Let $a,b$ be two positive, nonsquare integers such that $a>b$. What is the smallest number $\kappa$ such that $$\left\{(a,b):a\sqrt b+b\sqrt a\,\,\text{is within}\,\,\frac1{(ab)^\kappa}\,\,\text{of an integer}\right\}$$ has finite cardinality?
Note that if $x$ has irrationality measure $\mu$ then $\mu$ is the smallest number (more precisely the infimum) such that $$0<\left|x-\frac pq\right|<\frac1{q^\mu}$$ has finitely many solutions for $p,q$ integers. Here, the problem can be rephrased as determining the smallest number $\kappa$ such that, $$0<\left|a\sqrt b+b\sqrt a-K\right|<\frac1{(ab)^\kappa}$$ has finitely many solutions for an integer $K(a,b)$ which is either $\lfloor a\sqrt b+b\sqrt a\rfloor$ or $\lceil a\sqrt b+b\sqrt a\rceil$. This formulation is close to the definition of an irrationality measure (which is $2$ in this case), but is not directly related since the parameters $p,q$ cannot be matched.
From empirical results, I believe that $\kappa\in(1,2)$, as letting $\kappa=2$ yielded no solutions for a long time. The code in PARI/GP is
squar(k)=for(a=2,+oo,for(b=2,a-1,if((issquare(a)==0 && frac(a*sqrt(b)+b*sqrt(a))<1/((a*b)^k)) || 1-frac(a*sqrt(b)+b*sqrt(a))<1/((a*b)^k),print1([a,b]," "))))
While I recognise that the question posed at the beginning of this post is extremely difficult to determine exactly, I would appreciate proofs that $\kappa>1$ or $\kappa<2$ should they be true.
Interestingly, when $a=b$, I haven't managed to find any solutions when $\kappa=1$. In fact, in this case, I conjecture that $\kappa\in(1/2,1)$.
from Hot Weekly Questions - Mathematics Stack Exchange
TheSimpliFire
Post a Comment