$A, B, C, D$, and $E$ are relatively prime integers (i.e., have no single common factor) such that the polynomials $5 A x^{4}+4 B x^{3}+3 C x^{2}+2 D x+E$ and $10 A x^{3}+6 B x^{2}+3 C x+D$ together have 7 distinct integer roots. What are all possible values of $A$ ? Your team has been given a sealed envelope that contains a hint for this problem. If you open the envelope, the value of this problem decreases by 20 points. To get full credit, give the sealed envelope to the judge before presenting your solution.
Hint: Consider $A=1, B=D=0, C=750$, and $E=19845$.
Solution: Call the negatives of the roots of the first polynomial $a, b, c, d$, and the negatives of the roots of the second polynomial $e, f, g$ (using the negatives avoids negative signs for the rest of the proof, thus preventing the possibility of dropping a sign). Then $5 A x^{4}+4 B x^{3}+3 C x^{2}+2 D x+E=$ $5 A(x+a)(x+b)(x+c)(x+d)$ and $10 A x^{3}+6 B x^{2}+3 C x+D=10 A(x+e)(x+f)(x+g)$. Thus $B=\frac{5}{4} A(a+b+c+d)=\frac{5}{3} A(e+f+g), C=\frac{5}{3} A(a b+a c+a d+b c+b d+c d)=\frac{10}{3} A(e f+e g+f g)$, $D=\frac{5}{2} A(a b c+a b d+a c d+b c d)=10 A(e f g)$, and $E=5 A(a b c d)$. From these equations we see that since all the variables are integers, it must be the case that $A|B, A| 3 C, A \mid D$, and $A \mid E$, therefore $A, B, C, D$, and $E$ can only be relatively prime if $A$ is $\pm \mathbf{1}$ or $\pm \mathbf{3}$. Now we need numerical examples to show that both of these are possible. Without loss of generality let $g=0$, so $D=0$. Then $e$ and $f$ are $\frac{-3 B \pm \sqrt{9 B^{2}-30 A C}}{10 A}$, so $9 B^{2}-30 A C$ must be a perfect square. Let $B=0$ in the hope that solutions will still exist to this simplified problem. First let us try to find an example with $A=-1$, so we need $30 C$ to be a perfect square. This first happens for $C=30$, and in that case $e$ and $f$ are $\pm 3$. We need $a+b+c+d=0$, so let's try to look for $a=-b, c=-d$. This doesn't work for $C=30$ since $3 C / 5=18$ is not the sum of two distinct squares. For that we will need to try $C=5^{2} \cdot 30=750$, for which we get $e, f= \pm 15, a, b= \pm 3$, and $c, d= \pm 21$. Thus for $A= \pm 1$ we can use $B=D=0, C=\mp 750$, and $E= \pm 19845$. Similarly we can find $A= \pm 3, B=D=0, C=\mp 250$, and $E= \pm 735$, for which $(a, b, c, d, e, f, g)=(1,-1,7,-7,5,-5,0)$.

Note that these give us quintic polynomials with integer coefficients possessing 4 relative extrema and 3 points of inflection at lattice points, such as $3 x^{5}-250 x^{3}+735 x$.