# High school mathematical research

I am a grade 12 student. I am interested in number theory and I am looking for topics to research on.

Can you suggest some topics in number theory and in general that would make for a good research project?

I have self-studied certain topics in Abstract Algebra and Number Theory. I'm fascinated by primes (like most people are).

Preferably, suggest some unexplored problems so that new results can be obtained.

Thanks.

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How about finding integer solutions to equations of the form $x^2+y^2=z^2$ ? –  Weltschmerz Jun 18 '12 at 19:07
Pell's equation (en.wikipedia.org/wiki/Pell's_equation) –  user17762 Jun 18 '12 at 19:13
Investigate (and prove?) divisibility properties of Fibonacci numbers? –  Old John Jun 18 '12 at 19:15
I have flagged this to be made into a community wiki. –  user17762 Jun 18 '12 at 19:17
This might be too broad, but multiplicative functions and interesting/important. –  The Chaz 2.0 Jun 18 '12 at 19:34
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NOTE The OP didn't state "Preferably, suggest some open problems so that new results can be obtained." when this was answered.

I can provide you with Burton's Elementary Number Theory. It has a series of historical introductions and great examples you'll probably find worth of a research project. He has information and obivously theory about results from Fermat, Euler, Diophantus, Wilson, Möbius, and others. I can also provide you with the three volumes of the History of Number Theory, which might be a great source.

A few examples are

Fermat's Little Theorem If $p\not\mid a$ then$$a^{p-1} \equiv 1 \mod p$$

Wilson's Theorem If $p$ is a prime then

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

Möbius Inversion Formula If we have two arithmetical functions $f$ and $g$ such that

$$f(n) = \sum_{d \mid n} g(d)$$

Then

$$g(n) = \sum_{d \mid n} f(d)\mu\left(\frac{n}{d}\right)$$

Where $\mu$ is the Möbius function.

Maybe so interesting as the previous,

The $\tau$ and $\sigma$ functions

Let $\tau(n)$ be the number of divisors of $n$ and $\sigma(n)$ its sum. Then if $$n=p_1^{l_1}\cdots p_k^{l_k}$$

$$\tau(n)=\prod_{m=1 }^k(1+l_m)$$

$$\sigma(n)=\prod_{m=1 }^k \frac{p^{l_m+1}-1}{p-1}$$

Legendre's Identity

The multiplicty (i.e. number of times) with which $p$ divides $n!$ is

$$\nu(n)=\sum_{m=1}^\infty \left[\frac{n}{p^m} \right]$$

However odd that might look, the argument is somehow simple. The multiplicity with which $p$ divides $n$ is $\left[\dfrac{n}{p} \right]$, for $p^2$ it is $\left[\dfrac{n}{p^2} \right]$, and so forth. To get that of $n!$ we sum all these values to get the above, since each of $1,\dots,n$ is counted $l$ times as a multiple of $p^m$ for $m=1,2,\dots,l$, if $p$ divides it exactly $l$ times. Note the sum will terminate because the least integer function $[x]$ is zero when $p^m>n$.

Perfect numbers

A number is called a perfect number is the sum if its divisors equals the number, this means

$$\sigma(n) =2n$$

Euclid showed if $p=2^n-1$ is a prime, then $$\frac{p(p+1)}{2}$$ is always a perfect number

Euler showed that if a number is perfect, then it is of Euclid's kind.

$n$ - agonal or figurate numbers.

The greeks were very interested in numbers that could be decomposed into geometrical figures. The square numbers are well known to us, namely $m=n^2$. But what about triangular, or pentagonal numbers?

Explicit formulas have been found, namely

$$t_n=\frac{n(n+1)}{2}$$

$$p_n=\frac{n(3n-1)}{2}$$

You can try, as a good olympiadish excercise, to prove the following:

$${t_1} + {t_2} + {t_3} + \cdots + {t_n} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$$

We can arrange the numbers in a pentagon as a triangle and a square:

$${p_n} = {t_{n - 1}} + {n^2}$$

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The proof is trivial if we proceed by induction. –  Anunay Kulshrestha Jun 18 '12 at 20:08
No proof is trivial enough, but, good for you. =) –  Pedro Tamaroff Jun 18 '12 at 20:09
@AnunayKulshrestha While that's true, you can still look at the derivation or intuition of choosing such a formula in the first place. Proceeding by induction once the formula to prove has been given is only running the second half of the race. –  Robert Mastragostino Jun 18 '12 at 20:10
@RobertMastragostino Maybe I shouldn't have given the result away, that'd have been more interesting. –  Pedro Tamaroff Jun 18 '12 at 20:11