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I need to solve a system of 36 polynomial equations of degree 2 with integer coefficients in 36 variables over integers. (I do not need rational, real or complex solutions)

I know that the system has at least one solution in integers.

I can use different methods such as eliminating variables one by one, using resultants or using Gröbner basis. None of these methods seems to be efficient for 36 variables and 36 equations.

Do you know a better method? How long should it take to solve it? Can I take as an advantage of knowing that one integer solution exist somehow?

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    It is a special case of the system of polynomial equations? – 2017-01-31
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    It is a general type of system of polynomial equations. What is "special" about it is only that it is of degree 2 and that it has an integer solution. – 2017-01-31
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    There are several other methods, using number theory, but unless there is something special about the equations, it is hopeless. Having an integer solution, or degree $2$ is not special. It is just a system of Diophantine equations. – 2017-01-31
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    For 3 equation and 3 variables it could be for example:$$\left\{3 x^2+4 y^2+5 z^2+2 x y-x z-2 y z+11 x+12 y+7 z-109=0,2 x^2-5 y^2+4 z^2+6 x y+2 x z+8 y z+9 x+4 y-z-98=0,3 x^2+8 y^2+7 z^2+4 x y-4 x z+2 y z+5 x+6 y+2 z-129=0\right\}$$ which has solution x=1,y=2,z=3 – 2017-01-31
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    I may have thought of a method, but need to ask for more information to be sure. 1. Am I correct in thinking you have $36$ logically ordered sets of $703$ (or thereabouts) coefficients in digital format? 2. Apart being integers, are there any restriction on the values of the variables. A maximum would be good. 3. What’s the (approximate) greatest absolute value within coefficients? 4. Within any restrictions, do you need some, or all, solutions? 5. You know that the system has at least one solution in integers, but do you know the values within that solution? – 2017-07-18
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    6. Is this a once-only problem, or recurring? – 2017-07-18
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    1. Yes could be 703, but typically there are only 73 nonzero coefficients in each equation. – 2017-07-18
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    1. Yes could be 703, but typically there are only 73 nonzero coefficients in each equation. 2. Max value is set to 255, but in general you can choose any max value. 3. Around 10000. 4. I think if I have one solution I can find others by parametrization since equations are only of second degree. (not sure) 5. I do not know values within solution, only that max values could be, say, 255. 6. Recurring. A cipher can be decrypted using found 36 values, so if key for the cipher is changed you have to solve another system of equations. – 2017-07-18
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    Correction for 4: I do not know a method to find other solution if I got one. So I would say, I need all solutions, but I do not expect to be there too many. – 2017-07-18
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    Thanks for your answers. Just one further question, if I may. 7. Is your expected value range for variables $0$ to $255$, or $-255$ to $255$? – 2017-07-19
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    7. Yes default range for the cipher key is 0 to 255, but also can be changed. Would it make any difference whether it is 0 to 255 or -255 to 255? – 2017-07-19
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    Yes, a vast difference. I’ll run some timing tests tomorrow, for I’m now having concerns that my method may take far too long. – 2017-07-19
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    Yes it is a vast difference if we use brute-force, but I was hoping for some method that would be independent on the size of numbers involved. – 2017-07-19
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    Pure brute-force is clearly out of the question, for this Earth does not have enough time left. – 2017-07-21
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    I was considering producing a list of $mod$ $2$ solutions, which I estimate to be obtainable in a few processor-hours. Then substituting either the likes of $2v_i$ or $2u_i+1$ for each variable, to produce a new set of equations. Then repeating, using the new equations. However, I see all sorts of problems arising, and am unsure if this would produce a solution or not. – 2017-07-21
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    I was able to solve the system of equations even for modulus 13. But that did not helped me in any way since there were still lots of possibilities to test. – 2017-07-24
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    If you wonder why modulus 13... it is because more of the character of the cipher than for some mathematical reason. – 2017-07-24

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