Beastie problem!

From $(x+y)^2=x^2+2xy+y^2$ ,

$\Rightarrow \left( \dfrac{5}{2} \right)^2 = \dfrac{13}{4} + 2xy$

$\Rightarrow 2xy = \dfrac{25}{4} - \dfrac{13}{4} = \dfrac{12}{4} = 3$

Now, $x(x+y)=(x) \dfrac{5}{2}$

$\quad \Rightarrow 2x^2 +2xy = 5x$

$\quad \quad 2x^2 +3 = 5x$

$\quad \quad 2x^2 - 5x +3 = 0$

$\quad \quad (2x-3)(x-1) = 0$

$\quad \Rightarrow x = \frac{3}{2}, 1 \quad \Rightarrow y = \frac{3}{2}, 1$

Note that $x$ and $y$ are interchangeable. $(x,y)=(\frac{3}{2},1), (1, \frac{3}{2})$ .

Therefore,

$x^5+y^5=1^5+ \left( \dfrac{3}{2} \right) ^5 = 1 + \dfrac{243}{32} = \dfrac{275}{32} = \dfrac{a}{b}$

$\Rightarrow a+b = \boxed{307}$

$e=2$

$x=\frac{c}{e} , y=\frac{d}{e}$

$c+d=5$

$c^{2}+d^{2}=13$

$c=3,d=2$

$c^{5}+d^{5}=3^{5}+2^{5}=243+32=275$

$b=e^{5}=2^{5}=32$

$a+b=275+32=\boxed{307}$

Add $2xy$ to both sides of $x^2+y^2=\dfrac{13}{4}$ :

$x^2+2xy+y^2=\dfrac{13}{4}+2xy$

$(x+y)^2-\dfrac{13}{4}=2xy$

$\left(\dfrac{5}{2}\right)^2-\dfrac{13}{4}=2xy$

$xy=\dfrac{3}{2}$

Now, from $(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ obtain $x^5+y^5$ :

$x^5+y^5=(x+y)^5-5xy(x^3+y^3)-10(xy)^2(x+y)$

$x^5+y^5=(x+y)^5-5xy(x+y)(x^2+y^2-xy)-10(xy)^2(x+y)$

$x^5+y^5=\left(\dfrac{5}{2}\right)^5-5\left(\dfrac{3}{2}\right) \left(\dfrac{5}{2}\right) \left(\dfrac{13}{4}-\dfrac{3}{2}\right)-10 \left(\dfrac{3}{2}\right)^2 \left(\dfrac{5}{2}\right)$

$x^5+y^5=\dfrac{275}{32}$

Hence, $a=275$ , $b=32$ and $a+b=\boxed{307}$ .

x+y=5/2

(x+y)²=x²+y²+2xy

25/4=13/4+2xy

xy=3/2

So x & y are 2 numbers whose sum=5/2 and product=3/2

So they will satisfy the equation ∅²-S∅+P=0 which is quadratic in ∅

∅²-(5/2)∅+(3/2)=0

2∅²-5∅+3=0

∅=1,3/2=(x,y) or (y,x)

$x^{5}+y^{5}=1+(243/32)$

=275/32≈a/b

a+b=275+32=307

x+y =5/2 .........(1)
x^2+y^2 = 13/4 ............(2)

from (1) x= 5/2 -y ...............(3)

put in (2) => (5/2-y)^2 +y^2 =13/4

25/4 +y^2 -5y + y^2 =13/4
=> 2y^2 -5y = 13/4 -25/4

=>2y^2-5y=-3

by solving this quadratic eq. we get roots y= 1 & 3/2 => y^5=243/32 &1

by putting the values of y in (3) we have x= 3/2 &1 => x^5 = 243/32 &1

=> x^5+y^5= 243/32+1= 275/32 =a/b

hence a+b = 307

(x+y)^2 = x^2+2xy+y^2 = 25/4 13/4+2xy = 25/4 xy = 3/2

x^3+y^3 = (x+y)(x^2+y^2-xy) = 5/2*(13/4-3/2) = 35/8

x^5+y^5 = (x^3+y^3)(x^2+y^2) - (xy)^2(x+y) = (35/8)(13/4) - (9/4)(5/2) = 275/32 = a/b so a+b =307

Am I the only one that tried to solve $a+b=\frac{a}{b}$ at first?