Hi! I hope you enjoyed the previous entry.

If you didn't understand something in the last entry, please leave me a comment, I'll do my best to get back to you in a timely manner.

With that said, we'll now start today's entry.

Pythagorean Triples

There are few people who haven't heard of the Pythagorean theorem. But if you haven't it basically states that the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. Mathematically, if the legs are a and b and the hypotenuse (the longest side) is c then,

a^2  + b^2 = c^2

Now, that's all fine and dandy but because we're studying number theory, then we ask ourselves "what are the numerical solutions to this equation?"

We write out solutions in the notation (a,b,c). For example, (3,4,5) is a solution since 3^2 + 4^2 = 5^2

Are there infinitely many solutions to this equation (a^2  + b^2 = c^2) ?

Intuitively, you would probably think yes, which is correct.

Here's how we know that there are infinitely many. Let's take a pythagorean triplet (a,b,c) and mulitply it by some constant "d" yielding the triplet (da,db,dc). We know that this is a triplet because (da)^2 + (db)^2 = d^2(a^2+ b^2) = d^2*c^2 = (dc)^2.

Obviously these new pythagorean triplets are not very interesting, they're just the regular ones multiplied by some quantity.

We shall term the pythagorean triplets we are interested in as "primitive pythagorean triplets" (PPTs) or those triplets, (a,b,c) where a, b and c have no common factors and satisfy the pythagorean theorem.

For example, (3,4,5) is a PPT while (12,16,20) isn't since the common factor between (12,16,20) is 4. (12,16,20) is still a pythagorean triple but it is not primitive.

So, now we ask “Can we find a formula for all PPTs?”

Yes.

We first begin by collecting data on the first few PPTs by finding them out by hand using the equation a^2  + b^2 = c^2.

This yields,

(3,4,5)

(5,12,13)

(8,15,17)

(7,24,25)

(20,21,29)

(9,40,41)

(12,35,37)

(11,60,61)

(28,45,53)

(33,56,65)

(16,63,65)

Why the long list? Well, looking for patterns in the numbers is a method we can use to develop a formula for finding all PPTs.

From this list, it is clear that one of a or b is odd and the other one even, and that c is always odd. Can we prove this? Well, if a and b are both even, then c would be even. This would leave a common factor of two between a, b, and c thus making the triplet non-primitive.

What if a and b were odd? This implies that c would have to be even. Does this mathematically work?

Well, if that’s the case there are numbers x, y, and z such that:

a = 2x + 1            b = 2y + 1       c = 2z

We can substitute this into the Pythagorean theorem for a, b, and c and see if we get a true statement.

(2x + 1)^2 + (2y + 1)^2 = (2z)^2

Expansion yields…

4x^2 + 4x + 4y^2 + 4y + 2 = 4z^2

Dividing by 2,

2x^2 + 2x + 2y^2 + 2y + 1 = 2z^2

The left side is odd for all natural numbers (a natural number is an integer greater than 0) while the right side is even. So basically, we’re saying that an odd number is equal to an even number… CONTRADICTION! This means that a and b cannot both be odd, and we also found that they both cannot be even. Thus, one of a and b must be odd while the other even.

So now, we are looking for all PPTs such that,

a^2  + b^2 = c^2.

With a odd, b even, and a,b,c having no common factors.

Well, if we solve for a^2 we get

a^2 = c^2 – b^2 = (c-b)(c+b)

Let’s plug in some values I gave above and see what pops out. Note, we’re always assuming a is odd and b is even for consistency.

(3,4,5) 3^2 = 5^2 – 4^2  = (5-4)(5+4) = 1*9
(8,15,17) 15^2 = 17^2 – 8^2  = (17-8)(17+8) = 9*25

(12,35,37) 35^2 = 37^2 – 12^2  = (37-12)(37+12) = 25*49

It looks like c-b and c+b are always squares. Can we prove this?

Let’s take a step back, is there anything else we see about c-b and c+b? They do not seem to have any common factors. Can we prove that?

(NOTE: Read the following two paragraphs very carefully, you may have to read them more than once. These are the trickiest paragraphs in this entire lesson.)

If we suppose that k is a common factor of c-b and c+b then k also divides their sum and difference, namely 2c and 2b respectively. Since we know that b and c have no common factors, then that leaves only 1 or 2 as the common factor. But since k must also divide the product of c+b and c-b namely, a^2 then k has to be 1. So c+b and c-b must have no other cofactors other than 1.

So we know that c+b and c-b have no common factor and that their product is a square (a^2). The only way this is possible is if c+b and c-b are themselves squares. Now that we armed with this new tidbit of information, we have the following,

c+b = s^2            c-b = t^2       s > t >= 1

>= means greater than or equal to

Solving the two equations above treating s and t as constants, we yield the following.

c = (s^2 + t^2) / 2        b = (s^2 - t^2) / 2

Plugging this into the Pythagorean yields a = st

So after all that work, given any s and t which are odd integers without common factors we can produce any Pythagorean triplet.

We can produce every Primitive Pythagorean Triplet (a,b,c) with a odd and b even by using the formulas

a = st               b = (s^2 - t^2) / 2              c = (s^2 + t^2) / 2

where s > t >= 1 are chosen to be any odd integers with no common factors.

Thank you for reading this short lesson on number theory.

If you haven’t already taken a look at the previous entry, please do so.

Also, if you have any questions or comments please leave them here, I shall do my best to get back to you promptly.

Thanks again, and come back again next time when we cover…

>>>THE UNIT CIRCLE --  BEYOND YOUR BASIC TRIGONOMETRY CLASS<<<

Welcome!

This is a xanga I setup for you kids who want to do some math that's not so much proof based, but interesting and even fun. Enjoy!!

It is assumed that the reader knows what the terms odd, even, square, cube, prime, composite, triangular, perfect, and Fibonacci denote. These are all types of numbers. If you are unaware of their meaning, go to www.mathworld.com and type the term in their searchbar.

Let's move on to a more advanced topic which you may or may not be aware of.

>>>Modular arithmetic<<<

A number "x" is "y" modulo (abbreviated mod) "p" if "x" divided by "p" leaves a remainder "y."

That's the formal definition, it's kind of confusing. An example will help.

11 is 1 mod 10 because when 11 is divided by 10 we get a remainder of 1. Another 1 mod 10 number is 21 because when we divide it by 10 we get 1 as a remainder.

We will see this more later on, so keep it floating in the back of your head for now!

>>>Sums of integers<<<

(Note, it helps to draw out what I'm about to describe on a piece of paper... try using n=5)

Suppose we have n^2 pebbles arranged in an nXn square. Starting with the upper-left corner and going along the diagonal down to the lower-right corner we have n pebbles. If we a draw a line along this diagonal we seperate the square into two triangles (each with an equal number of pebbles.) Each triangle contains all of the pebbles on its side of the diagonal but not the diagonal itself.

Now, if we try to find the number of pebbles in the square we get the following equation.

2*(Number of pebbles in one triangle) + (number of pebbles along diagonal, "n") = n^2

Solving our equation for the (Number of pebbles in one triangle) variable we yield the following result--

Number of pebbles in one triangle = ( n^2 - n )/2

Let's examine the solution for a second. What have we found? Well, our results tells us that we have the number of pebbles in one of the triangles we created.

Well, looking at the triangle below the diagonal, we have one pebble in the first row, two in the second, three in the third... n-1 in the last row. If we add all those up, we get the sum of the numbers from 1 to n-1. What is this sum? We solved for it when we solved for the number of pebbles in our triangle back in the equation a few paragraphs above.

Number of pebbles in one triangle = ( n^2 - n )/2 = ( n(n-1) ) /2

So... the sum of the all the consecutive integers from 1 to n-1 is ( n(n-1) ) /2

We can write this in a more useful form... if we change n-1 to n, then we are increasing the limit of the sum by one. If we do this to each of the terms in the expression ( (n-1)*n ) /2 then we get ( n(n+1) ) /2.

So, after all that work, we get the following statement --

The sum of all the integers from 1 to n is ( n(n+1) )/2.

Now anyone could have told you that formula, but the real value of this formula comes from it's proof. There is a certain elegance in mathematics which I hope this has sparked in you. There are other proofs to our bolded statement using inductive methods (A very powerful method of proof we will discuss in a future outline.)

And PLEASE! This is meant to be easy to understand, if you have any questions at all -- POST THEM BELOW!! I will do my best to make a quick response.

Thank you for reading this.

>>>NEXT-- The Pythagorean Theorem: Super-sized!<<<