20191113, 05:49  #12 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}×149 Posts 
LaurV,
do not believe that there is "a squarefree counterexample" 
20191113, 20:13  #13 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}×149 Posts 
This is a joke 
I give you Anderson conjecture # 23. for every square free positive integer r, there does not exist any nonzero integer s and t such that sqrt(r) = s + sqrt(t). I believe I saw a theorem about this. 
20191114, 10:56  #14  
Dec 2012
The Netherlands
2^{5}·3^{2}·5 Posts 
Quote:
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. 

20191114, 13:09  #15  
Feb 2017
Nowhere
23×151 Posts 
Quote:
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 = s^{2} 2*sqrt(r*t) = s^{2}  r  t The right side is an integer, hence r*t is the square of an integer. Thus, t = r*v^{2}, v could be rational, but t is an integer [The denominator of v^{2} 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 = s^{2}  r  r*v^{2} r*v^{2}  2*r*v + r = s^{2} r*(v  1)^{2} = s^{2} 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. Uhoh, trouble... 
