Proving the existence of integers $a, b, c$ such that $\left\lvert a + b\sqrt{2} + c\sqrt{3}\right\rvert < 10^{-5}$

Question

Prove that there exist integers $a, b, c$ such that $|a|, |b|, |c| \leq 1000$ where not all of them are zero and $\left\lvert a + b\sqrt{2} + c\sqrt{3}\right\rvert < 10^{-5}$

I am stuck on this problem.

My Attempt :

I've tried the Pigeonhole Principle. I tried making the decimal values of $\left\lvert b\sqrt{2} + c\sqrt{3}\right\rvert$ as pigeons in $\left[0, 10^{-5}\right), \left[10^{-5}, 2 \cdot 10^{-5}\right), ... , \left[\left(10^5 - 1\right) \cdot 10 ^ {-5}, 1\right)$. Since there are $10^3 \cdot 10^3 = 10^6$ pigeons in $10^5$ holes, there must be a hole such that it has at least two pigeons.

If we add/subtract them, then $x = \left\lvert(b_1 \pm b_2)\sqrt{2} + (c_1 \pm c_2)\sqrt{3}\right\rvert$ will have decimal point less than $10^{-5}$. So, we need to make the integer value of $x$ to be zero which we can subtract as $a$.
One Problem is that $x$ may have an integer value on the orders of $1000\cdot \sqrt{2} + 1000 \cdot \sqrt{3} = 3146$ which highly exceeds the maximum for $a$.
Second Problem is that the part about adding/subtracting seems complicated in terms of ensuring that $\left\lvert b_1 \pm b_2\right\rvert \leq 1000$ and $\left\lvert c_1 \pm c_2\right\rvert \leq 1000$.

So, I am stuck!

Any help is appreciated.

Thanks!

Answer 1

DISCUSSION :

There are a couple of Issues with your nice approach. When those Issues are rectified , the Solution will work out.

(1) The range you have taken is $10^3$ , though the Correct range is $2001$ , because we can have Negative Integers too.
Over-all , we have $2001\times2001\times2001 \approx 8\times10^9$ Possible triplets $(a,b,c)$
(2) Coming to Pairs $(b,c)$ , we have $4\times10^6$ Pigeons , which is too much.
(3) You are considering "adding/subtracting" the Pairs , though that is wrong. When we add , the "Decimal Part" might increase beyond the $10^{-5}$ range. We must always subtract , whereby the "Decimal Part" will reduce within the $10^{-5}$ range.
(4) When we subtract , the Pair Components might go out of range eg $[-600\sqrt{2}+700\sqrt{3}]-[+600\sqrt{2}-700\sqrt{3}]=[-1200\sqrt{2}+1400\sqrt{3}]$. We have to avoid that.
(5) Solving those Issues will necessitate that we should not use the whole range available. We should use a restricted range.

Solution , incorporating those Points :

Let $b$ be Positive ($+1,...,+317$) & $c$ be Negative ($-1,...,-317$) & Consider the Number $X=b\sqrt{2}+c\sqrt{3}$

The Integer Part we need not worry about at the moment. We can see that the fractional Part can have $317\times317=100489$ Possibilities , hence Pigeon Hole Principle says that there are at least 2 Possibilities within $10^{-5}$
Let that be $X_1=b_1\sqrt{2}+c_1\sqrt{3}$ & $X_2=b_2\sqrt{2}+c_2\sqrt{3}$

Make the number $X_3=X_1-X_2=b_3\sqrt{2}+c_3\sqrt{3}$ , where $b_3$ is between $1-317=-316$ & $317-1=+316$ , like wise for $c_3$.

Now , the Integer Part of $X_3$ might not be $0$.
No matter , we can use $a_3$ to bring it to $0$ !
The Integer Part could have been maximum $\pm 317\sqrt{2} \pm 317\sqrt{3} \approx \pm 997.36581 \approx \pm 998$ , hence $a_3$ can easily neutralize that.

DONE !

Answer 2

The simplest one is to write a computer program that iterates through all b from -1000 to +1000, then through all c from -1000 to 1000 where $|b \sqrt 2 + c \sqrt 3| < 1000.5$, round the result to the nearest integer and check if it is close enough. (Has nothing to do with the pigeon hole principle).

Since -a, -b, -c is also a solution, you can safe half the time by taking b >= 0 only.

Or as you did you count how many values there are with 0 <= b <= 1000, then c with $c \ge \max(-1000, (-1000 -b \sqrt 2) / \sqrt 3)$ and $c \le \min(1000, (1000 -b \sqrt 2) / \sqrt 3)$.