## Tag Archives: right angled triangles

### Pythagorean Triples

The Pythagorean Theorem, that “beloved” formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the hypotenuse. In symbols,

$a^{2}+b^{2}=c^{2}$ where a and b are the sides and c is the hypotenuse.

Since we are interested in number theory, that is, the theory of the natural numbers, we will ask whether there are any Pythagorean triangles all of whose sides are natural numbers. There are many such triangles. The most famous has side 3,4 and 5. Here are the first few examples:

$3^{2}+4^{2}=5^{2}$

$5^{2}+12^{2}=13^{2}$

$8^{2}+15^{2}=17^{2}$

$28^{2}+45^{2}=53^{2}$

Our first naive question is whether there are infinitely many Pythagorean triples, that is, triples of natural numbers $(a,b,c)$ satisfying the equation $a^{2}+b^{2}=c^{2}$

The answer is “YES” for a silly reason. If we take a Pythagorean triple $(a,b,c)$ and multiply it by some other number d, then we obtain a new Pythagorean triple

$(ad,bd,cd)$. This is true because

$(da)^{2}+(db)^{2}=d^{2}(a^{2}+b^{2})=d^{2}c^{2}=(dc)^{2}$

Clearly, these new Pythagorean triples are not very interesting. So we will concentrate our attention on triples with no common factors. We will even give them a name:

A primitive Pythagorean triple (or PPT for short) is a triple of numbers $(a,b,c)$ so  that a,b, and c have no common factors and satisfy

$a^{2}+b^{2}=c^{2}$.

To investigate whether are infinitely many PPT’s is the same as asking whether there is a formula to find as many PPT’s as we want — the formula would contain relationship(s) between a, b and c.

As explained in the previous blog, the first step is to accumulate some data. I used a computer to substitute in values for a and b and checked if $a^{2}+b^{2}$ is a square. Here are some PPT’s that I found:

$(3,4,5)$; $(5,12,13)$; $(8,15,17)$; $(7,24,25)$;

$(20,21,29)$; $(9,40,41)$; $(12,35,37)$; $(11,60,6)$;

$(28,45,53)$; $(33,56,65)$; $(16,63,65$.

A few conclusions can be easily drawn even from such a short list. For example, it certainly looks like one of a and b is odd and the other is even. It also seems that c is always odd.

It is not hard to prove that these conjectures are correct. First, if a and b are both even, then c would also be even. This  means that a,b and c would have a common factor of 2, so the triple would not be primitive. Next, suppose that a and b are both odd, which means that c would have to be even. This means that there are numbers x,y and z so  that

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

We can substitute this in the equation $a^{2}+b^{2}=c^{2}$ to get

$(2x+1)^{2}+(2y+1)^{2}=(2z)^{2}$

Hence, $2x^{2}+2x+2y^{2}+2y+1=2z^{2}$

This last equation says that an odd number is equal to an even number, which is impossible, so a and b cannot both be odd. Since, we have just checked that they cannot both be even and cannot both be odd, it must be true that one is even and the other is odd. It’s then obvious from the equation $a^{2}+b^{2}=c^{2}$ that c is also odd.

We can always switch a and b, so our problem now is to find all solutions in natural numbers to the equation

$a^{2}+b^{2}=c^{2}$ with a odd, b even and a,b,c, having no common factors.

The tools we will use are factorization and divisibility.

Our first observation is that if $(a,b,c)$ is a primitive PPT, then we can factor

$a^{2}=c^{2}-b^{2}=(c-b)(c+b)$

Here are a few examples from the list given earlier, where note that we always take n to be odd and b to be even:

$3^{2}=5^{2}-4^{2}=(5-4)(5+4)=1.9$

$15^{2}=17^{2}-8^{2}=(17-8)(17+8)=9.25$

$35^{2}=37^{2}-12^{2}=(37-12)(37+12)=25.49$

$33^{2}=65^{2}-56^{2}=(65-56)(65+56)=9.121$

It looks like $c-b$ and $c+b$ are themselves always squares. We check this observation with a couple more examples:

$21^{2}=29^{2}-20^{2}=(29-20)(29+20)=9.49$

$63^{2}=65^{2}-16^{2}=(65-16)(65+16)=49.81$

How can we prove that $c-b$ and $c+b$ are squares? Another observation apparent from our list of examples is that $c-b$ and $c+b$ seem to have no common factors. We can prove this last assertion as follows. Suppose that d is a common factor of $c-b$ and $c+b$, that is, d divides both $c-b$ and $c+b$. Then, d also divides

$(c+b)+(c-b)=2c$ and $(c+b)-(c-b)=2b$

Thus, d divides $2b$ and $2c$. But, b and c have no common factor because we are assuming that $(a,b,c)$ is a primitive Pythagorean triple. So, d must equal 1 or 2. But, d also divides $(c-b)(c+b)=a^{2}$ and a is odd, so d must be 1. In other words, the only number dividing both $c-b$ and

$c+b$ is 1, so $c-b$ and $c+b$ have no common factor.

We now know that $c-b$ and $c+b$ have no common factor, and that their product is a square since $(c-b)(c+b)=a^{2}$. The only way that this can happen is if $c-b$ and $c+b$ are themselves squares. So, we can write

$c+b=s^{2}$ and $c-b=t^{2}$ where $s>t \geq 1$ are odd integers with no common factors. Solving these two equations for b and c yields

$c=(s^{2}+t^{2})/2$ and $b=(s^{2}-t^{2})/2$ and then

$a=\sqrt{(c-b)(c+b)}=st$

We have finished our first proof of elementary number theory! The following theorem records our accomplishment:

Theorem (Pythagorean Triple Theorem) You will get every primitive Pythagorean triple $(a,b,c)$ with a odd and b even by using the formulas

$a=st$ and $b=(s^{2}-t^{2})/2$ and $c=(s^{2}+t^{2})/2$ where

$s>t \geq 1$ are chosen to be any odd integers with no common factors.

More later…

Nalin Pithwa