Reducing the given equations by their respective modulii, the first four equations can be rewritten as:

1919 (a)(b) (mod 5107) = 1

1919 (ab-a+b) (mod 5108) = 1858

1919 (ab-2a+2b) (mod 5109) = 3715

1919 (ab-3a+3b) (mod 5110) = 2318

In the fifth given equation, we observe that: gcd (1919, 5111, 2280) = 19, so that upon division by 19, the equation now becomes:

101 (a+4)(b-4) (mod 269) = 120

or, 101(ab-4a+4b) (mod 269) = 122

We now denote the respective modular inverses of 1919 in mod 5107, mod 5108, mod 5109, mod 5110 in turn by A(1), A(2), A(3), A(4) and the modular inverse of 101 mod 269 by A(5).

By way of a little trial and error, we obtain:

A(1) = 165; A(2) = 2007; A(3) = 2657; A(4) = 3949, and A(5) = 8.

Accordingly, we must have:

(ab) (mod 5107) = 165

(ab-a+b) (mod 5108) = (1858*2007) (mod 5108) = 166

(ab-2a+2b) (mod 5109) = (3715*2657) (mod 5109) = 167

(ab-3a+3b) (mod 5110) = (462*3949) (mod 5110) = 168

(ab-4a+4b) (mod 269) = (122*8) (mod 269) = 169............(I)

Adding 5107(a-b), 5108(a-b), 5109(a-b), 5110(a-b), 5111(a-b) in turn to the lhs of the system of equations in (I), subtracting 5107, 5108, 5109, 5110, 5111 in turn from each of the rhs in (I), and upon subsequent rearrangement, we obtain:

(ab+5107(a-b) + 4942) (mod 5107) = 0

(ab+5107(a-b) + 4942) (mod 5108) = 0

(ab+5107(a-b) + 4942) (mod 5109) = 0

(ab+5107(a-b) + 4942) (mod 5110) = 0

(ab+5107(a-b) + 4942) (mod 269) = 0 ................(II)

Accordingly, we must have:

((ab+5107(a-b) + 4942) (mod L) = 0,

where: L = LCM(5107, 5108, 5109, 5110, 269)

= 91,600,075,916,256,180

or, (a-5107)(b+5107) (Mod L) = -26086391, and http://www.virtuescience.com/prime-factor-calculator.html

gives the prime factorization of 26086391 as 4241*6151 apart from the trivial factorization 1*N, where N = 26086391.

Thus, (a-5107, b+5107) (mod L) = (-4241, 6151), (4241, -6151), (6151, -4241), (-6151, 4241), (1, -N),

(-1, N), (N, -1), (-N, 1)

The only positive pair (a,b) occurs whenever (a-5107, b+5107) (mod L) = (-4241, 6151), (-1, N): giving:

(a, b) (Mod L) = (866, 1044), (5106, 26081284)

Now, each of 866, 1044, 5106 and 26081284 is less than L.

Consequently, (a, b) = (866, 1044), (5106, 26081284) are the only possible solutions to the given problem.

__Note__: There were some typing errors in this post, and these anomalies have been subsequently edited in the light of the comments in the subsequent posts.

*Edited on ***August 1, 2008, 6:41 am**