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As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?

$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$

$(2n)^2+(n^2-1)^2=(n^2+1)^2$ $(2n)^2+(n^2-1)^2=(n^2+1)^2$ $(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2*0)^2+(0^2-1)^2=(0^2+1)^2$ $(2*1)^2+(1^2-1)^2=(1^2+1)^2$ $(2*2)^2+(2^2-1)^2=(2^2+1)^2$
$(2*0)^2+(0-1)^2=(0+1)^2$ $(2*1)^2+(1-1)^2=(1+1)^2$ $(2*2)^2+(4-1)^2=(4+1)^2$
$0^2+1^2=1^2$ $2^2+0^2=2^2$ $4^2+3^2=5^2$
$0+1=1$ $4+0=4$ $16+9=25$
$1=1$ $4=4$ $25=25$
YGranja
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    Not a visual proof, hence I will leave it as a comment: Just expand both side of the equation to see that $4m^2+m^4-2m^2+1=m^4+2m^2+1=(m^2+1)^2$. – F. Conrad Sep 15 '22 at 11:22
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    @F.Conrad Is it possible to derive this formula from $(a+b)^2$, an old proof consisting of 4 right triangles inside a big square? – YGranja Sep 15 '22 at 11:28
  • That is actually a good idea. You got can get rectangles with (I donote $area=a\cdot b$, one side times the other) respective sizes $2m \cdot 2m$, $m^2 \cdot m^2$ and so on, one corresponding to each term in the equation. Then you can possible rerrange them in a nice way. The minus sign on the left side corresponds to splitting the rectangle $2m \cdot 2m$ in half. – F. Conrad Sep 15 '22 at 11:52

5 Answers5

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Here's a visual proof:

enter image description here

(This space intentionally left blank.)

Blue
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We have the identity $a^{2}- b^{2}= \left ( a- b \right )\left ( a+ b \right )\!,$ so $$\left ( m^{2}+ 1 \right )^{2}- \left ( m^{2}- 1 \right )^{2}\!=\!2m^{2}\cdot 2= \left ( 2m \right )^{2}$$

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The formula that you are referring is a sub-case of the Euclid's formula
Accroding to the Euclid's formula it is true that: $$\text{Given an arbitrary pair of integers m and n with m$\gt$n and m,n$\gt$0}$$ $$a=m^2-n^2 \text{ , } b=2mn \text{ , } c=m^2+n^2 \text{ form a pythagorean triple}$$ $\text{In your case for:} n=1 \text{ and } m\epsilon\mathbb{N} $

You can found more about this forumla in this links:
proof of euclid's formula MathExchange
Pythagorean triple-wikipedia

Constantinos Pisimisis
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This formula can be derived in various ways, and with connections to different areas of mathematics. One way of seeing why it generates all the triples is to note that the unit circle $$x^2+y^2=1$$ in polar co-ordinates has $y=\sin \theta, x=\cos \theta$ so this takes in all the points of the circle (for angles $-\pi\le \theta \lt \pi$ it traverses the circle once).

Now, with $t = \tan \frac {\theta}2$ we have the half angle formulae $y=\dfrac {2t}{1+t^2}, x=\dfrac {1-t^2}{1+t^2}$ which parametrise the unit circle with rational functions (the point at $x=-1, y=0$ corresponds to the singularity in the tangent function).

If we substitute back into the original equation for the circle, and multiply through by $(1+t^2)^2$ we get a pythagorean relationship $$(1-t^2)^2+(2t)^2 = (2t)^2+(t^2-1)^2=(t^2+1)^2$$

We can do this for every rational point on the unit circle, and for each integer $t$ we recover such a rational point.

There are "more elementary" ways of showing how pythagorean triples work, but this is worth noting as an example of the rational parametrisation of a curve - a topic which has its own theory, generalising the case of the circle. You can draw a diagram to show how a pythagorean triangle with integer sides corresponds to a rational point on the unit circle.

Mark Bennet
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  • You mentioned "more elementary" ways of showing this. Where can I find such proofs? – YGranja Sep 15 '22 at 11:58
  • @YGranja Hardy and Wright "Introduction to the Theory of Numbers" does this. But basically you eliminate common factors and note that one of the squares must be even in this case. Rearrange to $c^2-a^2=b^2$ with $a,c$ odd and $b$ even. Then $c-a$ and $c+a$ have difference 2a and common factor 2 (they are both even and have no greater common factor). So $c-a=2n^2$ and $c+a=2m^2$ whence $c=m^2+n^2, a =m^2-n^2$. Your case is with $n=1$. – Mark Bennet Sep 15 '22 at 12:49
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All Pythagorean triple have a "leg" that is a multiple of $\space4.\space$ Your formula generates a subset of these.

We begin with Euclid's formula shown here as $$A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$$ If we let $\space k=1,\space$ we have $\quad A=m^2-1\quad B=2m\quad C=m^2+1$

This formula also generates trivial triples where one term is zero. A variation of this replaces $\space m\space$ with $\space(2n-1+k)=2n\space$ and produces a formula that generates non-trivial triples for any natural number $n$. $$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1$$

By inspection, we can see that the middle term will always be even and that $\space C-A=2\space$ for all values of $\space n.\space$ Algebraically below, we can also "see" that the formula is valid.

\begin{align*}A^2+B^2=&(4n^2-1)^2+(4n)^2\\ =&(16 n^4 - 8 n^2 + 1)+(16n^2)\\ =&(16 n^4 + 8 n^2 + 1)\\ =&(4n^2+1)^2=C^2\\ \end{align*}

poetasis
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