Solving a System of Congruences with A Changing Modulus

drosser

I'm working on some basic number theory. I came across an idea and I'm having trouble finding a general solution.

A == X mod (L-k)
k^2 == Y mod (L-k)

It is the equivalent of:
( k^2 - A ) / ( L - k ) = integer

A and L are two different pre-chosen numbers. I know L-k is a prime number, I just don't know which prime number from L-1 to 2. Is there a way to find a value for k such that X=Y?

An example is A=129 L=130.
I used my calculator to find k=99.
A=1613 L=1940 k=333

I also used the quadratic formula to find a lower bound

[ 4L - sqrt ( (4L)^2 - 4( 2(L^2) + A ) ) ] / 2

I have found the greater:

( k^2 – A ) / ( L – k ) + 2k

Becomes, the less accurate the lower bound is.

I have tried to apply solving systems of congruences but the method was for linear equations. And the changing modulus made it even harder.

Also, when you derive an equation from a previously proven equation, do you have to prove the new equation?

I have a copy of Gareth Jones and Mary Jones “Elementary Number Theory” and George Andrews “Number Theory.” Maybe you can point me to something I missed.

drosser

I realize there is an easy algebraic equation for the remainder of the linear congruence, but for the quadratic congruence I don't recognize any pattern other than the standard 2n-1 from the series for calculating the square of a number.

ramsey2879

I am not sure why L-k must be a prime but if that is what you need then OK. If A is not pre-selected the problem is trite. Simply subtract any prime from L to get k and then select A = = k^2 mod L-k. But the other way around, i.e. preselecting A is more of a problem. I suggest getting a table of quadratic residues of primes and simply checking first if A == a quadratic residue. The lower primes are more apt to make A == a residue. Once you find A == a residue, say for prime = 31 as in your first example, you have your value for L-k!

ramsey2879

Opps I forgot that k relates both k^2 and L-k