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 Register FAQ Search Today's Posts Mark Forums Read  2019-11-13, 05:49 #12 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 23·3·52 Posts LaurV, do not believe that there is "a square-free counterexample"   2019-11-13, 20:13 #13 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 23×3×52 Posts This is a joke - I give you Anderson conjecture # 23. for every square free positive integer r, there does not exist any non-zero integer s and t such that sqrt(r) = s + sqrt(t). I believe I saw a theorem about this.   2019-11-14, 10:56   #14
Nick

Dec 2012
The Netherlands

26518 Posts Quote:
 Originally Posted by MattcAnderson for every square free positive integer r, there does not exist any non-zero integer s and t such that sqrt(r) = s + sqrt(t).
You need to rule out r=1 as well.

The key - as LaurV suggested - is that the square root of any integer is either an integer or it is irrational.
Take any fraction $$\frac{a}{b}$$ where a & b are integers (with b≠0).
We can choose them so that gcd(a,b)=1.
Suppose $$\left(\frac{a}{b}\right)^2=n$$ for some integer n.
Then $$a^2=nb^2$$ so every prime number dividing b also divides $$a^2$$ so it also divides a.
But gcd(a,b)=1 so there are no such prime numbers.
Hence b is 1 or -1 and the fraction $$\frac{a}{b}$$ was an integer all along.   2019-11-14, 13:09   #15
Dr Sardonicus

Feb 2017
Nowhere

1101110111102 Posts Quote:
 Originally Posted by MattcAnderson This is a joke - I give you Anderson conjecture # 23. for every square free positive integer r, there does not exist any non-zero integer s and t such that sqrt(r) = s + sqrt(t). I believe I saw a theorem about this.
With r = 1, we have, e.g.

1 = -1 + sqrt(4)

If we assume r is a positive integer, and not the square of an integer, and t is a positive integer, we can argue as follows:

sqrt(r) = s + sqrt(t)

sqrt(r) - sqrt(t) = s

r - 2*sqrt(r*t) + t = s2

-2*sqrt(r*t) = s2 - r - t

The right side is an integer, hence r*t is the square of an integer. Thus,

t = r*v2, v could be rational, but t is an integer

[The denominator of v2 must divide the largest square factor of r. If you assume r is square free and r > 1, then v must be an integer]. Then

-2*r*v = s2 - r - r*v2

r*v2 - 2*r*v + r = s2

r*(v - 1)2 = s2

The left side, though an integer, is not the square of an integer, because r is not the square of an integer. The right side is the square of an integer. Uh-oh, trouble...   Thread Tools Show Printable Version Email this Page

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