# Proof of Pythagorean theorem without using geometry for a high school student?

There are some proofs of Pythagoras theorem which don't even require high school maths to understand, but they all are using shapes to prove of the theorem. However, I am trying to find some proofs of Pythagoras theorem that don't use shapes in their proofs, for example a purely algebraic proof. Besides, they would still be easy enough to teach for students with high school level of knowledge in mathematics.

My question: Prove that for $x^2+y^2=z^2$ there are infinity many triples in $\mathbb Z^+$ fit into the equation? - which requires high school maths knowledge to understand.

• You're trying to prove a geometrical theorem without using geometry? – user117644 Aug 2 '15 at 10:33
• How do you define a right-angled triangle in an analytic/algebraic way? – peterwhy Aug 2 '15 at 10:35
• Let $F$ be a field of characteristic $0$, let $V$ be a vectors space over $F$ with inner product $\langle,\rangle$. Then for $a,b\in V$ with $\langle a,b\rangle = 0$ we immediately verify $\langle a-b,a-b\rangle=\langle a,a\rangle+\langle b,b\rangle$. -- Seriously, such is hardly enlightening and can perhaps even be considered circular. Personally, I like the geometric proof with the square of side length $a+b$ best. – Hagen von Eitzen Aug 2 '15 at 10:43
• @LuckyGuy How is your reply to perterwhy's comment a reply to peterwhy's comment? – Hagen von Eitzen Aug 2 '15 at 10:45
• So do you want to prove Pythagoras theorem or that you have infinite many solutions for $x^2+y^2=z^2$? Because those are two different things. – Hirshy Aug 2 '15 at 10:50

Take triplets $(t^2-1)^2+(2t)^2=(t^2+1)^2$,

in which $t\ge 2$ in order that $x,y,z\in \mathbb Z^+$

The question:

Prove that there are infinitely many solutions in positive integers to $x^2+y^2=z^2$

is not Pythagoras' theorem, because Pythagoras' theorem is a theorem about side lengths of right-angled triangles. Consequently, it doesn't make sense to talk about a non-geometrical proof: it's a statement about geometry.

If you want to prove your statement, it's quite easy:

For any pair of integers $r>s$, let

\begin{align} x&=r^2-s^2\\ y&=2rs\\ z&=r^2+s^2 \end{align}

Then $x,y,z$ are all positive integers, and you can check for yourself that $x^2+y^2=z^2$. It should be easy to convince yourself that this yields infinitely many distinct solutions.

But this statement is unrelated to Pythagoras' theorem. If you know in advance that there are infinitely many non-congruent right angled triangles with integer side lengths then you can prove your statement. But why should that be the case?

Here's a way of thinking about it. To be clear however, this does not prove the Pythagorean theorem.

Given two 2-dimensional vectors $\mathbb{x},\mathbb{y} \in \Bbb R^2,$ with entries $\mathbb{x} = (x_1,x_2)$ and $\mathbb{y} = (y_1,y_2),$ we define a norm as, $$|\mathbb{x}| = \sqrt{x_1^2 + x_2^2}$$ We say two vectors are orthogonal if, $$x_1y_1 + x_2y_2 = 0.$$

Theorem If $\mathbb{x}, \mathbb{y} \in \Bbb R^2$ are orthogonal, then $$|\mathbb{x} + \mathbb{y}|^2 = |\mathbb{x}|^2 + |\mathbb{y}|^2$$

Proof: We have, \begin{align*} |\mathbb{x} + \mathbb{y}|^2 &= | (x_1+y_1,x_2+y_2)|^2 \\ &= \left( \sqrt{ (x_1+y_1)^2 + (x_2+y_2)^2 } \right)^2 \\ &= (x_1^2 + 2x_1y_1 + y_2^2) + (x_2^2 + 2x_2y_2 + y_2^2) \\ &= (x_1^2 + x_2^2) + 2 ( x_1y_1 + x_2y_2 ) + (y_1^2 + y_2^2) \\ &= |\mathbb{x}|^2 + |\mathbb{y}|^2 \end{align*} As required. Note that the simplification $(\sqrt{x})^2 = x$ only holds if $x\geq 0,$ which holds since for any real number $x,$ $x^2 \geq 0.$

The geometric interpretation of the result is that if two vectors are orthogonal, then they are perpendicular (so their angles form a right angle). The norm corresponds to the length of the vector and $|\mathbb{x}+\mathbb{y}|$ corresponds to the length of the vector obtained by adding the two together - so the hypotenuse of the triangle formed by the two vectors.

It must be emphasized however that this does not prove the Pythagorean theorem. Rather, I have just proved an result about 2-dimensional real vectors and gave a geometrical interpretation. To complete this proof, you would need to prove this correspondence, which requires geometry.

On a side-note, this result holds for a large class of spaces, known as inner product spaces using a suitable norm and definition for orthogonality.

You could try to use something like this, but as you'll see, it would be a circular argument, because in the derivation of these results the Pythagorean theorem is already used. You could just Define arc-length to be the linked expression, but then I think you'll loose your high school students.