Show that the products of the lengths of the three sides is divisible by 60.
Suppose is a pythagorean triple
Then
Note that , and (these are the options)
Suppose none of them is divisible by then , and . But this is absurd since
Thus at least one must be divisible by 3.
Go over the other cases working in this way and you will get that at least one is divisible by 4 and 5, so the product is divisble by 60
I will prove that at least one of a, b and c must be a multiple of 4,5 and 3 and so their product must be a multiple of 60
recall that =>
c=
b=2rs
a= =(r+s)(r-s)
for some natural r,s
4.
suppose that a, b and c have no common factor. Since b is even, either c or a must be odd. Since b^2 is also even, a^2 is odd iff c^2 is odd therefore both a and c are odd. Since c = r^2+s^2, one of r and s is odd and the other even. So 4|b since b=2rs.
Since every triple is a multiple of a triple with no common factor, 4|b in general.
3.
if 3|r or 3|s then 3|b
if mod 3 then 3|(r-s) => 3|a
This leaves the possibilities mod 3 and s\equiv1 mod 3 or vica versa. Either way, 3|(r+s) => 3|a
5.
if 5|r or 5|s then 5|b
if r \equiv s mod 5 then 5|(r-s) => 5|a
mod 5, mod 5: r = 5x+4, s = 5y+3.
=
= mod 5, so 5|c
The previous approach can be used on the following possible combinations:
mod 5
mod 5
mod 5
mod 5
mod 5
If mod 5 then mod 5, so 5|a
The previous approach can also be used on the following
4 mod 5
mod 5
mod 5
These cover all possible combinations of equivalences modulo 5
thus, the product abc will have factors of 4, 5 and 3 and so 60|abc
Edit: how could I possibly have missed seeing PaulRS's post? :s