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Consider $5x+3y=4$ and $3x+6y=1.$ List the set of primes for which this system of linear equations does not have a solution in the field $Z_p.$

Mirunalini_UML
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  • What have you tried? Have you checked for a few simple cases like $p = 2, 3, 5$? – John Hughes Sep 15 '14 at 13:13
  • since these two are non-parallel lines, they meet at a point$(1,-1/3).$ sorry, I don't know how to proceed here after, – Mirunalini_UML Sep 15 '14 at 13:17
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    That's true in the real plane, but you're working over $Z/pZ$, and perhaps you've assumed something that's not quite true. Have you checked those simple cases? You may be enlightened by at least one of them. – John Hughes Sep 15 '14 at 13:20
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    The determinant of the matrix associated to the system is $5\cdot 6 - 3 \cdot 3 = 21$, so it is non zero (the system has a solution) for $p\neq 3,7$. Then study separately the two cases $p=3$ and $p=7$. – Crostul Sep 15 '14 at 13:22
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    In $\Bbb{Z}_3$ you have $3=0$, so dividing by three is.... In $\Bbb{Z}_7^*$ what is the inverse of $3$? – Jyrki Lahtonen Sep 15 '14 at 15:40
  • @Jyrki Lahtonen. I got the first statement. In $\Bbb{Z}_3$ we have $3=0,$ so the determinant become zero, so solution does not exist. In $\Bbb{Z}_7^*$ the inverse of 3 is 5. but what is the hint behind this? sorry i do not understand, – Mirunalini_UML Sep 16 '14 at 05:17
  • @Crostul. thank you very much i got some idea... – Mirunalini_UML Sep 16 '14 at 05:19
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    Ok. If $1/3=5$, then $(1,-1/3)=(1,-5)=(1,2)$ is a solution. Check it out! – Jyrki Lahtonen Sep 16 '14 at 05:20
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    And observe that when $p=7$, the former equation can be rewritten as $5x+10y=4$, so (divide by five) $x+2y=4/5$. This is parallel to the other line as $3x+6y=1$ is equivalent to $x+2y=1/3$ (divide by three). The lines actually coincide, because modulo seven $4\cdot3=1\cdot5$, so $1/3=4/5$. Modulo three the situation is worse in the sense that the latter equation does not define a line in the first place. – Jyrki Lahtonen Sep 16 '14 at 19:39
  • So when $p=7$ this system has infinitely many solution. Isn't it? Thanks a lot. – Mirunalini_UML Sep 18 '14 at 11:03

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