If the equation $x ^ { 2 } + 21 a x + 4 = 0$ has integral roots, then minimum value of $a$ is ?

If the equation x^2 + 21ax + 4 = 0 has integral roots, let the roots be m and n. By Vieta's formulas, the product of the roots mn = 4, meaning the possible integer pairs for (m, n) are (1, 4), (2, 2), (-1, -4), and (-2, -2). The sum of the roots m + n must equal -21a, so we test the sums: for 1+4=5 we get a=-5/21, for 2+2=4 we get a=-4/21, for -1-4=-5 we get a=5/21, and for -2-2=-4 we get a=4/21. The minimum value among these possible values of a is -5/21, which is not listed among the provided options, making None of these the correct choice.

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