Sunday, September 28, 2014

Quantitative Interview Brain Teaser: Computer Assistance

In a previous post I discussed a recent brain teaser I had come across:
Find a 10-digit number, where each digit represents the number of that ordinal number in the whole number. So, the first digit represents the number of 0's in the whole 10 digits. The second digit represents the number of 1's in the whole 10 digits. And so on. The first digit is not a 0.
As I promised at the end of the "brain only" post, we'll do better than simply finding an answer, we'll find all of them (with the aid of a computer).

Making the Problem Smaller

Without any thinking, there are 9 billion (9109) total choices. This is not only intractable for the human brain, but becomes difficult for a computer too. A 3 GHz processor in the most optimal case can only perform 3 billion operations per second but there is a lot of counting, lists to create, and otherwise to find a number which fits the criteria.

To start, we write our number as
n=d0,d1d2d3,d4d5d6,d7d8d9.
Since there are 10 digits in total and each of the digits represent a subcount of occurrences, the total number of occurrences can be represented as the sum:
d0+d1+d2+d3+d4+d5+d6+d7+d8+d9=10
Additionally, (for example) since each 3 contributes a value of 3 to the digit sum, we must also have
d1+2d2+3d3+4d4+5d5+6d6+7d7+8d8+9d9=10
This second equation (which requires the first to make sense) limits our choices of digits a great deal:
0d5,d6,d7,d8,d91
The last four are obvious since (for example) 26>10. We can also say that d5<2. It is clearly no bigger than 2 but in the case that d5=2 we'd also have d2>0 meaning 2d2+5d5=2d2+10>10. Thus to brute force the problem we can choose digits 5 through 9 (5 digits in total) from the 32 (25) different possible ways to make them 0 or 1.

Listing All Choices

Now for the fun part: programming (in Python). We now have 90,000 (9104) choices for our first 5 digits and 32 choices for our last 5 digits.

We can use Python's range(10**4, 10**5) to represent the 5-digit numbers between 10,000 and 99,999 (inclusive). For the 32 choices for the last 5 digits, we use itertools.product. To see it in action on a smaller set of data: When we ask for 5 repeated copies of the tuple (0, 1) we get 32 possible 5-tuples as expected:

Checking Candidates

Before we can iterate through all of our combinations, we need a way to check if a given number fits the criterion. To do that we implement This takes a list of digits, so as we loop over all choices, we'll turn them into lists.

Exhaustive Search

Now we can use our accumulated tools, loop through all choices and print any matches As luck would have it, the output is simply In other words, the only number which fits the criteria is the one we found with our brains alone:
6,210,001,000
This serves to make the interview question that much more difficult, since there is a unique solution.

Monday, September 22, 2014

Quantitative Brain Teaser: Brain Only

I've recently been working some atrophied mental muscles and came across a brain teaser that was pretty nifty:
Find a 10-digit number, where each digit represents the number of that ordinal number in the whole number. So, the first digit represents the number of 0's in the whole 10 digits. The second digit represents the number of 1's in the whole 10 digits. And so on. The first digit is not a 0.

Example

If we shortened from 10 digits to 4 digit, the number
2,020
works since we have d0=2 and two 0's (in the second and fourth slots), d1=0 since the number has no 1's, d2=2 since the number has two 2's (in the first and third slots) and d3=0 since the number has no 3's.

Shorthand Notation

In order to refer to each digit, for search we name them all:
n=d0,d1d2d3,d4d5d6,d7d8d9.
We can see this in the above example when we refer to the digits in the four digit number
n=d0,d1d2d3.

A Practical Approach, Breaking Into Subproblems

Our search space is massive, and with only our wits, we need to quickly find a way to focus on a small space of possibilities. Since the first digit allows us to place a number of 0's we try to set it equal to values starting from the largest. By doing this we only have a little wiggle room to find the places which don't hold a zero.

First Case: d0=9

In this case our only choice is
9,000,000,000
since we must have nine 0's. However since we have one 9, d9=0 should not occur.

Thus we see none of our choices are possible when d0=9.

Second Case: d0=8

Here we must have eight 0's and d8>0 so our possible solutions must look like
8,000,000,00
But this leaves us with d8=1 as our only choice since we can't place any more 8's. But now the presence of a 1 in
8,000,000,010
can't coexist with d1=0 so we again see none of our choices are possible when d0=8.

Third Case: d0=7

Here we have seven 0's and know that d7=1. It must be at least 1 since the first digit is a 7. It can't be 2 because the presence of another 7 would mean another digit (other than 0) would occur 7 times, which is impossible since there are only 10 total digits.

Since we know d7=1 our possible solutions must look like
7,00,000,100
But again here we reach an impossible point. If we set d1=1 then that digit would contradict itself since it is the second occurrence of 1. If d1=2 it would contradict d2=0 and so on for higher values. In addition, we have used all our digits, so can't increase the value of d1 by placing more 1's in our number.

Thus we see none of our choices are possible when d0=7.

Fourth Case: d0=6

Here we have six 0's and must have d6=1 since (as above), two different digits can't occur six times among 10 digits.

Also as before we can't have d1=1 but now have some extra freedom (an extra digit which doesn't have to be 0) so consider the case d1=2. This corresponds to an occurrence of the digit 2, hence we set d2=1.

Now we have 4 non-zero digits along with six 0's to place:
6,210,001,000
Thus we have found a number which satisfies the criteria! The zero digits in the 3, 4, 5, 7, 8, and 9 places correspond to the absence of those digits. The non-zero digits in the 0, 1, 2, and 6 places also are the correct counts of each of those digits.

As a math nerd, I was still curious to know how to find every possible number that satisfies the criteria, but that task is too tedious to handle with the brain alone (or at least to be worth reading about when solved by hand). In my follow up to this, I'll show how a combination of smarts and programming can perform an exhaustive search in under 10 seconds.

Friday, August 22, 2014

Math for Humans, A Second Attempt

The morning after posting my latest blog post, I woke up still thinking about how to explain the concept.

More importantly, I realized that my goal of writing math for humans failed miserably.

So here is a second go at it.

First we're told we're in a world where 85% of cabs are Green and the rest are Blue. Humans love tables (and they are easy to understand). So we start off with a representative sample of 100 cabs:

Category Green Blue Total
Cabs 85 15 100

After this, we're told that a bystander correctly identifies a cab 80% of the time, or 4 out of every 5. Applying this to the 85 Green cabs (85 is 17 times 5), this bystander will mis-identify 17 as Blue (1 out of 5) and the other 68 will correctly be identified as Green:

Category Green Blue Total
Cabs 85 15 100
ID'd Green 68
ID'd Blue 17

Similarly, of the 15 Blue cabs (15 is 3 times 5), this bystander will mis-identify 3 as Green (1 out of 5) and the other 12 will correctly be identified as Blue:

Category Green Blue Total
Cabs 85 15 100
ID'd Green 68 3
ID'd Blue 17 12

Now Kahneman wants us to use the data at hand to determine what the probability is that a cab is actually Blue given the bystander identified the cab as Blue. To determine this probability, we simply need to consider the final row of the table:

Category Green Blue Total
ID'd Blue 17 12 29

This rows tells us that only 29 cabs will be identified as Blue, and among those, 12 will actually be Blue. Hence the probability will be \[\boxed{\frac{12}{29} \approx 0.413793103}.\] What this really shows is that even though the bystander has a large chance (80%) of getting the color right, the number of Green cabs is so much larger it overwhelms the correctly identified Blue cabs with incorrectly identified Green ones.

What I Overlooked

  • Dense text is always bad
  • Using colors and breaking up text makes reading easier (more modular)
  • Introducing mathematical notation is almost always overkill
  • Tables and samples are a good way to discuss probabilities

Tuesday, July 29, 2014

Conditional Probabilities in "Thinking Fast and Slow"

I'm currently reading Thinking Fast and Slow by Daniel Kahneman. (Thanks to Elianna for letting me borrow it.) I'm not finished yet, but 60% of the way through I definitely recommend it.

While reading the "Causes Trump Statistics" chapter (number 16), there is a description of a study about cabs and hit-and-run accidents. It describes a scenario where participants are told that 85% of cabs are Green, 15% are Blue and a given observer has an 80% chance of correctly identifying the color of a given cab. Given this data, the chapter presents a scenario where a bystander identifies a cab in an accident as Blue and Kahneman goes on to explain how we fail to take the data into consideration. I really enjoyed this chapter, but won't wreck the book for you.

Instead, I want to do some math (big surprise, I know). However, I want to make it accessible to non-mathematicians (atypical for my posts).

Given the data, Kahneman tells us that the true probability that the cab was Blue is 41% though we likely bias our thinking towards the 80% probability of the identification being correct. I was on the bus and it kept bothering me, to the point that I couldn't continue reading. Eventually I figured it out (when I got to the train) and I wanted to explain how this is computed using Bayes' Law. As a primer, I wrote a post using layman's terms explaining how we use Bayes' Law. (There is some notation introduced but I hope it isn't too confusing.)

Putting Bayes' Law to Use

We need to understand what 41% even corresponds to before we can compute it. What's actually happened is that we know the event \(IDB\) has occurred -- the cab has been identified (\(ID\)) as Blue (\(B\)). What we want is the probability that the cab is Blue given we know it has been identified -- we want:
\[\text{Pr}(B \mid IDB).\] Using Bayes' Law, we can write
\[\text{Pr}(B \mid IDB) = \frac{\text{Pr}(B \text{ and } IDB \text{ both occur})}{\text{Pr}(IDB)} \quad \text{and} \quad \text{Pr}(IDB \mid B) = \frac{\text{Pr}(B \text{ and } IDB \text{ both occur})}{\text{Pr}(B)}.\] We're told that a cab can be correctly identified 80% of the time hence
\[\text{Pr}(IDB \mid B) = 0.8\] (i.e. the probability of correct ID as Blue given it is actually Blue). We're also told that 15% of the cabs are Blue hence
\[\text{Pr}(B) = 0.15.\] We can combine these with the second application of Bayes' Law above to show that
\[\text{Pr}(B \text{ and } IDB \text{ both occur}) = \text{Pr}(IDB \mid B) \cdot \text{Pr}(B) = 0.8 \cdot 0.15 = 0.12.\] The only piece of data missing now to finish our computation is \(\text{Pr}(IDB)\).

Using the extended form of Bayes' Law, since we know that the events \(B\) and \(G\) (the cab is Blue or Green) are exclusive and cover all possibilities for the cab, we can say that
\[\text{Pr}(IDB) = \text{Pr}(IDB \mid B) \cdot \text{Pr}(B) + \text{Pr}(IDB \mid G) \cdot \text{Pr}(G).\] Since there is only an 80% chance of correct identification, we know that \(\text{Pr}(IDB \mid G) = 0.2\) (the probability of misidentifying a Green cab as Blue). We also know that 85% of the cabs are Green hence we can plug these in (along with numbers already computed) to get
\[\text{Pr}(IDB) = 0.8 \cdot 0.15 + 0.2 \cdot 0.85 = 0.12 + 0.17 = 0.29.\] Putting it all together we get our answer
\[\text{Pr}(B \mid IDB) = \frac{\text{Pr}(B \text{ and } IDB \text{ both occur})}{\text{Pr}(IDB)} = \frac{0.12}{0.29} = \boxed{\frac{12}{29} \approx 0.413793103}.\] Fantastic! Now we can get back to reading...

Bayes' Law Primer

I'm currently writing a blog post that uses Bayes' Law but don't want to muddy the post with a review in layman's terms. So I have something to link, here is a short description and a chance to flex my teaching muscles before the school year starts.

Bayes' Law

For those who aren't sure, Bayes' Law tells us that the probability event \(X\) occurs given we know that event \(Y\) has occurred can easily be computed. It is written as \(\text{Pr}(X \mid Y)\) and the vertical bar is meant like the word "given", in other words, the event \(X\) is distinct from the event \(X \mid Y\) (\(X\) given \(Y\)). Bayes' law, states that
\[\text{Pr}(X \mid Y) = \frac{\text{Pr}(X \text{ and } Y \text{ both occur})}{\text{Pr}(Y)}.\]
This effectively is a re-scaling of the events by the total probability of the given event: \(\text{Pr}(Y)\).

For example, if \(X\) is the event that a \(3\) is rolled on a fair die and \(Y\) is the event that the roll is odd. We know of course that \(\text{Pr}(Y) = \frac{1}{2}\) since half of the rolls are odd. The event \(X \text{ and } Y \text{ both occur}\) in this case is the same as \(X\) since the roll can only be \(3\) is the roll is odd. Thus
\[\text{Pr}(X \text{ and } Y \text{ both occur}) = \frac{1}{6}\]
and we can compute the conditional probability
\[\text{Pr}(X \mid Y) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}.\]
As we expect, one out of every three odd rolls is a \(3\).

Bayes' Law Extended Form

Instead of considering a single event \(Y\), we can consider a range of \(n\) possible events \(Y_1, Y_2, \ldots, Y_n\) that may occur. We require that one of these \(Y\)-events must occur and that they cover all possible events that could occur. For example \(Y_1\) is the event that H2O is vapor, \(Y_2\) is the event that H2O is water and\(Y_3\) is the event that H2O is ice.

In such a case we know that since the \(Y\)-events are distinct
\[\text{Pr}(X) = \text{Pr}(X \text{ and } Y_1 \text{ both occur}) + \text{Pr}(X \text{ and } Y_2 \text{ both occur}) + \text{Pr}(X \text{ and } Y_3 \text{ both occur}).\]
Using Bayes' law, we can reinterpret as
\[\text{Pr}(X \text{ and } Y_j \text{ both occur}) =  \text{Pr}(X \mid Y_j) \cdot \text{Pr}(Y_j)\]
and the above becomes
\[\text{Pr}(X) = \text{Pr}(X \mid Y_1) \cdot \text{Pr}(Y_1) + \text{Pr}(X \mid Y_2) \cdot \text{Pr}(Y_2) + \text{Pr}(X \mid Y_3) \cdot \text{Pr}(Y_3).\]
The same is true if we replace \(3\) with an arbitrary number of events \(n\).

Monday, November 25, 2013

Trigonometry and Nested Radicals

Early last month, I was chatting with one of my officemates about a curious problem I had studied in high school. I hadn't written any of the results down, so much of the discussion involved me rediscovering the results and proving them with much more powerful tools than I once possessed.

Before writing about the problem I had played around with, I want to give a brief motivation. For as long as humans have been doing mathematics, finding values of \(\pi\) has been deemed worthwhile (or every generation has just found it worthwhile to waste time computing digits).

One such way the Greeks (particularly Archmides) computed \(\pi\) was by approximating a circle by a regular polygon and letting the number of sides grow large enough so that the error between the area of the unit circle (\(\pi \cdot 1^2\)) and the area of the polygon would be smaller than some fixed threshold. Usually these thresholds were picked to ensure that the first \(k\) digits were fully accurate (for some appropriate value of \(k\)).

In many introductory Calculus courses, this problem is introduced exactly when the limit is introduced and students are forced to think about the area problem in the regular polygon:
Given \(N\) sides, the area is \(N \cdot T_N\) where \(T_N\) is the area of each individual triangle given by one side of the polygon and the circumcenter.

Call one such triangle \(\Delta ABC\) and let \(BC\) be the side that is also a side of the polygon while the other sides have \(\left|AB\right| = \left|AC\right| = 1\) since the polygon is inscribed in a unit circle. The angle \(\angle BAC = \frac{2\pi}{N}\) since each of the triangles has the same internal angle and there are \(N\) of them. If we can find the perpendicular height \(h\) from \(AB\) to \(C\), the area will be \(\frac{1}{2} h \left|AB\right| = \frac{h}{2}\). But we also know that
\[\sin\left(\angle BAC\right) = \frac{h}{\left|AC\right|} \Rightarrow h = \sin\left(\frac{2\pi}{N}\right).\] Combining all of these, we can approximate \(\pi\) with the area:
\[\pi \approx \frac{N}{2} \sin\left(\frac{2\pi}{N}\right) = \pi \frac{\sin\left(\frac{2\pi}{N}\right)}{\frac{2 \pi}{N}}. \] As I've shown my Math 1A students, we see that
\[\lim_{N \to \infty} \pi \frac{\sin\left(\frac{2\pi}{N}\right)}{\frac{2 \pi}{N}} = \pi \lim_{x \to 0} \frac{\sin(x)}{x} = \pi\] so these are indeed good approximations.

Theory is Nice, But I Thought We Were Computing Something

Unfortunately for us (and Archimedes), computing \(\sin\left(\frac{2\pi}{N}\right)\) is not quite as simple as dividing by \(N\), so often special values of \(N\) were chosen. In fact, starting from \(N\) and then using \(2N\), the areas could be computed via a special way of averaging the previous areas. Lucky for us, such a method is equivalent to the trusty half angle identities (courtesy of Abraham De Moivre). To keep track of these polygons with a power of two as the number of sides, we call \(A_n = \frac{2^n}{2} \sin\left(\frac{2\pi}{2^n}\right)\).

Starting out with the simplest polygon, the square with \(N = 2^2\) sides, we have
\[A_2 = 2 \sin\left(\frac{\pi}{2}\right) = 2.\] Jumping to the octagon (no not that "The Octagon"), we have
\[A_3 = 4 \sin\left(\frac{\pi}{4}\right) = 4 \frac{\sqrt{2}}{2} = 2 \sqrt{2}.\] So far, the toughest thing we've had to deal with is a \(45^{\circ}\) angle and haven't yet had to lean on Abraham (himnot him) for help. The hexadecagon wants to change that:
\[A_4 = 8 \sin\left(\frac{\pi}{8}\right) = 8 \sqrt{\frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}} = 8 \sqrt{\frac{2 - \sqrt{2}}{4}} = 4 \sqrt{2 - \sqrt{2}}.\]
To really drill home the point (and motivate my next post) we'll compute this for the \(32\)-gon (past the point where polygons have worthwhile names):
\[A_5 = 16 \sin\left(\frac{\pi}{16}\right) = 16 \sqrt{\frac{1 - \cos\left(\frac{\pi}{8}\right)}{2}}.\] Before, we could rely on the fact that we know that a \(45-45-90\) triangle looked like, but now, we come across \(\cos\left(\frac{\pi}{8}\right)\), a value which we haven't seen before. Luckily, Abraham has help here as well:
\[\cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{1}{2} \sqrt{2 + \sqrt{2}}\] which lets us compute
\[A_5 = 16 \sqrt{\frac{1 - \frac{1}{2} \sqrt{2 + \sqrt{2}}}{2}} = 8 \sqrt{2 - \sqrt{2 + \sqrt{2}}}.\]

So why have I put you through all this? If we wave our hands like a magician, we can see this pattern continues and for the general \(n\)
\[A_n = 2^{n - 2} \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{\cdots + \sqrt{2}}}}}\]
where there are \(n - 3\) nested radicals with the \(\oplus\) sign and only one minus sign at the beginning.

This motivates us to study two questions, what is the limiting behavior of such a nested radical:
\[\sqrt{2 + s_1 \sqrt{2 + s_2 \sqrt{ \cdots }}}\] as the signs \(s_1, s_2, \ldots\) takes values in \(\left\{-1, 1\right\}\). Recasting in terms of the discussion above, we want to know how close we are to \(\pi\) as we increase the number of sides.

When I was in high school, I just loved to nerd out on any and all math problems, so I studied this just for fun. Having heard about the unfathomable brain of Ramanujan and the fun work he had done with infinitely nested radicals, I wanted to examine which sequences of signs \((s_1, s_2, \ldots)\) produced an infinite radical that converged and what the convergence behavior was.

I'm fairly certain my original questions came from an Illinois Council of Teachers of Mathematics (ICTM) contest problem along the lines of
\[\text{Find the value of the infinite nested radical } \sqrt{2 + \sqrt{2 + \cdots}}\] or maybe the slightly more difficult \[\text{Find the value of the infinite nested radical } \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \cdots}}}}.\] Armed with my TI-83, I set out to do some hardcore programming and figure it out. It took me around a month of off-and-on tinkering. This second time around as a mathematical grown-up, it took me the first half of a plane ride from SFO to Dallas.

In the next few weeks/months, I hope to write a few blog posts, including math, proofs and some real code on what answers I came up with and what other questions I have.

Tuesday, September 10, 2013

Calculating a Greatest Common Divisor with Dirichlet's Help

Having just left Google and started my PhD in Applied Mathematics at Berkeley, I thought it might be appropriate to write some (more) math-related blog posts. Many of these posts, I jotted down on napkins and various other places on the web and just haven't had time to post until now.

For today, I'm posting a result which was somewhat fun to figure out with/for one of my buddies from Michigan Math. I'd also like to point out that he is absolutely kicking ass at Brown.

While trying to determine if
\[J(B_n)_{\text{Tor}}\left(\mathbb{Q}\right) \stackrel{?}{=} \mathbb{Z}/2\mathbb{Z} \] where \(J(B_n)\) is the Jacobian of the curve \(B_n\) given by \(y^2 = (x + 2) \cdot f^n(x)\) where \(f^n\) denotes \(\underbrace{f \circ \cdots \circ f}_{n \text{ times}}\) and \(f(x) = x^2 - 2\).

Now, his and my interests diverged some time ago, so I can't appreciate what steps took him from this to the problem I got to help with. However, he was able to show (trivially maybe?) that this was equivalent to showing that
\[\gcd\left(5^{2^n} + 1, 13^{2^n} + 1, \ldots, p^{2^n} + 1, \ldots \right) = 2 \qquad (1)\] where the \(n\) in the exponents is the same as that in \(B_n\) and where the values we are using in our greatest common divisor (e.g. \(5, 13\) and \(p\) above) are all of the primes \(p \equiv 5 \bmod{8}\).

My buddy, being sadistic and for some reason angry with me, passed me along the stronger statement:
\[\gcd\left(5^{2^n} + 1, 13^{2^n} + 1\right) = 2 \qquad (2)\] which I of course struggled with and tried to beat down with tricks like \(5^2 + 12^2 = 13^2\). After a few days of this struggle, he confessed that he was trying to ruin my life and told me about the weaker version \((1)\).

When he sent me the email informing me of this, I read it at 8am, drove down to Santa Clara for PyCon and by the time I arrived at 8:45am I had figured the weaker case \((1)\) out. This felt much better than the days of struggle and made me want to write about my victory (which I'm doing now). Though, before we actually demonstrate the weaker fact \((1)\)  I will admit that I am not in fact tall. Instead I stood on the shoulders of Dirichlet and called myself tall. Everything else is bookkeeping.

Let's Start the Math

First, if \(n = 0\), we see trivially that
\[\gcd\left(5^{2^0} + 1, 13^{2^0} + 1\right) = \gcd\left(6, 14\right) = 2\] and all the remaining terms are divisible by \(2\) hence the \(\gcd\) over all the primes must be \(2\).

Now, if \(n > 0\), we will show that \(2\) divides our \(\gcd\), but \(4\) does not and that no odd prime can divide this \(\gcd\). First, for \(2\), note that
\[p^{2^n} + 1 \equiv \left(\pm 1\right)^{2^n} + 1 \equiv 2 \bmod{4}\] since our primes are odd. Thus they are all divisible by \(2\) and none by \(4\).

Now assume some odd prime \(p^{\ast}\) divides all of the quantities in question. We'll show no such \(p^{\ast}\) can exist by contradiction.

In much the same way we showed the \(\gcd\) wasn't divisible by \(4\), we seek to find a contradiction in some modulus. But since we are starting with \(p^{2^n} + 1 \equiv 0 \bmod{p^{\ast}}\), if we can find some such \(p\) with \(p \equiv 1 \bmod{p^{\ast}}\), then we'd have our contradiction from
\[0 \equiv p^{2^n} + 1 \equiv 1^{2^n} + 1 \equiv 2 \bmod{p^{\ast}}\] which can't occur since \(p^{\ast}\) is an odd prime.

With this in mind, along with a subsequence of the arithmetic progression \(\left\{5, 13, 21, 29, \ldots\right\}\), it seems that using Dirichlet's theorem on arithmetic progressions may be a good strategy. However, this sequence only tells us about the residue modulo \(8\), but we also want to know about the residue modulo \(p^{\ast}\). Naturally, we look for a subsequence in
\[\mathbb{Z}/\mathbb{8Z} \times \mathbb{Z}/\mathbb{p^{\ast}Z}\] corresponding to the residue pair \((5 \bmod{8}, 1 \bmod{p^{\ast}})\). Due to the Chinese remainder theorem this corresponds to a unique residue modulo \(8p^{\ast}\).

Since this residue \(r\) has \(r \equiv 1 \bmod{p^{\ast}}\), we must have
\[r \in \left\{1, 1 + p^{\ast}, 1 + 2p^{\ast}, \ldots, 1 + 7p^{\ast}\right\} .\] But since \(1 + kp^{\ast} \equiv r \equiv 5 \bmod{8}\), we have \(kp^{\ast} \equiv 4 \bmod{8}\) and \(k \equiv 4\left(p^{\ast}\right)^{-1} \bmod{8}\) since \(p^{\ast}\) is odd and invertible mod \(8\). But this also means its inverse is odd, hence \(k \equiv 4\cdot(2k' + 1) \equiv 4 \bmod{8}\). Thus we have \(1 + 4 p^{\ast} \in \mathbb{Z}/8p^{\ast}\mathbb{Z}\) corresponding to our residue pair. Thus every element in the arithmetic progression \(S = \left\{(1 + 4p^{\ast}) + (8p^{\ast})k \right\}_{k=0}^{\infty}\) is congruent to \(1 + 4 p^{\ast} \bmod{8p^{\ast}}\) and hence \(5 \bmod{8}\) and \(1 \bmod{p^{\ast}}\).

What's more, since \(5 \in \left(\mathbb{Z}/8\mathbb{Z}\right)^{\times}\) and \(1 \in \left(\mathbb{Z}/p^{\ast}\mathbb{Z}\right)^{\times}\), we have \(1 + 4 p^{\ast} \in \left(\mathbb{Z}/8p^{\ast}\mathbb{Z}\right)^{\times}\) (again by the Chinese remainder theorem). Thus the arithmetic progression \(S\) satisfies the hypothesis of Dirichlet's theorem. Hence there must at least one prime \(p\) occurring in the progression (since there are infinitely many). But that also means \(p\) occurs in \(\left\{5, 13, 29, 37, \ldots\right\}\) hence we've reached our desired contradiction. RAA.

Now What?

We still don't know if the strong version \((2)\)
\[\gcd\left(5^{2^n} + 1, 13^{2^n} + 1, \ldots, p^{2^n} + 1, \ldots \right) = 2\] By similar arguments as above, if any odd prime \(p^{\ast}\) divides this \(\gcd\), then we have
\[5^{2^n} \equiv -1 \bmod{p^{\ast}}\] hence there is an element of order \(2^{n + 1}\). This means the order of the multiplicative group \(\varphi\left(p^{\ast}\right) = p^{\ast} - 1\) is divisible by \(2^{n + 1}\). Beyond that, who knows? We're still thinking about it (but only passively, more important things to do).