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Mar 12, 2014 at 17:05 history edited Sasha CC BY-SA 3.0
added 17 characters in body
Mar 12, 2014 at 17:03 comment added Sasha You are right, sorry.
Mar 12, 2014 at 16:46 comment added Achim Krause You somehow assume that $f_5$ has a rational root, whereas in general a nonconstant polynomial can very well be irreducible over Q. Joro's example works perfectly well, since $x^2+3$ only has complex roots.
Mar 12, 2014 at 16:35 comment added joro Thanks what do you mean by any degree? I get (3-x^2)^2+(2*x)^2+(2*x)^2+(2*x)^2=(x^2+3)^2
Mar 12, 2014 at 16:22 history answered Sasha CC BY-SA 3.0