Sum of roots over each other

$\color{#3D99F6}{\frac{p}{q}+\frac{q}{p}=\frac{p^2+q^2}{pq}}$ From the quadratic equation we get $\color{#D61F06}{p+q=61.pq=m\rightarrow(p+q)^2=61^2\rightarrow p^2+q^2=3721-2pq=3721-2m}$ .So we can write the fraction as $\color{#20A900}{\frac{3721-2m}{m}}$ If $\color{#EC7300}{p,q=Prime\;Numbers}$ then one of p and q must be 2 because if p and q were both odd primes then $\color{#69047E}{p+q=Even\;Number=61(A\;contradiction)}$ .Let p=2 then q=61-2=59 and m=pq=(2)(59)=118.So we can write the fraction as $\color{#624F41}{\frac{3721-2(118)}{118}=\frac{3721-236}{118}= \frac{3485}{118}\rightarrow gcd(3485,118)=1\rightarrow a+b=3485+118=3\boxed{603}}$