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Let $(x,y) \in \{1,\dots, 100\}^2$. Find the ordered pair such that $x-y\sqrt{2}$ is close to $1/3$ as possible.

First, I noted that this is minimizing $|x-y\sqrt{2}-1/3|$, but I can't use Arithmetic-Geometric mean because $-1/3, -y\sqrt{2}$ are not positive, and I can't see Cauchy being useful either. Any inequality that could be helpful here?

  • Actually, it is minimizing $\left|x-y\sqrt{2}-\frac13\right|.$ Don’t forget the absolute value. – Thomas Andrews Apr 30 '21 at 17:02
  • One approach would be to pick a very good rational approximation for $\sqrt 2,$ like $3363/2378.$ – Thomas Andrews Apr 30 '21 at 17:12
  • I don't think a generic inequality will be useful here, as for unconstrained $x, y$, one can get arbitrarily close. On a computer this is done very quickly. If I had to do this by hand right now, the best way I can do it is to consider when $y \sqrt{2} + \frac{1}{3}$ is approximately an integer. This quickly shows that $y \leq 71$. But then this became a game of making $y \cdot 1.4142$ approximately equal to $0.666$ mod $1$. I could do this pretty rapidly, it turns out, but this isn't particularly systematic. – davidlowryduda Apr 30 '21 at 17:13
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    You could use a standard method for a rational approximation of $3\sqrt2$ and hope the numerator is $2\pmod3$ – Empy2 Apr 30 '21 at 17:14
  • It would be quick with a computer but I've been trying to do it by hand. –  Apr 30 '21 at 17:20
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    Does the answer happen to be $(x,y) = (47,33)$? –  Apr 30 '21 at 17:30
  • Whoops, I was wrong, $\frac{140}{33}$ also is an approximation for $3\sqrt 2.$ Then $x=\frac{141}3=47$ and $y=33$ gives a better solution, and $47-33\sqrt{2}\approx0.331.$ @Empy2 – Thomas Andrews Apr 30 '21 at 17:56

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