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"Everybody knows" that such coincidences as

$$2\times2\times\overbrace{41\times41} = 6724 \approx 6728 = 2\times2\times2\times\overbrace{29\times29}$$

(And why did I bother with the first two factors of $2$ on each side? Be patient.)

are "explained" by the fact that $\dfrac{41}{29}$ is a convergent in the simple continued fraction expansion of $\sqrt 2$. and maybe

$$2\times2\times2\times\overbrace{29\times29} = 6728 \approx 6727 = 7\times\overbrace{31\times31}$$

has a similar "explanation", as presumably would the fact that

$$2\times2\times\overbrace{41\times41} = 6724 \approx 6727 = 7\times\overbrace{31\times31}.$$

Is there some such "explanation" of the simultaneous proximity of all three of these numbers to each other?

Michael Hardy
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  • interesting, I will see if I can come up with such a thing. Your three quadratic forms are $\pm$ the principal forms of three discriminants, rather equivalent to. – Will Jagy Mar 23 '16 at 01:52
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    I don't know whether this is relevant, but $41^2-31^2=720=6!$ and $31^2-29^2=120=5!$ – Jaume Oliver Lafont Mar 24 '16 at 11:27
  • I think Tito is right, in his answer, but a similar set of correspondences can be obtained from the Pell-like equations $x^2 = \frac{8}{7}y^2+1$ (which yields the solutions $(15, 14), (449, 420), \ldots$) and $x^2 = \frac{7}{4}y^2+1$ (which yields the solutions $(8, 6), (127, 96), \ldots$). – Brian Tung Mar 24 '16 at 17:52
  • @BrianTung: I've edited my answer for clarity. Can you give examples of proximate triples $x_1,x_2,x_3$ similar to $29,,31,,41$ using your pair of Pell-like eqns? – Tito Piezas III Mar 24 '16 at 19:11
  • Would you consider $24\approx25\approx27$ similar? $$24=6\times\overbrace{2\times2}$$ $$25=1\times\overbrace{5\times5}$$ $$27=3\times\overbrace{3\times3}$$ – Jaume Oliver Lafont Mar 24 '16 at 22:02
  • @TitoPiezasIII: I'm not sure that you can come up with triples like that. I haven't given it much thought, though. – Brian Tung Mar 24 '16 at 22:13
  • @JaumeOliverLafont: If, as in my answer, we assume that Hardy required *both* $x_1 \approx x_2 \approx x_3$ and $ax_1^2 \approx bx_2^2 \approx cx_3^2$, then $2,3,5$ would be such a triple. But it's more non-trivial to find those with two, three, or four digits. – Tito Piezas III Mar 24 '16 at 23:08

3 Answers3

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I think it's just the law of small numbers. I'm assuming you want distinct triples $x_1,x_2,x_3$ such that $x_1\approx x_2\approx x_3$ as well as $ax_1^2\approx bx_2^2\approx cx_3^2$ with distinct $a,b,c$. If so, yours was the first of several examples and Mathematica quickly finds,

$$\begin{aligned} 6724&= 4\times41^2\\ 6727&= 7\times31^2\\ 6728&= 8\times29^2\\ \end{aligned}\tag1$$

$$\begin{aligned} 7935 &= 15\times23^2\\ 7938 &= 18\times21^2\\ 7942 &= 22\times19^2\\ \end{aligned}\tag2$$

$$\begin{aligned} 18490&= 10\times43^2\\ 18491&= 11\times41^2\\ 18496&= 16\times34^2\\ \end{aligned}\tag3$$

$$\begin{aligned} 55223&= 23\times49^2\\ 55225&= 25\times47^2\\ 55233&= 17\times57^2\\ \end{aligned}\tag4$$

And that's just the results by using certain assumptions, such as two of the numbers $x_i$ being squared are "twin numbers", i.e $47,49$, which differ by $2$. More generous assumptions and relaxing $x_1-x_2 = 2$ would probably net more results.

Tito Piezas III
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$$2\times2\times\overbrace{41\times41} = \overbrace{82\times82} \approx 7\times\overbrace{31\times31}.$$

This is due the fact that $$\cfrac{82}{31}=2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4+\cfrac{1}{1+\cfrac{1}{1}}}}}}$$

is a convergent of the simple continued fraction of $\sqrt{7}$.

Those two can be combined to get the result.

Michael Hardy
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wythagoras
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  • But again, that's just one of the three unordered pairs of proximate numbers. Is there a similar reason why all three are simultaneously close together? That was the question. $\qquad$ – Michael Hardy Mar 24 '16 at 18:38
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    I changed $$\frac{82}{31}=2+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{1}}}}}}$$ to $$\cfrac{82}{31}=2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4 + \cfrac{1}{1 + \cfrac{1}{1}}}}}}.$$ That is why \cfrac exists. $\qquad$ – Michael Hardy Mar 24 '16 at 18:39
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Because $$\sqrt{2}=\sqrt{\frac{8}{7}}\sqrt{\frac{7}{4}}\approx\frac{31}{29}\times\frac{41}{31}=\frac{41}{29}$$

We may group the three numbers into one single expression $$4+6724\times6728=(6727-1)^2$$ that can be written as $$\left(\frac{6727-1}{2}\right)^2-\left(\frac{6724}{2}\times\frac{6728}{2}\right)=1 $$ or $$\left(\frac{7\times31^2-1}{2}\right)^2-2\left(2\times 29 \times 41\right)^2=1 $$

This is Pell's equation $$X^2-dY^2=1$$

with $X=\frac{7\times31^2-1}{2}$, $Y={2\times29\times41}$ and $d=2$,

so the corresponding approximation to $\sqrt{2}$ is given by $$\sqrt{2}\approx\frac{7\times31^2-1}{4\times29\times41}=\frac{(31\sqrt{7}+1)(31\sqrt{7}-1)}{4\times29\times41}=\frac{3363}{2378}$$ which is the tenth convergent of the continued fraction expansion.

Factoring the numerator shows that the square in 6727 is related to $\sqrt{7}$, as in the answer by wythagoras.

A simpler example is given by

$$\begin{align}98&=2\times7^2\\ 99&=11\times3^2\\ 100&=1\times10^2 \end{align}$$ with $$99^2-98\times100=1$$

and $$\sqrt{2}\approx\frac{11\times3^2}{7\times10}=\frac{99}{70}$$

Approximating $\sqrt{2}$ with the sixth convergent explains the squares of $7$ and $10$, but we also need $$\sqrt{\frac{11}{1}}\approx\frac{10}{3}$$ and / or $$\sqrt{\frac{11}{2}}\approx\frac{7}{3}$$ to justify the square of $3$.

In this example, the approximation for $\sqrt{2}$ can be obtained by direct multiplication of the approximations for $\sqrt{\frac{2}{11}}$ and $\sqrt{\frac{1}{11}}$, but this is not the case in the example from the question.

$$\sqrt{2}=11\sqrt{\frac{2}{11}}\sqrt{\frac{1}{11}}\approx11\times\frac{3}{7}\times\frac{3}{10}=\frac{99}{70}$$

However, dividing the approximations implied by the equations involving number $7$, the convergent $\sqrt{2} \approx \frac{41}{29}$ is obtained.

$$\sqrt{2}=\sqrt{\frac{8}{7}}\sqrt{\frac{7}{4}}\approx\frac{31}{29}\times\frac{41}{31}=\frac{41}{29}$$

Community
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Jaume Oliver Lafont
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