Question 1208064
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Quadrilateral PQRS is cyclic and side PS = u is a
diameter of the circle. If PQ = QR = v, RS = w, and u, v,
and w are integers such that v does not equal w, prove that u cannot be a
prime number.
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        I will give another solution,  different from  Edwin,  and more compact.



<pre>
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

    {{{d^2}}} = {{{v^2 + v^2 - 2v*v*cos(Q)}}}

or

    {{{d^2}}} = {{{2v^2 - 2v^2*cos(Q)}}}.    (1)


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

    {{{d^2}}} = {{{u^2 - w^2}}}.             (2)


From (1) and (2), we can exclude  {{{d^2}}}  and write

    {{{2v^2 - 2v^2*cos(Q)}}} = {{{u^2 - w^2}}}.    (3)


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


Hence, from (3) we have

    {{{2v^2 + 2v^2*cos(S)}}} = {{{u^2 - w^2}}}.    (4)


Next,  cos(S) = {{{w/u}}}  (from the right triangle PRS).

Thus

    {{{2v^2 + 2v^2*(w/u)}}} = {{{u^2 - w^2}}}.      (5)


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

    {{{2v^2*u + 2v^2*w}}} = {{{u^3 - w^2*u}}}.


Combine the terms with "u" on the right side; keep the remaining term  {{{2v^2*w}}}  on the left side

    {{{2*v^2*w}}} = {{{u^3 - w^2*u - 2*v^2*u}}}.   (6)


In (6), right side is a multiple of "u".  So, if "u" is a prime number, 
then left side  {{{2v^2*w}}}  is a multiple of "u".


Hence, EITHER "u" divides 2,  OR  "u" divides  {{{v^2}}},  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 {{{v^2}}}, 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.
</pre>

Solved.