Prove

$latex x^{2} \equiv 3411 \mod 3457 $
has no solution?

Legendre Symbol:

$latex \displaystyle
x^{2} \equiv a \mod p
\iff
\boxed{
\left( \frac {a}{p} \right)
= \begin{cases}
-1, & \text{if 0 solution} \\
0 , & \text{if 1 solution} \\
1, & \text{if 2 solutions} \\
\end{cases}
}
$

Hint: prove $latex \left( \frac{3411}{3457} \right) = -1$

Using the Law of Quadratic Reciprocity, without computations, we can prove there is no solution for this equation.

Solution:

1.
3411 = 3 x 3 x 379 = 9 x 379

$Latex \displaystyle
\boxed{
\left(\frac{a}{p}\right)
\left(\frac{b}{p} \right)=
\left(\frac{ab}{p}\right)
}
$

$latex \displaystyle
\left(\frac{3411}{3457} \right)=
\left(\frac{9}{3457} \right).\left(\frac{379}{3457} \right)=
\left(\frac{379}{3457} \right)
$
since
$latex \displaystyle\left(\frac{9}{3457} \right)=1 $
because 9 is a perfect square, 3457 is prime.

2. By Quadratic Reciprocity,
$latex \displaystyle
\boxed{
\text{If p or q or both are } \equiv 1 \mod 4 \implies
\left(\frac{p}{q} \right)=
\left(\frac{q}{p} \right)}
$

Since
$latex…

The famous Singapore Math for children in primary schools is based on  visual models.

The Singapore Ministry of Education has published a new 2013 Math syllabus for primary and secondary schools, which will roll out in examinations within 4 to 6 years. Todate only Primary 1 and Secondary 1 Math syllabuses are published here:

http://www.moe.gov.sg/education/syllabuses/sciences

Who wins?

This comic video illustrates Singapore Math’s Arithmetics Polya-style problem solving process vs Algebra’s mechanical method.

The problem is as follow:
R is 3 times older than S two years ago. From now 2 years later, their total age is 32. How old is R now ?

See my previous blog (search “Monkey”) the Nobel Physicist Paul Dirac’s problem “The Monkeys and Coconuts“, 3 methods are used: 2 adanced modern math (by Sequence, eigenvector & eigenvalue), and the easiest & intuitive method (by Singapore Modelling Math). High-school Algebra method is impossible, if not cumbersome, to solve the Monkey problem !

I find Khan Linear Algebra video excellent. The founder / teacher Sal Khan has the genius to explain this not-so-easy topic in modular videos steps by steps, from 2-dimensional vectors to 3-dimensional, working with you by hand to compute eigenvalues and eigenvectors, and show you what they mean in graphic views.

If you are taking Linear Algebra course in university, or revising it, just go through all the Khan’s short (5-20 mins) videos on Linear Algebra here:

In 138 lessons sequence:

http://theopenacademy.com/content/linear-algebra-khan-academy

or random revision:

Relationship-Mapping-Inverse (RMI)
(invented by Prof Xu Lizhi 徐利治 中国数学家 http://baike.baidu.com/view/6383.htm)

Find Z = a*b

By RMI Technique:
Let f Homomorphism: f(a*b) = f(a)+f(b)

Let f = log
log: R+ –> R
=> log (a*b) = log a + log b

1. Calculate log a (=X), log b (=Y)
2. X+Y = log (a*b)
3. Find Inverse log (a*b)
4. ANSWER: Z = a*b

Prove:

$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$

1. Take f = log for Mapping:
$latex \log\sqrt{2}^{\sqrt{2}^{\sqrt{2}}} $
$latex = \sqrt{2}\log\sqrt{2}^{\sqrt{2}}$
$latex = \sqrt{2}\sqrt{2}\log\sqrt{2} $
$latex = 2\log\sqrt{2} $
$latex = \log (\sqrt{2})^2 $
$latex = \log 2$

2. Inverse of log (bijective):
$latex \log \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= \log 2$
$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$

Prof Su 苏步青, the founding pioneer Math professor of the China’s top universities (Zhejiang 浙江大学 and Fudan 复旦大学), was one of the few mathematicians who had longevity above 100 years old (the other was French Mathematician Hadammard).

http://en.m.wikipedia.org/wiki/Su_Buqing

Two men A and B are 100 km apart, walking towards each other, A at speed 6 km/hour and B at 4 km/hour.
A brings a dog which runs at 10 km/hour between them,  starting from A towards B, upon reaching B it runs back to reach A, then back to B again, and so on…

Find total distance the dog has covered when A and B finally meet ?

[Note: the content of this post is standard number theoretic material that can be found in many textbooks (I am relying principally here on Iwaniec and Kowalski); I am not claiming any new progress on any version of the Riemann hypothesis here, but am simply arranging existing facts together.]

The Riemann hypothesis is arguably the most important and famous unsolved problem in number theory. It is usually phrased in terms of the Riemann zeta function $latex {\zeta}&fg=000000$, defined by

$latex \displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}&fg=000000$

for $latex {\hbox{Re}(s)>1}&fg=000000$ and extended meromorphically to other values of $latex {s}&fg=000000$, and asserts that the only zeroes of $latex {\zeta}&fg=000000$ in the critical strip $latex {\{ s: 0 \leq \hbox{Re}(s) \leq 1 \}}&fg=000000$ lie on the critical line $latex {\{ s: \hbox{Re}(s)=\frac{1}{2} \}}&fg=000000$.

One of the main reasons that the Riemann hypothesis is so important to number theory is that the zeroes of…