A curious equality: Where do these numbers come from?

Question

This identity, which was shared on math.stackexchange and seem curious at first sight, caught my attention. Here is the equality:

$$\color {red} {\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\color{blue}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}$$

The truth is that even the almost all students can easily prove this.

But, what I'm wondering about isn't the proof that actually uses a cross multiplication.

Goal : Knowing where these numbers come from

Maybe, here something is hiding from us. Has a complicated fraction been simplified to an easy fraction?

I have reasons not to believe this.

I will try to show that the numbers are not randomly chosen and moreover, where they come from.

Let, $$m,n,r,s \in\mathbb {Z}.$$

We have:

$$\frac{m-n\sqrt{5}} {r-s\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}}$$

$$\implies ( r-s\sqrt{5})(17+15\sqrt{5})=(m-n\sqrt{5})(7+15\sqrt {5})$$

$$\implies (17r-75s)+\sqrt {5}(15r-17s)=(7m-75n)+\sqrt {5}(15m-7n)$$

Then we have,

$$\begin{cases} 15r-17s=15m-7n \\ 17r-75s=7m-75n \end{cases}$$

There are probably good methods for solving such equations in number theory. But since I don't know, I usually solve such equations by methods I know myself.

$$\begin{cases} 255r-289s= 255m-119n \\ 255 r -1125s =105 m - 1125 n \\ 1125r-1275s=1125 m - 525 n \\ 289r-1275s= 119 m - 1275 n \end{cases}$$

$$ \implies \begin{cases} r=\dfrac {503m+375n}{418} \in\mathbb Z \\ s=\dfrac {75m+503n}{418} \in\mathbb Z \end{cases}$$

First restriction

$$\begin {align} r=\dfrac {503m+375n}{418} \in\mathbb Z \Longrightarrow r=\dfrac {418m+85m+418n-43n}{418} \Longrightarrow \dfrac{85m-43n}{418}=k_1 \Longrightarrow m=\dfrac{418k_1+43n}{85}=\dfrac{425k_1-7k_1+43n}{85} \Longrightarrow \dfrac {43n-7k_1}{85}=k_2 \Longrightarrow k_1=\dfrac{42n-84k_2+(n-k_2)}{7} \Longrightarrow \dfrac {n-k_2}{7}=k_3 \Longrightarrow n=7k_3+k_2 \end {align} $$

$$\implies k_1=\dfrac{43(7k_3+k_2)-85k_2}{7}=43k_3-6k_2 $$

$$\implies m= \dfrac { 418(43k_3-6k_2)+43(7k_3+k_2)}{85}=215k_3-29k_2$$

$$\implies n=7k_3+k_2$$

$$\implies r=\dfrac {503m+375n}{418} =\dfrac {503(215k_3-29k_2)+375(7k_3+k_2)}{418}=265k_3-34k_2, ~ \text {where} ~ k_2, k_3 \in\mathbb Z.$$

$$\color {gold}{\boxed {\color{black} {r=265k_3-34k_2}}}$$

Second restriction

$$\begin{align} s=\dfrac {75m+503n}{418} \in\mathbb Z \Longrightarrow s=\dfrac {75(215k_3-29k_2)+503(7k_3+k_2)}{418}=47k_3-4k_2, ~ \text {where} ~ k_2, k_3 \in\mathbb Z. \end{align}$$

$$\color {gold}{\boxed {\color{black} {s=47k_3-4k_2}}}$$

Thus, no new restrictions are required.

Here for simplicity, I am changing $k_2 = v, k_3 = u$, where $u,v\in\mathbb Z$. Then we have,

$$\color {gold}{\boxed {\color{black} {\begin {align} r=265u-34v\\ s=47u-4v \\m=215u-29v \\n=7u+v \end{align}}}}$$

Now we have to solve any of the equations we obtain. Therefore, solving the first equation will be sufficient:

$$r=265u-34v.$$

$$\begin{align} v=\dfrac{265u-r}{34}=8-\dfrac{7u+r}{34}\Longrightarrow \dfrac {7u+r}{34}=k_4 \Longrightarrow u=5k_4-\dfrac{k_4+r}{7} \Longrightarrow k_4+r=7k_5 \Longrightarrow k_4=7k_5-r, ~ \text{where} ~ k_4, k_5 \in\mathbb Z \end{align}$$

$$\implies u=34k_5-5r$$

$$\implies v=\dfrac {265(34k_5-5r)-r}{34}=265k_5-39r$$

For simplicity, I accept $k_5 \longmapsto k$. So, we have

$$\begin{align} u=34k-5r \\ v=265k-39r \end{align}$$

Finally we have,

$$\color {gold}{\boxed {\color{black} {\begin{align} \color{blue}{r=r} \\ \color {red}{s=-79 r+538k} \\ \color{green} {m=56 r- 375 k} \\ \color{blueviolet}{n=- 74 r+503k} \end{align}}}}$$

Let's go back to the original problem.

If $r=2641$, then we have

$$\begin{align} r=2641 \\ s=-208639+538k \\ m=147896- 375 k \\ n=−195434+503k \end{align}$$

Now we can choose $k$ as we wish.

Let, $k=390$, then we have

$$\begin {align} m=1646 \\ n=736 \\ r=2641 \\ s=1181.\end{align}$$

Thus, we obtain the required equality:

$$\color {red}{\boxed {\color{black} {\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}}$$

And now here we can make the numbers as difficult as we want. Let's try one example.

If $m,n,r,s \in\mathbb {Z^{+}}$, then we must solve the following system of equations:

$$\begin{cases} r>0 \\ -79 r+538k>0 \\ 56 r- 375 k>0 \\- 74 r+503k >0 \end{cases} \Longrightarrow k>0, \dfrac {375k}{56}<r<\dfrac {503k}{74}$$

Let, $k=17639735$ and $r=118123483$ ( a prime number), then we have

$$\begin{align} r=118123483 \\ s = 158422273 \\m=14423 \\ n=131648963 \end{align}$$

Finally, our new "magical" identity will be as follows:

$$\color {gold}{\boxed {\color{blue} {\dfrac{14423-131648963\sqrt{5}} {118123483-158422273\sqrt{5}} \color{red}{=\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}}}$$

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Short proof

$$\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{445\sqrt 5-995} {1646-736\sqrt{5}} =\dfrac{10} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{89\sqrt 5-199} {1646-736\sqrt{5}} =\dfrac{2} {17+15\sqrt{5}}$$

$$ \Longleftrightarrow \dfrac{1845-825\sqrt 5} {1646-736\sqrt{5}} =\dfrac{15+15\sqrt{5}} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{123-55\sqrt 5} {1646-736\sqrt{5}} =\dfrac{1+\sqrt 5} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac {1646-736\sqrt{5}} {123-55\sqrt 5}=\dfrac {17+15\sqrt{5}}{1+\sqrt 5}=15+\dfrac{2}{1+\sqrt 5}$$

$$\Longleftrightarrow \dfrac{89\sqrt 5-199} {123-55\sqrt{5}} =\dfrac{2} {1+\sqrt{5}}$$

$$\Longleftrightarrow \dfrac {246-110\sqrt 5}{89\sqrt 5-199} =1+\sqrt 5$$

$$\Longleftrightarrow \dfrac {246-110\sqrt 5-89\times 5+199 \sqrt 5}{89\sqrt 5-199}=\dfrac {89\sqrt 5-199}{89\sqrt 5-199}=1.$$

Observation

• We always have the chance to choose at least one of the numbers $m, n, r, s$ as a prime number. According our formulas, we have

$$\color {red}{\boxed {\color{black} {\begin{align} r=r \\ s=-79 r+538k \\ m=56 r- 375 k \\ n=- 74 r+503k \end{align}}}}$$

So, we can choose the number $\color{blue}{r}$ always as a prime.

Let's look at the following equality, which is completely randomly selected.

$$\color {red} {\dfrac{179277361-8263\sqrt{5}} {9273317-5193\sqrt{5}} =\color{blue}{\dfrac{a+b\sqrt{5}} {c+d\sqrt{5}}}}$$

We have no good reason for solving this equation that "small" values ​​of $ a, b, c, d$ exist.

Therefore, I came to the following conclusion.

The expression $\color {red}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}$ is not taken from the expression $\color{blue}{\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}}}.$ On the contrary, the expression $\color{blue}{\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}}}$ is taken from the expression $\color{red}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}.$

Generalization:

For any $p\in\mathbb Z^{+}$ and small $a,b,c,d$, there exist infinitely many enough large pozitive integers $m,n,r,s$ (we can choose at least one as a prime number) such that,

$$\frac{m-n\sqrt{p}} {r-s\sqrt{p}} =\frac{a+b\sqrt{p}} {c+d\sqrt{p}}$$

Questions

• Could these numbers have come from a completely different place that has nothing to do with the method I use?

• Does mathematics allow for the techniques I have applied above? Are the things I've done correct?

Thank you.

Answer 1

We go from the blue fraction to the red one by multiplying the numerator & denominator by $\tfrac{-199+89\sqrt{5}}{2}$. More generally $\tfrac{17+15\sqrt{5}}{7+15\sqrt{5}}=\tfrac{(17+15\sqrt{5})(a+b\sqrt{5})}{(7+15\sqrt{5})(a+b\sqrt{5})}=\tfrac{17a+75b+(15a+17b)\sqrt{5}}{7a+75b+(15a+7b)\sqrt{5}}$. Now take $a=-\tfrac{199}{2},\,b=\tfrac{89}{2}$. These satisfy $a^2-5b^2=-1$, but we can do the above with arbitrary $a,\,b$.