3

Let $a,b$ be two positive integer numbers such that $a\sqrt{3}>b\sqrt{7}$. Prove that $(a\sqrt{3}-b\sqrt{7})(a+b)>1$.

Attempt. From the assumption we have $3a^2-7b^2\geq 1$. We try to argue that $3a^2-7b^2\geq 3$ to get $$ a\sqrt{3}-b\sqrt{7}=\frac{3a^2-7b^2}{a\sqrt{3}+b\sqrt{7}}\geq\frac{3}{a\sqrt{3}+b\sqrt{7}}>\frac{1}{a+b}. $$

Michael Rozenberg
  • 1
Blind
  • 1,076
  • 1
    You've almost there. To conclude the desired inequality, just think of the possible values of $3a^2-7b^2$ modulo 7. – Phil. Z Dec 25 '17 at 12:12
  • @Phil.Z Thank you for your guidance. How about the solution for b)? – Blind Dec 25 '17 at 12:36
  • For part b, look for solutions among convergents of the continued fraction for$\sqrt{\frac73}$. Not hard to find $(a,b)=(2,1)$ gives minimum, though formally showing the lower bound is more difficult. – Macavity Dec 26 '17 at 11:38
  • @Macavity $(2,1)$ is not even the minimum among those convergents. Try $(55,36)$, or $(6049,3960)$, or $(665335,435564)$, ... – Phil. Z Dec 26 '17 at 11:57
  • @Phil.Z Yes $(73180801, 3*15969360)$ gets better. – Macavity Dec 26 '17 at 12:10

1 Answers1

3

$3a^2=7b^2+1$ is impossible for $b$ by modulo $3$.

$3a^2-2=7b^2$ is impossible for $a$ by modulo $7$.

Thus, $3a^2-7b^2\geq3$ and we are done!

The equality in the last inequality occurs for $a=55$ and $b=36$, which gives $$S\geq91(55\sqrt3-36\sqrt7)=1.43....$$ Since $(55,36)$ is a minimal solution of the Pell $3a^2-7b^2=3$,

it seems that we got a minimal value of $S$. I have no a proof for this statement.

Michael Rozenberg
  • 1
  • 1
    Could you give solution for part b)? – Blind Dec 25 '17 at 21:02
  • Thank you for your guide. Could you make clear why $S\geq 91(55\sqrt{33}-36\sqrt{7})$? – Blind Dec 27 '17 at 06:41
  • By the solution $(55,36)$ we can get all solutions of the Pell. But why for $3a^2-7b^2=3$ we'll get a minimal value of $S$, I don't see. – Michael Rozenberg Dec 27 '17 at 06:55
  • I think that part b) is not easy. Hence I move it to another question in the following link: https://math.stackexchange.com/questions/2581498/extremal-problem-with-positive-integer-numbers?noredirect=1#comment5328365_2581498 – Blind Dec 27 '17 at 07:10
  • 2
    Unfortunately $(55,36)$ does not give the minimum. Check out the pairs listed in the comment. – Phil. Z Dec 27 '17 at 07:29
  • Thank you, Phil! What do you think about the answer for the part $b$? What you can say about this? – Michael Rozenberg Dec 27 '17 at 07:36
  • @MichaelRozenberg You're welcome. I don't know the exact answer either. I've studied the convergents of the continued fraction of $\sqrt{21}/3=[1;\overline{1,1,8,1,1,2}]$ and it shows that the convergents $a/b=[1;\cdots,1,1,8,1,1]$ with more and more periods $(1,1,8,1,1,2)$ within the dots converges to $(3\sqrt{7}+3\sqrt{3})/(2\sqrt{21})\approx 1.432972$, so at least the minimum(if exists) has to be less than or equal to this value. – Phil. Z Dec 27 '17 at 08:47