I have a system of equations as follow:
For $k,i,j=0,1,2,3,4..... $ $$1) k=3+18ij+11i+5j$$ $$2) k=5+18ij+13i+7j$$ $$3) k=16+18ij+17(i+j)$$ $$4) k=18ij+19i+19j+20$$
the $k$,$i$,$j$ all are integers .. I am interested in all $k$ in a given range where range always begins from 0 up to a known integer (say 1000). Evidently these equations may deliver the same $k$, i.e. we may get duplicates. here are two questions: 1-How could I count all such $k$ excluding repetitions? 2- is there a way to unite all four equations?
Thanks
The four equations might have a universal common equation they arise from. The first two transform as follows:
$$ k=18ab+11a+5b+3\\ 18k=18^2ab-7\cdot 18a+5\cdot 18b-36\\ 18k+1=(18a+5)(18b-7) $$
$$ k=18ab+13a+7b+5\\ 18k=18^2ab-5\cdot 18a+7\cdot 18b-36\\ 18k+1=(18a+7)(18b-5) $$
These two come together to make
$$18k+1=(18a\pm 5)(18b\mp 7)$$
With the other two equations, we do some of the same processes and we get
$$18k+1=(18a\pm 1)(18b\pm 1)$$
The universal equation they come from is along the lines of
$$18k+1=(2a+1)(2b+1)$$
The problem with using this equation is that you get more results on the RHS than on the LHS.
The easy approach is to note that this set of numbers is the count of values of the form $18k+1$ which are not prime, and Dirichlet's theorem on primes in arithmetic progressions can give a good approximation for the value.
To find the complete count of such results, you probably have to manually (with a program) count the set of $k$ results which are unique. With no more than $k=1000$, it may even be possible to count all the $18k+1$ primes by hand.