Re: Rational and irrational numbers

Counterexample: a=8,b=7,k=7


> Let a,b be relatively prime, non-square positive integers, and let k
> be an odd integer, k>5.
>
> If u,v are real and such that:
>
> sqrt(a)+sqrt(b)=u^k
>
> sqrt(a)-sqrt(b)=v^k
>
> u*v=1
>
> Then u,v can be expressed in the form:
>
> u = A*sqrt(g) + B*sqrt(h)
>
> v = A*sqrt(g) - B*sqrt(h)
>
> for some positive integers A,B,g,h and where g,h are non-square.
>
>

u*v=1 implies (u^k)*(v^k)=1, so we have
(sqrt(a)+sqrt(b))*(sqrt(a)-sqrt(b))=1, that is a-b=1.

So, can we say that u = (sqrt(b+1)+sqrt(b))^(1/k)
must be of the form A*sqrt(g) + B*sqrt(h) ?

No. (sqrt(8)+sqrt(7))^(1/7) is not of that form.
It has degree at least 28 over the rationals, since it is a root of the
degree 28 irreducible polynomial x^(28) - 30*x^(14) + 1 .
But A*sqrt(g) + B*sqrt(h) has degree at most 4.
.

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