If $b$ and $c$ are odd integers, then the equation $x^2+bx+c=0$ has:
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two integers roots, one even and one odd
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no integers roots
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two odd roots
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none of these
If b and c are odd, the discriminant D = b^2 - 4c. Since b is odd, b^2 is odd. Since c is odd, 4c is even. D = odd - even = odd. For integer roots, D must be a perfect square. However, an odd perfect square is always congruent to 1 mod 4. Here, b^2 is 1 mod 4 (if b is odd) and 4c is 0 mod 4, so D is 1 mod 4. This doesn't rule out roots, but testing parity shows that if x is even, x^2+bx+c is odd+even+odd = even (not 0), and if x is odd, x^2+bx+c is odd+odd+odd = odd (not 0). Thus, no integer roots exist.
Assume the equation x^2 + bx + c = 0 has integer roots p and q, which means their sum is -b and their product is c. Because the product pq = c is an odd integer, both roots p and q must be odd integers. The sum of two odd integers is always an even integer, which implies that b must be even, but this contradicts the given fact that b is an odd integer. Therefore, the equation cannot have integer roots.