; I first rather lazily tried Magma's command EliminationIdeal(I,2); The result was a degree-55 polynomial in x3, which factors as x3^3*(quartic)^3*(irreducible degree-40). So we expect maybe 45 or so points in the intersection. But that's missing the "exceptional" points (where several have the same x3). So really we have to Do The Right Thing. The question here is the number of components in a variety of (we hope) dimension zero, determined by an ideal generated by three polynomials in (say) Z[x1, x2, x3]. The algebraic geometry involved can be expressed ring-theoretically: we need to know the decomposition of the radical of that ideal as an intersection of prime ideals. High-flying talk? Fear not. Magma also has a command RadicalDecomposition(I); to replace the EliminationIdeal. It computes exactly these primes, and as a bonus determines the number of points in each of the corresponding varieties. With a few more hours of machine time we find the variety has four points with x3=0, four with x2=0, and 40 in an irreducible component with neither x2=0 nor x3=0. (Actually I got this result faster first and then corroborated more slowly: Declaring instead p:=157526249; (or some other large prime) and K:=GF(p); gives the decomposition into prime ideals fairly quickly; one notices the x2=0 and x3=0 patterns quickly, and verifies them integrally, and then finds the remaining factor must be irreducible to be consistent with the data from a few such primes. Of course, it _could_ happen that I just happened to choose primes from among the set over which the number of points decreases from the integral picture, so I had to verify the whole calculation integrally, which took much longer. OK, I'm lying: I killed it after about an hour.) Since we're going more carefully here, it's worth noting that the cases x2=0 or x3=0 don't translate so nicely to the original 5-variable problem. One can re-run the program starting with a 5-variable ring, and do it all properly. Each of the 40 points indicated above does indeed lead to a (unique) point in C^5; those with x2=0 or x3=0 are not in the image of the variety in C^5. So, after many wasted machine cycles, we are led to the conclusion that there are 40 points in this solution set in C^5. Of course, this all pertains only to the _particular_ choices I made for L2 L5 L7 S1 S2, and assumes L1, L3, and L6 are linearly independent. I tried another set of choices: p:=10000000019; K:=GF(p); P:=PolynomialRing(K,5,"lex"); L1:=x1;L4:=x2;L6:=x3; c0:=1; c1:=1; c2:=2; c3:=13; L2:=c0+c1*x1+c2*x2+c3*x3; L7:=1-4*x1+10*x2+16*x3; L5:=4-x1-8*x2+6*x3; S1:=2+x1-x2+3*x3-11*x1^2+2*x2^2+5*x3^2-6*x1*x2+7*x1*x3-17*x2*x3; S2:=+1-2*x1+99*x2+0*x3+7*x1^2-111*x2^2+13*x3^2-17*x1*x2+11*x1*x3-23*x2*x3; I:=ideal
;
RadicalDecomposition(I);
Conclusion: another set of 40 points mod p, presumably 40 points over
the rationals too. Taking another leap of faith, we expect 40 points
in C^5 for almost all choices of the L's and S's.
(I wanted to be a little more general, so I tried replacing the first Magma
ring (and leaving c0...c3 variable):
L:=GF(p); K