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.

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