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A system of modular equations (Posted on 2008-07-20) Difficulty: 3 of 5
Solve the following set of equations for positive integers a and b:

1919(a)(b) mod 5107 = 1
1919(a+1)(b-1) mod 5108 = 5047
1919(a+2)(b-2) mod 5109 = 1148
1919(a+3)(b-3) mod 5110 = 3631
1919(a+4)(b-4) mod 5111 = 2280

See The Solution Submitted by pcbouhid    
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re(2): Solution ------ can you check? | Comment 3 of 8 |
(In reply to re: Solution ------ can you check? by pcbouhid)

True, there has been an anomaly in the placement of signs.

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  

Thus, there was a typing error, and it should have been '+5107(a-b)' instead of
'-5107(a-b)' in the lhs of (II) in the previous post.

This anomaly is now corrected in the original post.

However, we will still obtain:

(a-5107)(b+5107) (mod L) = -26086391, so that:

(a, b) = (866, 1044), (5106, 26081284) as was originally stated.

Edited on August 1, 2008, 7:21 am
  Posted by K Sengupta on 2008-08-01 06:32:41

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