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Pythagorean Identity

This year we will start a blog series about mathematical identities. Let us begin today with the Pythagorean Identity $$a^2 + b^2 = c^2$$

In mathematics, the Pythagorean theorem is probably the most well known theorem. Undoubtedly, anyone who has studied geometry would remember this famous theorem.

The Pythagorean theorem asserts that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

 Pythagorean theorem: $a^2 + b^2 = c^2$.

At school, we learn about the Pythagorean theorem around year 8. After introducing the theorem, most teachers would probably use the following example:

That is why most of us would remember the Pythagorean identity: $$3^2 + 4^2 = 5^2$$ $$9 + 16 = 25$$

Me and a classmate of mine once challenged one another to write down as many as possible the Pythagorean triples like $$3^2 + 4^2 = 5^2,$$ in 5 minutes, whoever got the most identities would win.

Both of us finished the writing in much less than 5 minutes and looking at one another, each of us was sure that we would win.

We had a good laugh when the papers were opened, we both got the same idea $$30^2 + 40^2 = 50^2$$ $$300^2 + 400^2 = 500^2$$ $$3000^2 + 4000^2 = 5000^2$$ $$\vdots$$

So we changed the rule. The new rule was that the identities of this type $$(3n)^2 + (4n)^2 = (5n)^2$$ were forbidden. With this new rule, the game was harder!

Now let us pause here for 5 minutes, would you like to have a try to see how many Pythagorean equations you can come up with?!

Let us now get back to the Pythagorean equation $$a^2 + b^2 = c^2$$ how can we find $a$, $b$ and $c$ ?!

Looking at  $$3^2 + 4^2 = 5^2$$ the first thing we may try is to check if $4^2 + 5^2$ is equal to $6^2$ or not? $$4^2 + 5^2 = 16 + 25 = 41$$ $$6^2 = 36$$ So they are not equal $$4^2 + 5^2 \neq 6^2$$
(4, 5, 6) is not a Pythagorean triple.

Now, don't lose heart! We can try $b=n$ and $c = n+1$ and see what the value of $a$ would be.

We have $$a^2 + n^2 = (n+1)^2 = n^2 + 2n + 1$$ so $$a^2 = 2n + 1$$
Ah-hah!!!, $a$ is an odd number, $a = 2k + 1$, and we have
$$a^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2n +1$$
So $n = 2k^2 + 2k$.

With this, we have found general formula for $a$, $b$, $c$: $$a = 2k + 1$$ $$b = n = 2k^2 + 2k$$ $$c = n+1 = 2k^2 + 2k + 1$$ and we obtain the Pythagorean identity $$(2k + 1)^2 + (2k^2 + 2k)^2 = (2k^2 + 2k + 1)^2$$
With $k = 1$ we have the familiar $$3^2 + 4^2 = 5^2$$ With $k=2$, we have $$5^2 + 12^2 = 13^2$$ $$25 + 144 = 169$$ And with $k=3$, we get $$7^2 + 24^2 = 25^2$$ $$49 + 576 = 625$$
Fantastic !!!

Let us stop here for now. Next time we will learn more mathematical identities.

In the meantime, you can read about the Pythagorean triples here. The Pythagorean theorem is proved here and here.

Hope to see you again soon.

Homework.

1. Find other ways to generate Pythagorean identities.