# Mere coincidence? (prime factors) [closed]

Whether some things in mathematics are mere coincidences might keep philosophers busy for 100,000 aeons, but maybe when such a coincidence gets exploited then it's not a "mere" coincidence any more.

So time for a somewhat imprecise question: a list of prime factorizations shows us this: $$1445=5\cdot17\cdot17$$ and if you were doing factorizations of consecutive numbers one by one, maybe you'd be just a teensy bit surprised to see $17$ twice in a row; maybe you'd even stir a bit before descending back into deep sleep. But then the very next number (if you're going downward) is: $$1444=2\cdot2\cdot19\cdot19$$ Two squares of somewhat....um....big...primes in a row!

Does this fit into some grander design that the god of mathematics dreamed up when he wasn't busy distributing zeros of the zeta function as if they were eigenvalues of random matrices? LATER EDIT: In view of some of the comments, here's a more prosaic version of the question: Is there some point of view from which this is more meaningful than it is from the point of view from which I recited certain facts above?

After noticing that I noticed this: $$\sqrt{\frac54} = 1+\cfrac{1}{8+\cfrac{1}{2+\cfrac{1}{8+\cfrac{1}{2+\cfrac{1}{8+\cfrac{1}{2+\ddots}}}}}}$$ whereas $$\frac{19}{17} = 1+\cfrac{1}{8+\cfrac{1}{2}}$$

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## closed as unclear what you're asking by Gerry Myerson, BlueRaja - Danny Pflughoeft, Brandon Carter, T. Bongers, MicahAug 27 '13 at 5:34

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

I am... – Pedro Tamaroff Aug 17 '13 at 22:29
Given your reputation, I feel somewhat silly for asking this but... you do know that questions asked here are supposed to be specific, right? – hasnohat Aug 17 '13 at 22:34
$1\cdot 3^2 = 2\cdot 2^2+1$; $13\cdot 5^2=36\cdot 3^2+1$; $24\cdot 7^2 = 47\cdot 5^2+1$. – celtschk Aug 17 '13 at 22:42
What is the question ? What is so weird about those continued fractions ? – mick Aug 17 '13 at 23:11
This gives an example of numbers with $Ax^2-By^2=1$ for which the continued fractions of $\sqrt{A/B}$ and $y/x$ are related (the first is a periodic continuation of the second). Given the deep structure behind that Diophantine equation and its relation to continued fractions, I am not quick to dismiss this as trivia sans deeper explanation, though I haven't thought about it. – anon Aug 17 '13 at 23:22

As anon beat me to it, it seems Michael Hardy observed that,

$$Ax^2-By^2=1\tag{1}$$

$$\frac{y}{x} = 1+\cfrac{1}{a+\cfrac{1}{b}}\tag{2}$$

$$\sqrt{\frac{A}{B}} = 1+\cfrac{1}{a+\cfrac{1}{b+\cfrac{1}{a+\cfrac{1}{b+\ddots}}}}\tag{3}$$

However, there are an infinite number, given by the identity,

$$(n+1)(4n+1)^2-n(4n+3)^2=1\tag{4}$$

$$\frac{4n+3}{4n+1} = 1+\cfrac{1}{2n+\cfrac{1}{2}}\tag{5}$$

$$\sqrt{\frac{n+1}{n}} = 1+\cfrac{1}{2n+\cfrac{1}{2+\cfrac{1}{2n+\cfrac{1}{2+\ddots}}}}\tag{6}$$

where Hardy's was just the case $n=4$.

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I am reminded of Greg Muller's answer to this question: math.stackexchange.com/questions/2694/… where he says "Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization.", but this question is evidence that it's not random and has a lot of mappable structure. – deoxygerbe Aug 18 '13 at 0:09
Maybe some explicitness about just which person named Hardy is reasonable in this particular instance. – Michael Hardy Aug 18 '13 at 0:23
Has been rectified. Every time I see your last name here and wikipedia, I always get reminded of the other Hardy (Ramanujan's). – Tito Piezas III Aug 18 '13 at 0:30

This is not a full answer, just too large for a comment. It looks like you can rest assured this isn't just coincidence! I did a large search with Mathematica, sifting through every number whose factorization had the largest prime occurring precisely twice, and then of those taking only pairs of the form $n$, $n+1$, and then finding the continued fraction expansions of the ratio and roots after splitting the number into parts as described in the original post. I got the following dataset:

{
{{{0, 2, {1, 1, 2, 20, 2, 1, 1, 4}}, {0, 2, 1, 1, 2}, 675}},
{{{1, {8, 2}}, {1, 8, 2}, 1444}},
{{{3, {3, 2, 32, 2, 3, 6}}, {3, 3, 2}, 2645}},
{{{1, {14, 2}}, {1, 14, 2}, 6727}},
{{{0, 1, {1, 1, 3, 254, 3, 1, 1, 2}}, {0, 1, 1, 1, 3}, 9800}},
{{{0, 2, {1, 5, 2, 56, 2, 5, 1, 4}}, {0, 2, 1, 5, 2}, 13689}},
{{{1, {20, 2}}, {1, 20, 2}, 18490}},
{{{1, {4, 2}}, {1, 4, 2, 4, 2}, 23762}},
{{{9, {2, 2, 206, 2, 2, 18}}, {9, 2, 2}, 24299}},
{{{0, 2, {5, 1, 2, 82, 2, 1, 5, 4}}, {0, 2, 5, 1, 2}, 26010}},
{{{1, {1, 2, 2, 4, 44, 4, 2, 2, 1, 2}}, {1, 1, 2, 2, 4}, 36517}},
{{{3, {9, 2, 86, 2, 9, 6}}, {3, 9, 2}, 48734}},
{{{0, 8, {1, 2, 2, 278, 2, 2, 1, 16}}, {0, 8, 1, 2, 2}, 59535}},
{{{2, {3, 1, 3, 1, 184, 1, 3, 1, 3, 4}}, {2, 3, 1, 4}, 75809}},
{{{3, {11, 2, 104, 2, 11, 6}}, {3, 11, 2}, 85697}}
}


"{1,{8,2}}" represents a continued fraction starting with 1, and with 8,2 repeating forever:

$\{1,\{8,2\}\} \to 1+\cfrac{1}{8+\cfrac{1}{2+\cfrac{1}{8+\cfrac{1}{2+\cfrac{1}{8+\cfrac{1}{2+\ddots}}}}}}$

and, {1,8,2} has no part which repeats forever.

$\{1,8,2\} \to 1+\cfrac{1}{8+\cfrac{1}{2}}$

The first element of each row represents the continued fraction of the root, $\sqrt{A/B}$ as in anon's comment, and the second element of each row represents the continued fraction of the ratio, $y/x$ in anon's comment. The last element is the lesser of the two numbers looked at. (for example, 1444 is listed, meaning that that row refers to the pair 1444 and 1445)

The continued fractions seem to always begin with the same sequences!(Note that in the second to last row, {2,3,1,4} can be written as {2,3,1,3,1})

I don't know the first thing about number theory so I can't provide much else!

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If you start off with any number $x$, it is possible to generate a series of the type where $a^2 - 4$ is the product of a series constant, and a very large square.

For example, suppose we start with $n=3$. The series starts off with $2, n$, and $n_{a+1}= n.n_a - n_{a-1}$

You get: $2, 3, 7, 18, 47, 123$ on one strand, and $1, 1, 4, 11, 29, 76$ on the other strand. What we're looking at here is on the first strand, $a^2-4 = (n+2)b^2$, and on the second strand, $a^2+4 = (n+2)b^2$.

The example here is $76^2 + 4 = 5 \cdot 34^2$, or dividing by 2, one gets $38^2+1 = 17^2 \cdot 5$.

Of course, one can use all sorts of numbers. Using for example, with $n=38$, and the two initial series, one through $1, 19$, and the other through $3, 3$, (where $6^2+2=38$), one finds the second series passes through $4443$. So we get eg $4443^2 = 3^2 \cdot 1489^2$, followed immediately by $2 \cdot 5^3 \cdot 281^2$. Numbers of the first series give the tenth-square as one smaller than the square.

It is also little cooincidence that these are distributed at the same density as the powers of numbers.

Grander still, but not recurring, is that $7^3 \cdot 17$ is followed by $18^3$.

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So, as you've shown, $\frac{19}{17}$ is a continued fraction convergent for $\sqrt{\frac54}$. Being that the next continuant is $8$, we have that $$\left|\,\frac{19}{17}-\sqrt{\frac54}\,\right|\le\frac18\cdot\frac1{17^2}$$ which implies that \begin{align} \left|\,\left(\frac{19}{17}\right)^2-\frac54\,\right| &\le\frac18\cdot\frac1{17^2}\left(\frac{19}{17}+\sqrt{\frac54}\right)\\ &=\frac14\cdot\frac1{17^2}\cdot\frac12\left(\frac{19}{17}+\sqrt{\frac54}\right)\\ &\lt\frac14\cdot\frac1{17^2}\cdot2 \end{align} Therefore, $$\left|\,4\cdot19^2-5\cdot17^2\,\right|\lt2$$ which, since the left-hand side is not $0$, implies that $$\left|\,4\cdot19^2-5\cdot17^2\,\right|=1$$ which means that $4\cdot19^2$ and $5\cdot17^2$ are consecutive integers.

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