John Cremona on Tue, 20 Jan 2015 10:20:13 +0100 |
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Mixing variables in Mod expressions |
I am trying to understand the following. The answer might well be something like "if you try to do something stupid then you must expect a stupid answer" ? Mod(x,x^2-3) + Mod(x,x^2-5) %1 = 0 -- but I was pretending to be a student trying to find the polynomial satisfied by sqrt(3)+sqrt(5) and doing what seemed natural. This version works: ? Mod(x,x^2-3) + Mod(y,y^2-5) %2 = Mod(x + Mod(y, y^2 - 5), x^2 - 3) ? a = Mod(x,x^2-3) + Mod(y,y^2-5) %3 = Mod(x + Mod(y, y^2 - 5), x^2 - 3) ? a^2-8 %4 = Mod(Mod(2*y, y^2 - 5)*x, x^2 - 3) ? (a^2-8)^2 %5 = Mod(Mod(60, y^2 - 5), x^2 - 3) ? (a^2-8)^2-60 %6 = 0 but that is not the point; result %1 is surely going to confuse people. John