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I recently posted a related question at Using GCF to Prove Pick's Theorem, but I accidentally intended the converse. Instead of revising a mostly coherent post, I'm making a new one with the backstory copied.

I am teaching high school geometry and I'm building between Euclid's number theory and his geometry. I want to prove Euler's formula in multiple ways, for much the same reason that Bonnie Stewart describes in "Adventures Among the Toroids" exposition. I know Euler's formula is not from Euclid, however I like that I can present it as absolutely true and then show a counterexample for genus $> 0$. It's an attempt to justify the significance of mathematical proof.

One proof of Euler's formula uses Pick's Theorem, so I'm also trying to offer a more gratifying/intuitive proof of Pick's. It inevitably comes down to proving the result for lattice triangles with area $= 1/2$. I believe this may tie nicely to GCD through the determinant formula for area. That's nice for the course because we can easily review GCD. I feel like this final missing step may be doable. Or, maybe there's another route to proving Pick's theorem from a GCD, number theoretic way.

Please help with a proof of the following:

Claim: Suppose $A,B,C,D\ge 0$ with $GCD(A,B)=GCD(C,D)=1$. If $0 \le m,n \le 1$ are so that $mA+nC$ and $mB+nD$ are both integers, then $|AD-BC|=1$ or $m,n\in\{0,1\}$.

It seems that considering only rational values for $m$ and $n$ would suffice and allow better wielding of the theory of integers.

Thanks!

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    $\begingroup$ "It seems..." Does that mean you can prove that $m,n$ are not irrational, or only that it looks like it is true? $\endgroup$ Commented Oct 4 at 21:43
  • $\begingroup$ From a more advanced point of view, this is about integer rmatrices. You might be able to take a linear algebra proof and transfer it back to number theory. Just a guess, though. $\endgroup$ Commented Oct 4 at 21:50
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    $\begingroup$ @RobArthan - that is what the previous question suggests. I believe greatest common divisor (gcd) or highest common factor (hcf) are used more frequently. $\endgroup$ Commented Oct 5 at 0:20
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    $\begingroup$ @AaronGoldsmith I meant that $m'A+n'C$ and $m'B+n'D$ are integers (this is a direct calculation). Unless one of $m'$ and $n'$ is an integer, one can shift them to the interval $(0,1)$. $\endgroup$ Commented Oct 5 at 11:08
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    $\begingroup$ Then, on that note, we can similarly say that $|AD-BC|$ divides $A$ and $C$ as well! So, it should follow easily. $\endgroup$ Commented Oct 5 at 13:29

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After richrow's comments, I have relaxed the claim just a bit and followed his advice. I simplify the claim to what I call a proposition, then the claim will follow as a corollary.

Proposition Suppose $A,B,C,D\ge 0$. If $0 \le m,n \le 1$ are so that $mA+nC$ and $mB+nD$ are both integers, then $m,n\in\{0,1\}$. Then $|AD-BC|$ divides each of $A,B,C,D$ without remainder.

Pf: Let $\langle \cdot\rangle$ denote fractional part.

Since

$$\left\langle\left\langle\frac{C}{|AD-BC|}\right\rangle\cdot A+\left\langle\frac{-A}{|AD-BC|}\right\rangle\cdot C\right\rangle=\left\langle\frac{AC-AC}{|AD-BC|}\right\rangle=0$$

and

$$\left\langle\left\langle\frac{C}{|AD-BC|}\right\rangle\cdot B+\left\langle\frac{-A}{|AD-BC|}\right\rangle\cdot D\right\rangle=\left\langle\frac{BC-AD}{|AD-BC|}\right\rangle=0,$$

then $|AD-BC|$ divides both $A$ and $C$. A similar argument shows $|AD-BC|$ divides both $B$ and $D$. QED

Once this is done, the $GCD$ condition easily leads to $|AD-BC|=1$.

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  • $\begingroup$ Nice proof, though your notation is nonstandard/wrong - you can’t just add fractional parts like that I.e. $<.5>+<.5>=.5+.5=1\neq 0=<1>=<.5+.5>$. It’s more correct to either work $\mod 1$ or just the numerators $\mod AD-BC$ $\endgroup$ Commented Oct 5 at 20:09
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    $\begingroup$ Indeed, I forgot to put another fractional part on the outside LHS. I've revised the equations. Thanks for the comment. $\endgroup$ Commented Oct 6 at 20:34

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