Product of Roots I recently tutored a neighbor's son who is a $11$th grade honor student at an academy which is not far from my residence. One of the questions of the extra credit that his teacher gave his class was: 
Question:  Let $N > 2$ be the smallest integer such that $N^{N-1}$ is not a perfect square. Find the product of all the rational numbers that can be the roots of the polynomial: $5x^4 + bx^3 + cx^2 +dx + N$ where $a,b,c$ can be any integers. Luckily the answer sheet has $4.3$ as the approximate answer. I wrestled this puzzle and haven't got to any where close to the answer yet. I am leaning more and more toward Calculus than Algebra and my Algebra is not as sharp as I thought it could be. Any quick insight or wizards into the solution? Thank you in advance. 
 A: We first show that $N=6$.
For this, note that if $N>2$ is such that $N^{N-1}$ is not a perfect square, then

*

*$N$ cannot be a perfect square. So $4$ is ruled out.


*$N-1$ cannot be even i.e. $N$ cannot be odd. So $3,5$ are ruled out.
We get $N=6$ from here.

After figuring out $N=6$ you get that you must take the product over all rational roots of $5x^4+bx^3+cx^2+dx+6$.

The trick, though, is that you don't need $b,c,d$ at all!

Why? Note that if $\frac pq$ is a rational root (in lowest terms), then (I am just proving the rational root theorem, in case you already know that) substituting gives :
$$
5\frac{p^4}{q^4} + b\frac{p^3}{q^3} + c \frac{p^2}{q^2} + d \frac{p}{q} + 6 = 0\\\implies  5p^4 = -q(6q^3+dpq^2+cp^2q+bq^3) 
$$
So $q$ divides $5p^4$. Thus, since $q$ is coprime to $p$, it is also coprime to $p^4$. Therefore, $q$ divides $5$. This results in $q = \pm 1,\pm 5$.
Similarly, one deduces by transposing $p$ instead of $q$, that $p$ divides $6$. Thus, $p = \pm 1,2,3,6$.
Thus, the answer will be  the product of all these possible combinations. Some combinations repeat (for example $\frac {-5}2 = \frac {5}{-2}$ ) so we will be a little careful and keep all denominators positive.
The product of all the denominators is $25$. Each numerator appears $4$ times so we get $25^4$. The product of the numerators is $36$. Each one appears four times again (with sign, which cancels out) so we get $36^4$.
Thus, we get $36^4 / 25^4 \approx 4.3$.

Caveat : The rational root theorem only tells you what kind of rationals can be a root. But we have not shown that every candidate obtained above can be a rational root! For example, what would be $b,c,d$ for $\frac 15$? What about for $6$?(But this is easy, I leave you to see it)
