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I am an IB student (HS equivalent), and I have to write a 12+ page long project for my mathematics class.

I am greatly interested in number theory and was thinking of starting by exploring and proving primitive Pythagorean triplets, and then moving on to prove Fermat's last theorem for the case of $n=4$. Then I was thinking of using my exploration of the Pythagorean triplets to prove for example that the radius of an incircle to a Pythagorean triangle is always an integer (or so I've heard), or that the area of a Pythagorean triangle can never take the form of $2a$ where $a$ is a perfect square.

However, I am concerned about the lack of complexity of my project. The proof for $n=4$ of formats theorem is unfortunately surprisingly easy. Would anyone please be willing to give me some ideas to improve the complexity of the project or perhaps take it a different route? Maybe explore different aspects of Fermat's last theorem, attempt a different proof, perhaps explore some more complex aspects of Pythagorean triples that I may not know about or take a different route altogether? Thanks a lot for your time.

Toby Mak
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John Ntogias
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    $(1)$ What do you mean with the "radius of a triangle" ? $(2)$ If the case $n=4$ is "too easy" , you can as well show the proof for the case $n=3$. You can also mention Sophie Germains brillant and important partial result. – Peter Jul 02 '20 at 11:09
  • oh sorry for not being clear enough. I mean the radius of the incircle of a pythagorean triangle (Unfortunately this proof is pretty elementary too). My bad. – John Ntogias Jul 02 '20 at 11:12
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    Nevertheless, the proof of the $n=4$-case contains the nice method "infinite descent" which is useful for many other proofs, maybe you collect some other proofs using this technique to show how useful and important it is.. – Peter Jul 02 '20 at 11:14
  • I will explore the applications of infinite descent aswell, thanks. Hopefully ill find some more complex applications of pythagorean triples too – John Ntogias Jul 02 '20 at 11:17
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    Or what about the question which pythagorean triples have legs with difference $1$ , which is realated to the pell-equation $x^2-2y^2=1$ ? – Peter Jul 02 '20 at 11:17
  • I was not aware of that equation, thanks! What do you mean by legs? – John Ntogias Jul 02 '20 at 11:18
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    If I am not wrong, the two shortest sides in a triangle with some angle with degree $90$° are called legs. – Peter Jul 02 '20 at 11:19
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    Given that the cases 3 and 4 were proved before Wiles came along, It seems unnecessary for you to prove them again (unless you have a new and original method). Why not explain in detail why certain specific case are easy and others are hard. – chasly - supports Monica Jul 02 '20 at 11:20
  • Thanks for the clarification peter. – John Ntogias Jul 02 '20 at 11:39
  • And chasly, thanks for your feedback. I am just not too sure about how I would format such a project. – John Ntogias Jul 02 '20 at 11:39
  • From now on, you should try to come up with your IA/EE idea yourself, instead of getting others to do the thinking from you. It might not be possible to contact your supervisor and get a reply soon, but you can still get help from online resources and past students' work. – Toby Mak Jul 02 '20 at 11:46
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    I am sorry if it sounds like i am attempting to "outsource" my IA. I am just seeking assistance when it comes to developing ideas ive had in mind so i can decide. I do not consider it immoral or against the rules considering the actual IA is going to be written and developed by me. – John Ntogias Jul 02 '20 at 11:54
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    Perhaps something in Pythagorean Triangles by Sierpiński or Three lectures on Fermat's last theorem by Mordell (has $n=3$ case). Mordell's "phamplet" is reprinted in this book, which can be found in many university libraries, if you want a physical book to work with. @chasly from UK: What you suggest is probably too ambitious for all but a handful of high school students. I think what the OP suggests is a good start for an appropriate IB project. – Dave L. Renfro Jul 02 '20 at 13:01
  • is reprinted in this book --- I notice that amazon.com didn't pay much attention to the book when providing a preview of the table of contents, since the table of contents provided is only for the first monograph of the 4 monographs reprinted in that book. See this link for the book itself. – Dave L. Renfro Jul 02 '20 at 13:13
  • If you still need ideas, I suggest that you look further into Gaussian integers. It can be shown that, for any primitive Pythagorean triple $(a,b,c)$ where $a$ is odd, the Gaussian integer $a+b\text{i}$ is a square of another Gaussian integer, whose modulus equals $\sqrt{c}$. For example, if $(a,b,c)=(15,8,17)$, then we have $15+8\text{i}=(4+\text{i})^2$, with $|4+\text{i}|=\sqrt{17}$. If you have more space left, you can try to write about Gaussian integers or about integers that can be written as sums of two perfect squares. – Batominovski Jul 02 '20 at 17:21
  • I do have a hardbound 1962 copy of Pythagirean Triples and it is a worthwhile read. They have gone up a lot but the paperback might still be worthwhile. The one thing I got out of it was a formula for $(B-A)=1$. Starting with a seed $T_0=(0,0,1)$

    $$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2$$

     – poetasis Jul 04 '20 at 16:54

2 Answers2

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You might consider ways of finding "triples on demand" such as by side, perimeter, area, area/perimeter ratio, product, and side difference. I've been working on such a paper for 10 years and have finally cut it down to 14 pagers. Here is an example of something you might work with given Euclid's formula where

$$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$

To find a triple, we solve for $n$ and test a range of $m$ values to se which yield integers, for example:

Finding side A using $F(m,n)$ $$A=m^2-n^2\implies n=\sqrt{m^2-A}\qquad\text{where}\qquad \sqrt{A+1} \le m \le \frac{A+1}{2}$$ The lower limit ensures $n\in\mathbb{N}$ and the upper limit ensures $m> n$. $$A=15\implies \sqrt{15+1}=4\le m \le \frac{15+1}{2} =8\quad\text{ and we find} \quad m\in\{4,8\}\implies n \in\{1,7\} $$ $$fF4,1)=(15,8,17)\qquad \qquad f(8,7)=(15,112,113) $$

This kind of work is easy. but it gets harder when you get to area (a cubic equation) and product (a quintic equation) and side difference which is easy for $C-B\quad C-A\quad \text{and}\quad B-A=\pm1$ but not so for other $B-A$ differences. For primitives, if $X=B-A$,

$X$ can be any prime number $(p)$ where $p=\pm1\mod 8$, raised to any non-negative power.

Under $100$, $X\in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}$.

If you discover the patterns I did, it could be enough for the paper alone.

By the way, you might acquire and learn to use $LaTeX{} $ because it is much better than word in typesetting equations.

poetasis
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  • Thanks a lot! May i ask, what does k represent when you wrote k = sqrt(m^2 - A) ? – John Ntogias Jul 02 '20 at 13:02
  • I'm sorry, that was a typo. In my paper, I am comparing Euclid's formula to a formula I developed which generates all of the triples in the subset where. $GCD(A,B,C)=(2X-1)^2, X,\in\mathbb{N}$ in the paper, the variable $k$ is the same in both formulas where mine is $F(n,k)$ $$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$ and Euclid's formula is $F(m,k)$... and $m=2n-1+k$. It appears to be original as per my question here. – poetasis Jul 02 '20 at 17:10
  • I fixed the typo. If you are going to explore $B-A=\pm x$, you may consider solving for $m$ instead of $n$ and you will find a formula that generates successive Pell number needed for $(m,n)$ values. It's easy for $B-A=\pm1$ but the higher "differences" split into two series each. You'll figure it out. – poetasis Jul 02 '20 at 17:17
  • You may find the On-line Encyclopedia of Integer Sequences useful, for instance the series $(1,2,5,12,29,70,169,...)$ can be found here. – poetasis Jul 03 '20 at 01:17
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Some not so common elementary about Pythagorean triplets that you could possibly use:

  1. If two of the numbers in a Pythagorean triplet are primes $> 5000$ then the third number must have a prime factor $> 17$. Proof
  2. In a primitive Pythagorean triplet, the sum of the two perpendicular sides is never divisible by any of the following primes $2,3,5,11,13,19,29,37,43,53,59,61,67,83,101,107,109,131,\ldots$. Proof
Nilotpal Sinha
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