Maybe Ikleyn or Greenestamps can simplify my solution but here it is at last.

Since quadrilaterals are normally lettered counter-clockwise, and PS is a
diameter, the quadrilateral must be inscribed in a semi-circle. I will need
some right triangles so I will bisect everything including the diameter. The
green line segments are the perpendicular bisectors of the upper 3 sides and
the angles as well since the triangles are isosceles. So the radius of the
circle is u/2, and will be the hypotenuse of all 6 right triangles in this figure:



[eqs. A]   and 

The sum of all 6 angles is 180o



2α and β are complementary,

α and α+β are also complementary

So 

[eq. B]   

. Taking sines of both sides:





Using the associative law and since 2α and β are complementary,

Substituting from [eq. B] above





From [eqs. A]





Multiply through by u3





Since  are factors of the last term, we try both and
find that w=-u is a solution and so we factor it by synthetic 
division

-u | u      2v2   2uv2-u3
   |        -u2  -2uv2+u3
     u   2v2-u2         0



;  
;    
              
              
               
              
So the term  must be an integer, since the terms 
on the right are integers.

For contradiction, assume u is a prime number.

If u=2, then the radius is 1, and the semicircle is π.

So , that would make v=1, w=2. then triangle
SRO would have sides 1,1,2 which violates the triangle
inequality.  So u is not the prime number 2.

Thus u must divide evenly into v2.   But if u is a prime 
then u must divide evenly into v as well. This cannot be true 
because v < u, the diameter.

Thus we have a contradiction and u cannot be a prime number. 

Edwin

Make a sketch.


So, you have quadrilateral PQRS inscribed in the circle such that PS 
is a diameter of this circle; PQ = QR = v; RS = w; PS = u.


Draw the diagonal PR.  Let PR = d (the length).


We have isosceles triangle PQR.  Using the cosine law, we can write

     = 

or

     = .    (1)


From the other hand, triangle PRS is the right triangle, since angle PRS 
leans on the diameter PS.  Therefore,

     = .             (2)


From (1) and (2), we can exclude    and write

     = .    (3)


Since quadrilateral PQRS is inscribed in the circle, its opposite angles S and Q 
are supplementary:  S + Q = .  Therefore,  cos(Q) = -cos(S).


Hence, from (3) we have

     = .    (4)


Next,  cos(S) =   (from the right triangle PRS).

Thus

     = .      (5)


Multiply both sides of (5) by  "u".  You will get

     = .


Combine the terms with "u" on the right side; keep the remaining term    on the left side

     = .   (6)


In (6), right side is a multiple of "u".  So, if "u" is a prime number, 
then left side    is a multiple of "u".


Hence, EITHER "u" divides 2,  OR  "u" divides  ,  OR  "u"  divides  "w".


It leads to contradiction, since

    - if "u" divides 2, then u= 2;  but then from the right triangle PRS the leg w
      must be shorter than the hypotenuse u, i.e. w= 1; and side v must be shorter 
      than the diameter u= 2; so, in this case w = v, which is excluded in the problem.


    - if "u" divides , then "u" divides "v";  but it is impossible, since 
      "v" is less than "u" (the diameter).


    - if "u" divides "w", it is also impossible, since "w" is less than "u" (the diameter).


These contradictions prove that "u" can not be a prime number.