Mathematical Side Show - 4

Soroban · #1 $Old$ March 26th, 2009, 08:16 AM

Mathematical Side Show - 4

Complex Numbers and Pythagorean Triangles

The square of any complex number
yields the legs of a Pythagorean triangle.

$(4+i)^2 \:=\:15 + 8i \quad\Rightarrow\quad 15^2 + 8^2 \:=\:17^2$
$(5 - 2i)^2 \:=\:21 - 20i \quad\Rightarrow\quad 21^2 + 20^2 \:=\:29^2$

This is true because: . $(a+bi)^2 \:=\:(a^2-b^2) + 2abi$
and: . $(a^2-b^2)^2 + (2ab)^2 \:=\:(a^2 + b^2)^2$

Two squares can be found whose sum is an $n^{th}$ power.

$(2+i)^3 \:=\:2 + 11i \quad\Rightarrow\quad 2^2 + 11^2 \:=\:5^3$
$(3+2i)^4 \:=\:\text{-}119 - 120i \quad\Rightarrow\quad 119^2 + 120^2 \:=\:13^4$

Pythagorus extended

There are quadruples of the form: . $a^2 + b^2 + c^2 \:=\:d^2$

$a,b,c$ are dimensions of a box with integral diagonals.

The generating formulas are:
$a \;=\;p^2+q^2-r^2$
$b \;=\;2pr$
$c \:=\:2qr$
$d \:=\:p^2+r^2+r^2$

Some examples are: . $(1,2,2,3),\;(2,3,6,7),\;(1,4,8,9)$

This is used to create two points whose distance is an integer,
or a vector whose magnitude is an integer (in 3-D).

Multigrades

This remarkable equation:

$0^n + 5^n + 6^n + 16^n + 17^n + 22^n \:=\:1^n + 2^n + 10^n + 12^n + 20^n + 21^n$

is true for $n \,=\,1,2,3,4,5$