If $<a_0, b_0, c_0>$ is a Primitive Pythagorean Triple, then $<a_0, b_0, c_0>U_i$ generates a new primitive triple $<a_i, b_i, c_i>$ where

$U_1 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_2 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_3 = \left| \begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right|$.

Let's exam a simple example. We were told that $<3, 4, 5>$ is a primitive triple. So…

$<3 ,4, 5>U_1 = <5, 12, 13> ,$ $<3, 4, 5>U_2 = <21, 20, 29> ,$ $<3, 4, 5>U_3 = <15, 8, 17> ,$

…which can easily be verified to be valid!

]]>The smallest such number is 14.

My question is how would one prove for example that 14 is nontotient?

]]>http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html

]]>http://www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html

]]>\begin{align} M(n) = \sum_{1\le k \le n} \mu(k) \end{align}

where $\mu$ is the Moebius function

Merten's conjecture says that

(2)\begin{align} \left| M(n) \right| < \sqrt {n} \end{align}

Now in 1985, this conjecture was proved false and also that a counter example exists somewhere between $10^{14}$ and $e^{1.59 \cdot 10^{40}}$.

So this result kind of kills peoples argument that "Goldbach's Conjecture is true because we've seen it to be true for any number tested".

But the fact that it's true for so many numbers makes me think it's more than just a coincidence. What do you guys think?

]]>Define

(1)\begin{align} \psi = \sum_{k=1}^{\infty} \frac{1}{F_k} \approx 3.359885666243177553172011302918927179688905133731 \ldots \end{align}

the sum of the reciprocals of every Fibonacci number

It has been proved to be irrational which looks like a daunting task to me

heres a link for more info:

http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant

After doing some further research, I stumbled upon two other series that have convergence related to $\pi$:

**Ramanujan's Formula**

\begin{align} \pi = 2 \sqrt{3} \sum_{n=0}^{\infty}{\frac{(-1)^{n}}{(2n+1)3^{n}}} \end{align}

**Chudnovsky Algorithm**

\begin{align} \frac{1}{\pi} = 12\sum_{k=0}^{\infty}{\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+\frac{3}{2}}} \end{align}

The Chudnovsky Algorithm is based on a rapidly converging hypergeometric series. It was used to generate over a billion digits of $\pi$! Mathematica uses it today to calculate $\pi$.

]]>My boyfriend sent me this article a couple of weeks ago, and I had been meaning to post it to the forum. It seems only appropriate now! I don't know how much has changed since this was published in 2006, but it's still interesting anyway.

Here you go: Riemann

Miller-Rabin-http://en.wikipedia.org/wiki/Miller-Rabin_primality_test

Solovay-Strassen:-http://en.wikipedia.org/wiki/Solovay–Strassen_primality_test

AKS:http://en.wikipedia.org/wiki/AKS_primality_test

there are many many more if you're interested and wikipedia has many links.

]]>\begin{align} \left(\frac{a}{n}\right) = \left(\frac{a}{p_1}\right)^{b_1} \left(\frac{a}{p_2}\right)^{b_2}... \left(\frac{a}{p_r}\right)^{b_r} \end{align}

where $\left(\frac{a}{p_1}\right),\left(\frac{a}{p_2}\right),...,\left(\frac{a}{p_r}\right)$ are Legendre symbols.

Now, that's cool and all, but did you know:

- There is a crater on the Moon named after him?
- The phrase "Invert, always invert," is associated with Jacobi because he thought that problems were best addressed backwards?
- He contributed to a great number of mathematical areas, including number theory, vector theory, and partial differential equations.

C.J. Scriba once compared Jacobi to Euler, saying,

"Jacobi and Euler were kindred spirits in the way they created their mathematics. Both were prolific writers and even more prolific calculators; both drew a great deal of insight from immense algorithmical work; both laboured in many fields of mathematics (Euler, in this respect, greatly surpassed Jacobi); and both at any moment could draw from the vast armoury of mathematical methods just those weapons which would promise the best results in the attack of a given problem."

Wikipedia also mentioned that Jacobi was "the most inspiring teacher of his time" and I was wondering if there were any other mathematicians out there who were similarly inspiring and how so. Does anyone out there have a mathematician who inspires them to tackle tough problems and plow through mathematics?

I've always found Paul Erdős to be a huge inspiration because of his undying dedication to mathematics and his treatment of it as a social discipline and one that is always thoroughly interesting. Check out Paul Hoffman's The Man Who Loved Only Numbers for a great read on Erdős.

]]>Here is the conjecture take from wikipedia article Artin's conjecture:

Let a be an integer which is not a perfect square and not -1. Denote by S(a) the set of prime numbers p such that a is a primitive root modulo p. Then

1. S(a) has a positive Schnirelmann density inside the set of primes. In particular, S(a) is infinite.

2. under the condition that a be squarefree, this density is independent of a and equals the Artin constant which can be expressed as an infinite product

I would guess that the Schnirelmann density obtain is greater when a particular *a* has a large set of primes for which *a* is a primitive root.

For example:

$13 \equiv 1 \mod{12}$, 13 is prime

$25 \equiv 1 \mod{12}$, $25=5*5, 5 \equiv 5 \mod{12}$

$37 \equiv 1 \mod{12}$, 37 is prime

$49 \equiv 1 \mod{12}$, $49=7*7, 7 \equiv 7 \mod{12}$

$61 \equiv 1 \mod{12}$, 61 is prime

$85 \equiv 1 \mod{12}$, $85=17*5, 17 \equiv 5 \mod{12}$

$109 \equiv 1 \mod{12}$, 109 is prime

$121 \equiv 1 \mod{12}$, $121=11*11, 11 \equiv 11 \mod{12}$

$133 \equiv 1 \mod{12}$, $133=19*7, 19 \equiv 7 \mod{12}$

$157 \equiv 1 \mod{12}$, 157 is prime

$169 \equiv 1 \mod{12}$, $169=13*13, 13 \equiv 13 \mod{12}$

I have no idea if what this means (if it means anything at all), but I haven't found anything through Google on it yet. I also haven't tried to prove it yet, so it might not even hold true for all *k*, but I still thought it was worth posting.

This Wednesday, one of UIUC's best grad students will be giving a talk entitled "The Birth of Analytic Number Theory" to the local undergrad math club (MATRIX). It will be held at 7pm in Altgeld 245, and should be great. The abstract is below. I **highly** encourage people to attend.

ABSTRACT: In the mid-1700's Euler made the first significant advance in the study of the distribution of prime numbers that had been made since Euclid's proof that there are infinitely many prime numbers. He established what is now known as the Euler product formula and used it to show that the sum of reciprocals of prime numbers diverges. This last result can be interpreted as a statement about how slow the prime numbers thin out. The influence of Euler's paper has been vast and can be felt on much contemporary research in number theory and related areas. In this talk I will derive Euler's product formula and show how Euler proved that the sum of reciprocals of primes diverges. If time permits I will show how he subsequently used his method to establish that the prime numbers are in some sense equally split between the arithmetic progressions 4k + 1 and 4k + 3.

]]>http://www.ocf.berkeley.edu/~gagnanda/mathstuff/Touchard.pdf

]]>It is fairly easy to see that all integers are *quadratic* residues mod 2, and that 0 and 1 are both *quadratic* residues mod 2.

But what about for larger powers of 2?

For $k = 2$, that is, mod 4, the *quadratic* residues are also 0 and 1. ($0^2 \equiv 2^2 \equiv 0 \mod 4, 1^2 \equiv 3^2 = 9 \equiv 1 \mod 4$). For $k = 3$, the *quadratic* residues are 0, 1, 4; for $k = 4$, the *quadratic* residues are 0, 1, 4, 9. Do we have a pattern? Not quite.

For $k = 5$, the *quadratic* residues are 0, 1, 4, 9, 16, 25, and **17**.

For $k = 6$, the *quadratic* residues are 0, 1, 4, 9, 16, 25, 36, 49, and 17, 33, 41, 57.

For $k = 7$, we get 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, and 17, 33, 41, 57, and 65, 73, 89, 97, 105, 113, and 68.

We can surely generate the *quadratic* residues of $2^k$ from the first $2^{k-1}$ elements. But $\forall k \geq 2$, we can generate the *quadratic* residues from the first $2^{k-2}$ elements. (I cannot figure out an easy proof of this.)

According to Wikipedia, a non-zero number will be a *quadratic* residue $\mod 2^k, k > 3$ if it is in the form $4^n \times (8m+1)$ with $0 \leq n \leq \frac{k}{2}, 0 \leq m < 2^{k-3}$.

A rough sketch of a proof was offered; but my observations can confrim it (i.e., $68 = 4 \times 17 = 4 \times (2 \times 8 + 1)$).

Including the trival *quadratic* residue 0, for the given $k$, there are # residues:

$k$ = | # quadratic residues |

1 | 2 |

2 | 2 |

3 | 3 |

4 | 4 |

5 | 7 |

6 | 12 |

7 | 23 |

8 | 44 |

*edit: added "quadratic" to every instance of the word "residue" for Andy's benefit.*

The Kronecker symbol is an extension of the Jacobi Symbol to all integers.

For general n, we write n = $2^c m$, with m odd and define: (1)

\begin{align} \left( \frac {a}{n} \right) = \left( \frac {a}{2} \right) ^c \left( \frac {a}{m} \right) \end{align}

where $\left( \frac {a}{m} \right)$ is the Jacobi symbol.

And define:

\begin{align} \left( \frac {a}{2} \right) = \begin{cases} 0 &\text{if }a \equiv 0\pmod{4}\ & 1& \text{if } a \equiv 1\pmod{8}\ & -1& \text{if } a \equiv 5\pmod{8}\end{cases}\ \end{align}

]]>
$\zeta(s) =\sum_{n=1}^\infty \frac{1}{n^s}$ where $s \in \mathbb{R}$ and $s>1$.

This function can be extended to all $s \in \mathbb{C}, s \neq 1$ which leads to much more such as the Riemann hypothesis, but I'm just going to focus on $s \in \mathbb{R}$. This is my favorite function and I've seen the formula for its inverse but never seen a derivation. But I think I was able to prove it myself thanks to our homework. So here it is:

One cool identity is called the Euler product formula:

$\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$

-Proof*:

$\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+ \cdots$

$\frac{1}{2^s}\zeta(s) = \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+ \cdots$

Subtracting the second from the first we remove all elements that have a factor of 2:

$\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\frac{1}{13^s}+ \cdots$

Repeating for the next term:

$\frac{1}{3^s}\left(1-\frac{1}{2^s}\right)\zeta(s) = \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+\frac{1}{27^s}+\frac{1}{33^s}+ \cdots$

Subtracting again we get:

$\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+ \cdots$

where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely we get:

$\cdots \left(1-\frac{1}{11^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1$

Dividing both sides by everything but the $\zeta(s)$ we obtain:

$\zeta(s) = \frac{1}{\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{11^s}\right) \cdots }$

This can be written more concisely as an infinite product over all primes ''p'':

$\zeta(s)\;=\;\prod_{p} (1-p^{-s})^{-1}$

*Taken from Wikipedia

Now, we proved in our homework if $f(n)$ is multiplicative and $n = p_{1}^{a_{1}}... \;p_{k}^{a_{k}}$, then $\sum_{d \mid n} \mu(d)f(d)& = \prod_{i=1}^k (1-f(p_{i}))$ where $\mu$ is the Möbius function.

Now let $n = k!$ and $f(n) = n^{-s}$ (which is multiplicative).

So we obtain;

$\sum_{d \mid k!} \mu(d)d^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$, but since all $d \leq k!$ divide $k!$

$\sum_{j=1}^{k!} \mu(j)j^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$

$\lim_{k \to \infty}\sum_{j=1}^{k!} \mu(j)j^{-s}& = \lim_{r \to \infty}\prod_{i=1}^r (1-p_{i}^{-s})$, so

$\sum_{j=1}^{\infty} \mu(j)j^{-s}& = \prod_{p} (1-p^{-s}) = \frac{1}{\zeta(s)}$

Now we must test $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ for convergence, but this is easy to see since

$\mu(j)j^{-s} \leq j^{-s} \;\forall\; j \in \mathbb{N}$

By the direct comparison test, $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ converges if $\zeta(s)$ does as well.

SO…

$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}$

It's pretty interesting because the Jacobi Symbol have same properties as the Legendre Symbol, and then some.

]]>A pair of numbers m and n such that

(1)\begin{align} \sigma (m) = \sigma (n) = m + n - 1 \end{align}

It said that only 11 pairs of augmented amicable pairs have been found.

I was wondering if there were any other types of augmented pairs, maybe such that

\begin{align} \sigma (m) = \sigma (n) = m + n + 1 or =m + n - 2 \end{align}

]]>
\begin{align} f(n) = \left\{\begin{array}{ll} 1 & n = 1 \\ f\left(\frac{n}{2}\right) & n\mbox{ is even} \\ f(3n+1) & n\mbox{ is odd and }n>1\end{array}\right. \end{align}

An example would be $f(11)$:

(2)\begin{align} 11\to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1. \end{align}

So, 11 finally terminates at 1. The big (and unsolved) question is: does every positive integer terminate, i.e., does every input eventually reach 1?

It is easy to see that every $n = 2^a$ reaches 1 eventually, so in some sense, this is some measure of how close numbers are to powers of 2. As for determining whether the "$3x+1$ conjecture" is true, one method is to do some sort of strong induction argument: if $f(n) = f(k)$ for some $k < n$, then we say $f(n)$ terminates by our inductive hypothesis.

If we use this method, we don't need to look at any even numbers since $f(2m) = f(m)$. In addition, we don't need to examine numbers of the form $4m + 1$ since $3(4m + 1) + 1 = 12m + 4 = 4(3m + 1)$ and hence $f(4m + 1) = f(3m + 1)$ for $m > 1$. Thus, we've narrowed it down to numbers of the form $4x + 3$. However, this is still a rather harrowing task since there doesn't seem to be much of a pattern in the number of iterations (at least to me): $f(27)$ and $f(31)$ take over 90 iterations to dip below their starting points, but $f(35)$ takes less than 10.

Incidentally, there's a group in Portugal attempting to "prove" the $3x+1$ conjecture via computing "all" the possibilities. In September, they verified the conjecture for numbers through $19\cdot 2^{58} \approx 5.5 \cdot 10^{18}$. They also have done work on the similar $5x+1$ and $7x+1$ problems.

]]>I was reading the wiki on Sociable numbers and I found it interesting that there are no known social numbers of order 3 (i.e. there are three numbers in the cycle). Do you think there's a way to prove that social numbers do or do not exist for any known order?

Also, the wiki talked about the theorizing the possibility that all numbers (positive integers) are either part of a social cycle or terminate at 1 by summing up proper divisors to get numbers in the next sequence. While this seemed kind of mind blowing — that perhaps there might be a counterexample to this of a sequence of composite numbers that never reach a prime, since all primes determine the next number in the sequence to be 1 — it also reminded me of a similar conjecture that i enjoy: the Collatz Conjecture.

The Collatz Conjecture makes a statement about a certain sequence. Take any positive integer to be the first term of a sequence and consider the following operations to determine the next number in the sequence. If the number is even, divide it by two, but if it is odd, multiply it by three and add one. The Collatz Conjecture says that for any sequence starting at any positive integer, this sequence will always reach 1.

Pretty crazy, huh? I think so, anyway. Paul Erdős said of the problem, "Mathematics is not yet ready for such problems." He offered $500 for its solution, which if you know Paul (or know of him), is quite the sum.

]]>While reading up on the unlikelyhood of odd perfect numbers, I ran across the primorial function. Combine *prime* and *factorial*: **primorial**! That is:

\begin{align} J = \prod_{i=1}^{k}(p_i) \end{align}

For N primorial, we write N#. To calculate N#, we take $J$ such that $p_k \leq$ N.

Thus, 10# = 9# = 8# = 7# $= 2 \times 3 \times 5 \times 7$.

And of course, Wikipedia's article on primorials (O beloved reference!).

Summing up Pomerance's Heuristic that Odd Perfect Numbers are Unlikely (link above): We can calculate the probability that a large perfect number ($n: n = pm^2 > 10^{300}$) ($10^{300}$ chosen because no odd perfects have been found in the interval $[1,10^{300}]$) exists. For a large, arbitrary $m$, hence, $m^2$, the probability is really small. It so happens that for even perfect numbers we can get a much better estimate on $m$ than simply arbitrary (thus, increasing the probability we can find large even perfect numbers, which do exist. But, at the same time, we really have not gotten a good handle on $m$ for odd primes (see the other threads on Perfect numbers).

Primorials come in to play in getting an upper bound for $\sigma(m^2)$ (for $N$ to be perfect, $p| \sigma(m^2)$). The upper bound is then used as an estimate in the nasty probability integral done in the calculations.

$p_n$# +/- 1 is sometimes a prime.

$p_n$# $\pm 1$ are twin primes for $n = 3, 5$ (the OEIS lists were too short to determine if there were any more overlaps (for $n > 1829$, there are none $5 < n < 1829$)).

Highly composite numbers (in the sense that the HCN is the smallest integer such that $\nu(HCN)$ increases to a record) are products of primorials.

Many HCNs are formed by multiplying together smaller HCNs. New primes are introduced on some HCNs, such as 2 (2), 6 (3), 60 (5), and 840 (7). For some HCNs, a prime is "re-introduced", such as 1260 (7*180).

The smallest prime factor of N is less than (2k + 8) / 3

The largest prime factor of N is greater than 10^8

The second largest prime factor is greater than 10^4, and the third largest prime factor is greater than 100

N has at least 75 prime factors and at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors

An odd perfect number is not divisible by 105

Every odd perfect number is of the form 12m + 1 or 324m + 81 or 468m + 117

This website also has a few links regarding unsolved problems. In the interest of the topic, this includes odd perfect numbers.

]]>"Let $m$ and $n$ be positive integers with $m \nmid n$. Prove that $\phi(mn) = m\phi(n)$."?

Or could/should it be:

"Let $m$ and $n$ be non-positive integers with $m | n$. Prove that $\phi(mn) = m\phi(n)$."?

6 = (2)(3) = (2^(2-1))*(2^2 - 1)

28 = (4)(7) = (2^(3-1))*(2^3 - 1)

496 = (16)(31) = (2^(5-1))*(2^5 - 1)

8,128 = (64)(127) = (2^(7-1))*(2^7 - 1)

33,550,336 = (4096)(8191) = (2^(13-1))*(2^13 - 1)

8,589,869,056 = (65536)(131071) = (2^(17-1))*(2^17 - 1)

137,438,691,328 = (262144)(524287) = (2^(19-1))*(2^19 - 1)

Here is the link where I found the formula Perfect Numbers. There is also information on perfect odds.

]]>http://en.wikipedia.org/wiki/Abundant_number

http://en.wikipedia.org/wiki/Practical_number

http://en.wikipedia.org/wiki/Deficient_number

\begin{align} \sigma(n) = \sum_{d|n} d \end{align}

and

(2)\begin{align} \sigma_k(n) = \sum_{d|n} d^k, \end{align}

we also have a similar generalization for $\phi(n)$. Given $n=p_1^{a_1}\cdots p_k^{a_k}$, we can express Euler's totient function as

(3)\begin{align} \phi(n) = n\prod_{i=1}^k \left(1 - \frac{1}{p_i}\right). \end{align}

Then Jordan's totient function is

(4)\begin{align} J_m(n) = n^m\prod_{i=1}^k \left(1 - \frac{1}{p_1^m}\right). \end{align}

Thus, we have that $\phi(n) = J_1(n)$. Incidentally, we had that $\sigma_0(n) = \nu(n)$, which was pretty cool; $J_0(n)$ is unfortunately a bit less interesting:

(5)\begin{align} J_0(n) = \left\{\!\begin{array}{rl} 1 & \mbox{if }n = 1 \\ 0 & \mbox{if }n > 1 \end{array}\right. \end{align}

]]>
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]]>I think we can generalize for any n, the prime divisors of n! are the same as all primes p where 1$<$p$\le$n.

This is true because n, n-1, …, 3, 2 all possess prime divisors that are less than themselves, and thus must be less than the n = max{n, n-1, …, 3, 2}. Also, all p$\le$n are included because they are included as integers in the factorial of n!.

]]>$p = 2^{43112609} - 1$

The LA Times story about it.

]]>http://www.math.cornell.edu/~mec/2003-2004/cryptography/diffiehellman/diffiehellman.html

]]>http://www.springerlink.com/content/w81443j4h313r512/fulltext.pdf?page=1

]]>(Note, $a,b = 0 \forall n$ is trivial.)

For example:

$6 \times 9 = 54 \equiv 2 \mod 13$.

$6 + 9 = 15 \equiv 2 \mod 13$.

(Also, the familiar $2 \times 2 = (2+2) \mod 10$.

Related note:

From what I can see, modulo $n$ and base $n$ are related by the fact that the 0-th power term of $x$ base $n$ is equal to $x \mod n$.

But then for $4k+3$, we used a theorem that was never proven to us (Dirichlet's Theorem).

So here is my own attempt to prove the same statement. Tell me what you think.

Assume there are finitely ($n$) many primes of the form $4k+3$: $p_{1}, \dots, p_{n}$

Let $N = 4(p_{1}, \dots, p_{n}) - 1$

This odd number can be factored into prime divisors.

But, it cannot be that all prime factors have the form $4k+1$ because as we showed in class, the product of numbers of the form $4k+1$ still has the form $4k+1$.

Therefore, there must be a prime divisor of the form $4k+3$. Thus, $N$ is divisible by one of $p_{1}, \dots, p_{n}$ since these are all the primes in the form $4k+3$.

But this implies that $p_{i} \mid 1$ for $1 \leq i \leq n$.

And this is impossible. $\to \gets$

\begin{eqnarray} x \equiv a_1\mod n_1 \\ x \equiv a_2\mod n_2 \\ x \equiv a_3\mod n_3 \\ \vdots \end{eqnarray}

We could try to apply CRT to this problem, but we would run into problems with our system being infinite: we would get an infinite lcm as our modulus and our solution would be an infinite sum.

We can alternatively try an inspection method: if there is a solution $X$, then there must be some $N$ such that for all $i \geq N$, we get $X \equiv a_i \mod n_i$. Thus, we could try to look far ahead enough into the system to find such a solution. Aside from the fact that this is rather infeasible, the following system has no solution:

(2)\begin{eqnarray} x \equiv 1 \mod 2 \\ x \equiv 2 \mod 3 \\ x \equiv 2 \mod 5 \\ x \equiv 2 \mod 7 \\ \vdots \end{eqnarray}

Using this method, we would look at this and say "ooo, the solution must be 2" and be wrong. Thus, we could try to truncate the system after finding such a bound and apply CRT to it, then check the solution it gives against the supposed solution.

]]>A number in decimal is palindromic if the digits read the same forward and backward. Prove that:

a) Every palindromic integer with an even number of digits is divisible by 11.

b) Every integer whose base k representation is palindromic and has even length is divisible by k+1.

The proof for (a) is sort of trivial if you remember the divisibility rule for 11. And you can find slightly more interesting proofs online, or try it on your own!

(b) seems more exciting and I haven't thought too much about how that proof would work yet, but just to see it in action, we'll take a number in hexadecimal (base 16) and show that it's divisible by 17.

So take AC33CA which is 10*16^{5} + 12*16^{4} + 3*16^{3} + 3*16^{2}+12*16 + 10 = 11285450 = 17 * 663850.

$(p-1)! = (p-1)(p-2)! \equiv (-1)(p-2)! \mod{p} \equiv -1 \mod{p} \Rightarrow$

$(p-2)! \equiv 1 \mod{p}$

Using induction, one can show for $a>0$,

$(a-1)!(p-a)! \equiv (-1)^a \mod{p}$

so if $a$ is even, $(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p$

and if $a$ is odd, $-(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p \Rightarrow$

$(-1)^a(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p$ if $a>0$

Now when $a \leq 0, (p-a)! \equiv 0 \mod{p}$

This is because $p! \equiv 0 \mod{p}$, so for $k>0$,

$(p+k)! = p!(p+1) \dots (p+k) \equiv 0*(p+1) \dots (p+k) \mod{p} \equiv 0 \mod{p}$

That might not be practical for the average musician, but creating a shortcut method can be useful. There have been times when I'm not sure how many measures of rest I've counted. I find my place again by looking at a repeated figure the violins might be playing and realizing it will start again on beat 1 of a measure when I've counted 3n measures, where n is an integer greater than 0.

I'm not sure how many of you have a music background (I know someone out there is a music major!) but it'd be cool to see how number theory and performing music are related and if any useful applications could come out of that.

]]>http://www.ted.com/index.php/talks/arthur_benjamin_does_mathemagic.html

Thoughts??

]]>Maybe some of you will find this alternative much easier or harder than the method we learn in class.

Find the least non negative solution of each system of congruence below.

(1)\begin{align} $ x \equiv 2 \mod{5} $ \end{align}

(2) \begin{align} $ x \equiv 4 \mod{7} $ \end{align}

(3) \begin{align} $ x \equiv 3 \mod{9} $ \end{align}

Suppose we have a solution such that x=5a+2, $$ a \in \mathbb{Z}$

$5a+2 \equiv 4 \mod{7}$

$5a \equiv 2 \mod{7}$

$a \equiv 6 \mod{7}$

a = 7b +6, $$ b \in \mathbb{Z}$

So we have:

x=5a+2

x=5(7b+6)+2

x=35b+32

$35b+32 \equiv 3 \mod{9}$

$35b \equiv -29 \mod{9}$

$35b\equiv -2 \mod{9}$

$b \equiv 2 \mod{9}$

b= 9c+2, $$ c \in \mathbb{Z}$

Now we have:

x=35b+32

x=35(9c+2)+32

x=315c+102

So all solutions are:

{315c+102: $$ c \in \mathbb{Z}$}

The smallest positive solution is 102.

$\exists\ x,y,z \in \mathbb{Z} \ \ s.t.\ \ x^{n}+y^{n} = z^{n} ,\ where \ n \in \mathbb{N} \iff n=2$

There was a lot of complex and amazing math that went into the proof that involved a lot of number theory and complex analysis that went into this proof. I highly recommend watching this if you get a chance.

]]>Is there always a prime between $n^2$ and $(n^2+1)^2$?

(The fact that there is always a prime between n and 2n was called Bertrand's conjecture and was proved by Chebyshev.)

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Prime_numbers.html

If we first look at Bertrand's conjecture we can see:

n=1, gives us the interval [1,2] in which the prime number 2 resides

n=2, gives us [2,4] primes: 2,3

…

n=5, gives us [5,10] primes: 5,7

…

n=10, gives us [10,20] primes: 11,13,17,19

…

n=17 gives us [17,34] primes: 17,19,23,29,31

So, this would seem that as n gets bigger not only is there a prime number in between them but the set of primes is in fact larger than the n's that come before it.

But that's not really a proof.

The unsolved problem of the interval of [ $n^2$ , $(n^2+1)^2$ ]

If we let $n=n^2$ from the Bertrand conjecture and compare $2n$ and $(n^2+1)^2$

because $2n^2$ is less than $(n^2+1)^2$

So the interval of [ $n^2,2n^2$ ] is always contained inside of [ $n^2$ , $(n^2+1)^2$ ]

Right? so if the Bertrand conjecture has been proven so why has this one been elusive

]]>I also found another website full of a bunch of other almost integers. Though some of them are a bit ridiculous — almost integer (8) involves taking the cosine of the cosine of the cosine … of the cosine (I think I counted seven times) of 5 and just multiplying it by two — there are other simpler ones. The website is here if you're interested.

]]>\begin{align} \lim_{x \to \infty} \frac{ \pi (x) \log (x)}{{x}} = 1 \end{align}

Does this suggest that the frequency of primes decreases?

Meaning:

\begin{align} \frac{ \pi (k)}{{k}} > \frac{ \pi (k+1)}{{k+1}} \end{align}

]]>
Looking at even numbers, we need either to sum three even numbers or two odd numbers with an even number. The only number that matches the former description is 6 = 2 + 2 + 2. As for the latter, we can see that this boils down to the Goldbach's conjecture that we know today; the even number must be 2, forcing us to find two odd primes that add up to an even number.

]]>Looking at even numbers, we need either to sum three even numbers or two odd numbers with an even number. The only number that matches the former description is 6 = 2 + 2 + 2. As for the latter, we can see that this boils down to the Goldbach's conjecture that we know today; the even number must be 2, forcing us to find two odd primes that add up to an even number.

]]>