Now I understand your idea better. But you are possibly wrong: The main complexity of your approach is not because of the degree of the polynomials but because of the referencing of the literals (you called them 'variables') to each other.
At this point you have just two choices:
1. Try random input (you know this solution under the (wrong) name 'brute force')
2. Try to approximate to a good solution (but then your result would only be partly correct: Who is interested in a 90% correct nonce to the given block and targeted result?) and it is not proven in this case that an approximation exists.
I think there is some misunderstanding of the technique that I will try and clear up with an example. It should in fact give an exact solution, not an approximate one.
As mentioned, traditional bitcoin mining performs target=SHA256(SHA256(header+nonce)), looking for a very specific solution (top 4 bytes of 'target' are 0.) Mining today uses a brute force approach, iterating through every possible nonce value in the 32 bit address space, performing the double SHA256, and looking at the result to check if those top 4 bytes are 0.
In my example here, rather than the function target=SHA256(SHA256(header+nonce)), let's start with a very simple function:
target = (nonce + 0xAD) ^ 0x6E;
(Assume we're working with 8 bit numbers for this example.)
Just like mining, let's try to find a value for 'nonce' such that target = 0; Mining today would try and brute force try all possible values for 'nonce' until that condition is satisfied, e.g:
for(nonce = 0; nonce <= 255; nonce++)
{
target = (nonce + 0xAD) ^ 0x6E;
if(target == 0)
return nonce;
}
In my proposed approach, this is done symbolically. First, symbolically do the addition and Xor operations to produce a set of equations that represents each of the 8 bits of 'target' in terms of the 8 bits of 'nonce'
t0: (!n0)
t1: (Xor[1, n0, n1])
t2: (Xor[n2, n1 && n0])
t3: (Xor[1, !n3, n2 || (n1 && n0)])
t4: (Xor[n4, n3 || n2 || (n1 && n0)])
t5: (Xor[1, n4 && (n3 || n2 || (n1 && n0)), !n5])
t6: (Xor[1, n6, n5 || (n4 && (n3 || n2 || (n1 && n0)))])
t7: (Xor[n6 && (n5 || (n4 && (n3 || n2 || (n1 && n0)))), !n7])
Now we can solve the system of equations, which will give us values for n0...n7 which satisfy the condition that t0...t7 == 0; This will not be an approximation, we will get exact values for each bit.
As mentioned earlier, we convert the logical operations to arithmetic operations, modulo 2:
t0: (1+n0)
t1: (1+n0+n1)
t2: (n0*n1+n2)
t3: (n2+n0*n1*(1+n2)+n3)
t4: (n2+n3+n2*n3+n0*n1*(1+n2)*(1+n3)+n4)
t5: ((n2+n3+n2*n3+n0*n1*(1+n2)*(1+n3))*n4+n5)
t6: (1+n5+(n2+n3+n2*n3+n0*n1*(1+n2)*(1+n3))*n4*(1+n5)+n6)
t7: (1+(n5+(n2+n3+n2*n3+n0*n1*(1+n2)*(1+n3))*n4*(1+n5))*n6+n7)
And now we can solve (done using wolfram mathematica):
In[1] := Solve[(1 + n0) == 0 && (1 + n0 + n1) == 0 && (n0*n1 + n2) == 0 && (n2 + n0*n1*(1 + n2) + n3) == 0 && (n2 + n3 + n2*n3 + n0*n1*(1 + n2)*(1 + n3) + n4) == 0 && ((n2 + n3 + n2*n3 + n0*n1*(1 + n2)*(1 + n3))*n4 + n5) == 0 && (1 + n5 + (n2 + n3 + n2*n3 + n0*n1*(1 + n2)*(1 + n3))*n4*(1 + n5) + n6) == 0 && (1 + (n5 + (n2 + n3 + n2*n3 + n0*n1*(1 + n2)*(1 + n3))*n4*(1 + n5))*n6 + n7) == 0 && n0*n0 - n0 == 0 && n1*n1 - n1 == 0 && n2*n2 - n2 == 0 && n3*n3 - n3 == 0 && n4*n4 - n4 == 0 && n5*n5 - n5 == 0 && n6*n6 - n6 == 0 && n7*n7 - n7 == 0, {n0, n1, n2, n3, n4, n5, n6, n7}, Modulus -> 2]
Out[1] = {{n0 -> 1, n1 -> 0, n2 -> 0, n3 -> 0, n4 -> 0, n5 -> 0, n6 -> 1, n7 -> 1}}
You can see this produced the exact solution: 11000001 == 0xC1 == 193, which does in fact validate:
(0xC1 + 0xAD) ^ 0x6E == 0
Now we just expand this for bitcoin mining. Do everything I did above, but with the symbolic version of SHA256(SHA256(block_header)) rather than my simple example. Take the first 32 equations from this result set (corresponding to the 4 bytes we want to be 0), and fill in literal values from the block header for everything except for the 32 bits of nonce we want to solve for. Now solve the system of equations; we get the exact bits for nonce that produces a target with the first 4 bytes zeroed.
Again, the symbolic SHA256 stuff can all be computed offline and saved - the computation part of mining now will just be solving that system of 32 equations, which mathematica (or another symbolic math module) can do incredibly fast. Much faster than my GPU can brute-force every possible nonce combination.
the ram is ECC too 
