Series Within Parallel Resistance Networks

If electrical resistors r1 and r2 are connected in series they give
a total resistance of r1 + r2.  Likewise connecting r3 and r4 in 
series gives a total resistance of r3 + r4.  Putting these series 
combination in parallel gives a combined resistance of 

                   (r1 + r2)(r3 + r4)
            rx  =  ------------------
                    r1 + r2 + r3 + r4

Quentin Grady notes that if we connect the node between r1 & r2 to 
the node between r3 & r4 the circuit is transformed to a parallel 
within series connection, and the resistance becomes

                      r1 r3         r2 r4
            ry   =   -------   +   -------
                     r1 + r3       r2 + r4

He asks for positive integer solutions of this equation, i.e., 
integers values of r1, r2, r3, r4, rx, and ry, "preferably under 
100", that satisfy this equation.

There are a very large number of solutions to this.  Even if we
require r1 through r4 to be distinct, and add the restriction that 
the values of (r1*r3)/(r1 + r3) and (r2*r4)/(r2+r4) must both be 
integers, there are still an over-abundance of solutions.  Most of 
them have r1*r4 = r2*r3, which implies that rx=ry.  In other words, 
the overall resistance is unchanged by inserting the jumper.  An 
example is [r1,r2,r3,r4] = [3,9,6,18].  You can easily define
infinite families of such solutions.  For example, with

 r1 = 8j+2     r2 = 24j+8k+10      r3 = 24j+6     r4 = 72j+24k+30

we have rx = ry = 24j + 6k + 9.  On the other hand, there are also
many solutions with rx > ry.  An example is [5,99,95,9].

There are so many solutions to these equations that I think it's
more interesting to restrict the problem a little more by requiring
not only that Rx and Ry are integers but also the components of
Ry, i.e., R1R3/(R1+R3) and R2R4/(R2+R4).  In other words, we seek
four integers a,b,c,d such that each of the quantities

         ac              bd             (a+b)(c+d)   
    M = -----      N =  ----       K =  ----------
         a+c             b+d             a+b+c+d    

is an integer.  The expressions for M and N imply

           (a-M)(c-M) = M^2         (b-N)(d-N) = N^2

This means there are integers W,X,Y,Z such that WX=M^2, YZ=N^2, and

     a = M+W       b = N+X        c = M+Y        d = N+Z

Thus for any integers M,N we need only select factorizations of M^2 
and N^2 into two factors to generate values of a,b,c,d that satisfy
the equations for M and N.  Now we just need to ensure that the 
equation for K is also satisfied.

The equation for K implies the existence of integers m,n such that
mn=K^2 and
              (a+b) = K + m        (c+d) = K + n

Thus we have

                (a+b) - (c+d)  =  m - n                  (1)
          (a+b) + (c+d)  - 2K  =  m + n                  (2)
                          K^2  =  m n                    (3)

Letting Q denote a+b+c+d-2K, equations (2) and (3) imply that m
and n are the roots of

               m^2 - Qm + K^2  =  0

so we have
                  _________                     _________
             Q + /Q^2 -4K^2                Q - /Q^2 -4K^2
         m = --------------            n = --------------
                   2                             2

To make these integers it's clear that the quantity in the radical
must be a square, so there is an integer f such that

                  f^2 + (2K)^2  =  Q^2

By Pythagorean triples this implies the existence of integers q,r,s
such that

     f^2 = q(r^2 - s^2)        K = qrs       Q = q(r^2 + s^2)

and so we can express m and n as

           q(r^2 + s^2) + q(r^2 - s^2) 
     m  =  ---------------------------  =  qr^2
                       2

           q(r^2 + s^2) - q(r^2 - s^2)
     n  =  ---------------------------  =  qs^2
                      2

Substituting these expressions for m, n, and K back into equations 
(1) and (2) gives

              q(r^2 - s^2)  =  (a+b) - (c+d)                  (4)

               q(r + s)^2   =  (a+b) + (c+d)                  (5)

Now recall that we have

     a = M+W       b = N+Y        c = M+X        d = N+Z

where  WX=M^2  and  YZ=N^2.  It follows that there must be integers
e,f,w,x,y,z such that

       W = ew^2     X = ex^2      Y = fy^2      Z = fz^2

and so  M = ewx  and  N = fyz.  Substituting these expressions into
equations (4) and (5) gives

   q(r^2 - s^2)  =  (ewx+ew^2 + fyz+fy^2) - (ewx+ex^2 + fyz+fz^2)

    q(r + s)^2   =  (ewx+ew^2 + fyz+fy^2) + (ewx+ex^2 + fyz+fz^2)

Cancelling terms, these equations reduce to

        q(r+s)(r-s)  =  e(w^2 - x^2)  +  f(y^2 - z^2)

        q(r+s)(r+s)  =   e(w + x)^2   +  f(y + z)^2

Multiplying the first by (w+x) and the second by (w-x), and then 
subtracting one form the other gives

 q(r+s)[(w+x)(r-s) - (w-x)(r+s)] = f(y+z)[(w+x)(y-z) - (w-x)(y+z)]

which reduces to

                 q(r+s)(xr-ws) =  f(y+z)(xy-wz)

Similarly we can eliminate f from the two preceeding equations 
to give the relation

                 q(r+s)(zr-ys) =  e(w+x)(wz-xy)

This shows that there are two general classes of solutions to the 
overall problem.  The degenerate case is when

                 xr-ws = xy-wz = zr-ys = 0

which is true iff  (r/s) = (w/x) = (y/z).  In this case we can set
w=jy and x=jz for any rational j, and we have the parametric set of 
solutions

      a = ewx+ew^2  =  ey(y+z)j^2          M = y z e j^2
      b = fyz+fy^2  =  fy(y+z)             N = y z f
      c = ewx+ex^2  =  ez(y+z)j^2          K = y z (ej^2 + f)
      d = fyz+fz^2  =  fz(y+z)

On the other hand, if the quantities  xr-ws, xy-wz, and zr-ys  are 
not zero, we can still have solutions by solving the equations

                 q(r+s)(xr-ws) =  f(y+z)(xy-wz)

                 q(r+s)(zr-ys) =  e(w+x)(wz-xy)

for e, f, and q.  Clearly the first equation is satisfied if we set
q = (y+z)(xy-wz) and f = (r+s)(xr-ws).  Substituting this value of q
into the second equation gives

            (y+z)(r+s)(zr-ys)(xy-wz)  =  e(w+x)(wz-xy)
or
                  (y+z)(r+s)(ys-zr)
           e  =  -------------------
                        (w+x)

To ensure that e is an integer we can multiply e, f, and q by (w+x) 
to give the solution

             e = (r+s)(y+z)(ys-zr)
             f = (w+x)(r+s)(xr-ws)
             q = (y+z)(w+x)(xy-wz)

Any common factors can be divided out of these three numbers.  Then 
the final form of the solution (in the non-degenerate case, and 
without reducing to lowest terms) is

      a = w(ys-zr)J          M = (r+s)(y+z)(ys-zr)wx
      b = y(xr-ws)J          N = (r+s)(x+w)(xr-ws)yz
      c = x(ys-zr)J          K = (x+w)(y+z)(xy-wz)rs
      d = z(xr-ws)J

where J = (x+w)(y+z)(r+s).  By the way, notice that

    Rx - Ry   =   K - N - M    =   (ys-zr)(xr-ws)(xy-wz)

so this is the amount by which the two circuits differ.  The smallest
solutions are those for which each of the three values ys-zr, xr-ws,
and xy-wz have their minimum possible value, which is +1, so most of
the small solutions give Rx - Ry = +1.

There's also a connection with what are called "Farey sequences".  
Remember that in order to give all positive resistances we require 
each of the quantities ys-zr, xr-ws, and xy-wz to be positive, which 
is the case if and only if

                 (y/z)  >  (r/s)  >  (w/x)

Now, observe that the four basic resistors a,b,c,d will be proportional 
to w,y,x,z if we can ensure that the "commutators"  are all +1.  All we 
really need to do is take any four integers w,x,y,z such that xy-wz=1 
and then set r=w+y and s=x+z.  This automatically ensures not only that 
r/s is between y/z and w/x, but that the quantities xr-ws and ys-zr are 
both equal to 1.  This automatically gives a solution of the simple 
form
      a = wJ          M = (r+s)(y+z)wx
      b = yJ          N = (r+s)(x+w)yz
      c = xJ          K = (x+w)(y+z)rs
      d = zJ

where J = (x+w)(y+z)(r+s)  and the difference Rx-Ry equals 1.  Of course, 
the difference between Rx and Ry is also equal to

                (ad-bc)^2
             ------------------
             (a+c)(b+d)(a+b+c+d)

which shows (again) why the difference cannot be negative.