Vieta's is the easy part

$\frac { \phi (k^{ j }) }{ \phi (k) } =\frac { k^{ j }\prod _{ p|k^{ j } }^{ }{ (1-\frac { 1 }{ p } ) } }{ k\prod _{ p|k }^{ }{ (1-\frac { 1 }{ p } ) } }$ , where the product is taken over all distinct prime divisors of $k^{ j }$ and $k$ . Note that if j is a natural number, $k$ and $k^{ j }$ have the same distinct prime divisors. Thus the products cancel and we find that $\frac { \phi (k^{ j }) }{ \phi (k) } =\frac { k^{ j } }{ k } =k^{ j-1 }$ . Note that this is true even for $k =1$ . Thus, we have $\sum _{ j=1 }^{ n+1 }{ \sum _{ k=1 }^{ 5 }{ (k^{ j-1 }) } } =\sum _{ j=0 }^{ n }{ \sum _{ k=1 }^{ 5 }{ k^{ j } } }$ .

Define a function $f(j)=\sum_{ k=1 }^{ 5 }{ k^{ j } }$ . We then have the coefficient of $x^{ n }$ as $\sum _{ j=0 }^{ n }{ f(j) }$ . Calculate $f(0), f(1), f(2), f(3), and f(4)$ . You'll find that

$f(0)=5$

$f(1)=\frac { 5(6) }{ 2 } =15$

$f(2)=\frac { 5(6)(11) }{ 6 } =55$

$f(3)=15^{ 2 }=225$

$f(4)=1^{ 4 }+2^{ 4 }+3^{ 4 }+4^{ 4 }+5^{ 4 }=979$

Thus, the coefficient of $x^{ 0 }$ is 5, that of $x^{ 1 }$ is $5+15=20$ , that of $x^{ 2 }$ is $5+15+55=75$ , that of $x^{ 3 }$ is $5+15+55+225=300$ , and that of $x^{ 4 }$ is $5+15+55+225+979=1279$ .

So, the polynomial is $1279x^{ 4 }+300x^{ 3 }+75x^{ 2 }+20x+5$ . By vieta's, $X=\frac { -300 }{ 1279 }$ and $Y=\frac { 5 }{ 1279 }$ . Therefore, $Y-X=\frac { 305 }{ 1279 }$ . So, $p+q+145=305+1279+145=\boxed { 1729 }$