quadratic

For how many integer values of \(\color{red}a\) does the equation
\(x^2 + ax + 28a = 0\)
have integer solutions for x?

\(\text{Let the roots are $r_1$ and $r_2$}:\)

Vieta:

\(\begin{array}{|rcll|} \hline x^2 \underbrace{+a}_{=-(r_1+r_2)}x \underbrace{+ 28a}_{=r_1r_2} \\ \hline \mathbf{a} &=& \mathbf{-(r_1+r_2)} \\ 28a &=& r_1r_2 \\ 28\left(-(r_1+r_2)\right) &=& r_1r_2 \\ -28r_1-28r_2 &=& r_1r_2 \\ r_1r_2+28r_1+28r_2 &=& 0 \\ (r_1+28)(r_2+28) -28*28 &=& 0 \\ (r_1+28)(r_2+28)&=& 28*28 \\ \mathbf{(r_1+28)(r_2+28)}&=& \mathbf{784} \\ \hline \end{array}\)

The divisors of 784 are:
  1 |  2 |  4 |   7 |   8 |  14 |  16 | 28 |
 49 | 56 | 98 | 112 | 196 | 392 | 784  (15 divisors)

\(\begin{array}{|rcll|} \hline (r_1+28)(r_2+28)&=& 1*784 \\ &=& 2*392 \\ &=& 4*196 \\ &=& 7*112 \\ &=& 8*98 \\ &=& 14*56 \\ &=& 16*49 \\ &=& 28*28 \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|r|r|} \hline (r_1+28) & r_1 & (r_2+28) &r_2 & a=-r_1-r_2 \\ \hline 1 & 1-28 = -27 & 784 & 784-28=756 & 27-756=\color{red}-729 \\ 2 & 2-28 = -26 & 392 & 392-28=364 & 26-364=\color{red}-338 \\ 4 & 4-28 = -24 & 196 & 196-28=168 & 24-168=\color{red}-144 \\ 7 & 7-28 = -21 & 112 & 112-28= 74 & 21-74=\color{red}-53 \\ 8 & 8-28 = -20 & 98 & 98-28= 70 & 20-70=\color{red}-50 \\ 14 & 14-28 = -14 & 56 & 56-28= 28 & 14-28=\color{red}-14 \\ 16 & 16-28 = -12 & 49 & 49-28= 21 & 12-21=\color{red}-9 \\ 28 & 28-28 = 0 & 28 & 28-28= 0 & -0-0=\color{red}0 \\ \hline 49 & 49-28= 21 & 16 & 16-28 = -12 & -21+12=\color{red}-9 \\ 56 & 56-28= 28 & 14 & 14-28 = -14 & -28+14=\color{red}-14 \\ 98 & 98-28= 70 & 8 & 8-28 = -20 & -70+20=\color{red}-50 \\ 112 & 112-28= 74 & 7 & 7-28 = -21 & -74+21=\color{red}-53 \\ 196 & 196-28=168 & 4 & 4-28 = -24 & -168+24=\color{red}-144 \\ 392 & 392-28=364 & 2 & 2-28 = -26 & -364+26=\color{red}-338 \\ 784 & 784-28=756 & 1 & 1-28 = -27 & -756+27=\color{red}-729 \\ \hline \end{array}\)

Distinct Integer values of \(\color{red} a\) are 7 without \(a= 0\): \(\mathbf{\{ -9,~-14,~-50,~-53~-144,~-338,~-729\}}\)