All the solutions for this system 5x+33y = 6 (mod 13) and 7x + 2y = 9 (mod 13)

figoveck38

figoveck38

Answered question

2022-11-11

All the solutions for this system 5x+33y = 6 (mod 13) and 7x + 2y = 9 (mod 13)

Answer & Explanation

Sean Sutton

Sean Sutton

Beginner2022-11-12Added 17 answers

5 x + 33 y = 13 a + 6         ( 1 )
7 x + 2 y = 13 b + 9         ( 2 )
where a , b are integers
( 1 ) 7 ( 2 ) 5 221 y = 7 ( 13 a + 6 ) 5 ( 13 b + 9 ) = 13 ( 7 a 5 b ) 3
3 = 13 ( 7 a 5 b 17 y )
which is impossible as the Right Hand Side is divisible by 13 unlike the Left one
Ty Gaines

Ty Gaines

Beginner2022-11-13Added 3 answers

Doing arithmetic modulo 13 all the time:
I 5 x + 33 y = 6 5 x + 7 y = 6 I I 7 x + 2 y = 9 7 x + 2 y = 9
Multiply I by 7 and II by 5 :
I 4 x + 3 y = 10 I I 9 x + 10 y = 6 I + I I 0 = 1 contradiction
and thus the equation has no solution.
Another way: solve as over the reals (again, multiplying I by 7 and etc.)
I 5 x + 33 y = 6 I I 7 x + 2 y = 9
I 35 x 231 y = 42 I I 35 x + 10 y = 45 221 y = 3
Yet the above is already 0 = 3 since 221 = 13 17 = 0 ( mod 13 )

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