(a) Let two consecutive integers $n$ and $(n + 1)$ be the roots of $x^{2} - bx + c = 0$. Then, $n + (n + 1) = b$ and $n(n + 1) = c$ $\therefore\ \ b^{2} - 4c = (2n + 1)^{2} - 4n(n + 1) = 1$
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