Congruences: Show by congruences that q^2 + 3r^2 + 6s^3 – 9p^5 = 2 has no solutions in Z.
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Did you try reading the equation modulo three?
Originally Posted by Isomorphism Did you try reading the equation modulo three? How would i do that?
In the equation becomes because 3,6 and 9 are in the equivalence class of 0. What are the squares in ?
Originally Posted by vincisonfire In the equation becomes because 3,6 and 9 are in the equivalence class of 0. What are the squares in ? Would it be plus or minus root 2?
Originally Posted by vincisonfire and these are not solutions in the set Z? sorry im new to number theory
No problem. We all have to start somewhere. It will seem simple to you in few weeks from now.
Originally Posted by vincisonfire No problem. We all have to start somewhere. It will seem simple to you in few weeks from now. I hope so. Would i not be able to show it has no solutions in Z without using congruences??
Originally Posted by vincisonfire In the equation becomes because 3,6 and 9 are in the equivalence class of 0. What are the squares in ? Here when you say q^2=2, do you mean q^2 is congruent to 2(mod3)?
Yes, because there is no elements in mod3 such that its square is 2.
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