## Tag Archives: Pythagoras theorem

### Who was Pythagoras?

We recognize the name “Pythagoras” because it is attached to a theorem, one that most of us have grappled with at school.  The square on the Pythagoras of a right angled triangle is equal to the sum of squares on the other two sides. That is, if you take any right-angled triangle, then the square of the longest side is equal to the sum of the squares of the other two sides. Well known as his theorem may be, the actual person has proved rather elusive, although we know more about him as a historical figure, than we do for, say, Euclid. What we don’t know is whether he proved his eponymous theorem, and there are good reasons to suppose that even if he did, he wasn’t the first one to do so.

But more of that story later.

Pythagoras was Greek, born around 569BC on the island of Samos in the north-eastern Aegean. (The exact date is disputed, but this one is wrong by at most 20 years.) His father, Mnesarchus was a merchant from Tyre; his mother, Pythais, was from Samos. They may have met when Mnesarchus brought corn to Samos during a famine, and was publicly thanked by being made a citizen.

Pythagoras studied philosophy under Pherekydes. He probably visited another philosopher, Thales of Miletus. He attended lectures given by Anaximander, a pupil of Thales, and absorbed many of his ideas on cosmology and geometry. He visited Egypt, was captured by Cambyses, II, the King of Persia, and taken to Babylon as a prisoner. There he learned Babylonian mathematics and musical theory. Later he found the school of Pythagoreans in the Italian city of Croton (now, Crotone), and it is for this that he is best remembered. The Pythagoreans were a mystical cult. They believed that the universe is mathematical, and that various symbols and numbers have a deep spiritual meaning.

Various ancient writers attributed various mathematical theorems to  the Pythgoreans, and by extension to Pythagoras — notably, his famous theorem about right-angled triangles. But we have no idea what mathematics Pythagoras himself originated. We don’t know whether Pythagoreans could prove his theorem, or just believed it to be true. And there is evidence from the inscribed clay tablet known as Plimpton 322 that the ancient Babylonians may have understood the theorem 1200 years earlier — though they probably didn’t possess a proof, because Babylonians didn’t go much for proofs anyway.

More later,

Nalin Pithwa

### 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