Bashing Available Part 1

Let's use Newton sums. Let $S_k=a_1^k+a_2^k+a_3^k+a_4^k$ and let $P(x)=x^4+mx^3+nx^2+px+q=0$ whose roots are $a_1$ , $a_2$ , $a_3$ and $a_4$ . We know that $S_1=1$ , $S_2=2$ , $S_3=3$ and $S_4=4$ , and:

$S_1=-m \Longrightarrow m=-1$

$S_2=-mS_1-2n \Longrightarrow n=-\dfrac{1}{2}$

$S_3=-mS_2-nS_1-3p \Longrightarrow p=-\dfrac{1}{6}$

$S_4=-mS_3-nS_2-pS_1-4q \Longrightarrow q=\dfrac{1}{24}$

So, $P(x)=x^4-x^3-\dfrac{1}{2}x^2-\dfrac{1}{6}x+\dfrac{1}{24}$ .

Now, by recursion we have:

$S_k=-mS_{k-1}-nS_{k-2}-pS_{k-3}-qS_{k-4}$

$S_k=S_{k-1}+\dfrac{1}{2} S_{k-2}+\dfrac{1}{6} S_{k-3}-\dfrac{1}{24} S_{k-4}$

Obtain $S_5$ , which is the value we want:

$S_5=S_4+\dfrac{1}{2} \left(S_3\right) + \dfrac{1}{6} \left(S_2\right)-\dfrac{1}{24} \left(S_1\right)$

$S_5=4+\dfrac{3}{2}+\dfrac{1}{3}-\dfrac{1}{24}$

$\boxed{S_5=\dfrac{139}{24}}$

Hence our final answer is $139+24=\boxed{163}$ .

There is a neat way to do this with Newton's Sums:

Let:

$P_1 = a_1+a_2+a_3+a_4 \quad S_1 = a_1 + a_2 + a_3 + a_4$

$P_2 = a_1^2+a_2^2+a_3^2+a_4^2 \quad S_2 = a_1a_2 + a_1a_3 + a_1a_4+a_2a_3+a_2a_4+a_3a_4$

$P_3 = a_1^3+a_2^3+a_3^3+a_4^3 \quad S_3 = a_1a_2a_3 + a_1a_2a_4+a_1a_3a_4+a_2a_3a_4$

$P_4 = a_1^4+a_2^4+a_3^4+a_4^4 \quad S_4 = a_1a_2a_3a_4$

Then:

$P_1=S_1$

$P_2=S_1P_1-2S_2$

$P_3=S_1P_2-S_2P_1+3S_3$

$P_4=S_1P_3-S_2P_2+S_3P_1-4S_4$

$P_5=S_1P_4-S_2P_3+S_3P_2-S_4P_1$

Substitute in the known values:

$P_1=1$

$2=(1)(1)-2S_2\quad \Rightarrow S_2 = -\frac {1}{2}$

$3=2+\frac {1}{2} (1) +3S_3 \quad \Rightarrow S_3 = \frac {1}{6}$

$4=3+\frac {1}{2}(2)+\frac {1}{6}(1)-4S_4 \quad \Rightarrow S_3 = \frac {1}{24}$

$P_5=4+\frac {1}{2}(3)+\frac {1}{6}(2)-\frac {1}{24}(1) = \frac {139}{24} = \frac {p}{q}$

$\Rightarrow p+ q = 139+24 = \boxed {163}$