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I was trying to find a solution to problem 1 of the 1959 IMO that didn't use the Euclidean Algorithm and I was wondering if someone could check my answer.
pi³
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hi pi_cubed
I hadn't come across these problems so many thanks for that. The algebra is ok so your argument seems to depend upon
<the re-arranged expression is irreducible> implies <the original is irreducible>
I think that is ok (maybe it's even a theorem) but it's not something I've encountered before, so I'm having a think about that.
LATER EDIT.
It's trivial to show that 2/3 = 4/6
But <2/3 is irreducible> does not imply <4/6 is irreducible>
So a re-arrangement alone is in-sufficient; something more is needed I think.
Bob
Bob
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