for b/. Such polynomial is reducible if and only if it has a form:
x² + ax + b = (x - α)(x - β) for some α, β, belonging to Zp. How many ways are to select the roots α and β? Taking α=0 we can combine with p values for β /0,1,2,. . p-1/, then α=1 can be combined with p-1 values for β /1,2, etc./ and so forth, so the number of ways is
p + (p-1) + . . + 2 + 1 = p(p+1)/2 = (p+1)_C_2
This is the number of the reducible monic polynomials over Zp, so the irreducible are
p² - p(p+1)/2 = p(p-1)/2 = p_C_2.

b) The required number is p-1 times greater than the answer of a) because the leading coefficient A of the polynomial
Ax² + Bx + C
can be any of 1, 2, . . , p-1 /not 0 only/.