1. ## pythagorean triples

Show that the products of the lengths of the three sides is divisible by 60.

2. $\displaystyle 60=2^2\cdot{3}\cdot{5}$

Suppose $\displaystyle (x, y, z)$ is a pythagorean triple

Then $\displaystyle x^2+y^2=z^2$

Note that $\displaystyle x^2\equiv{0;1}(\bmod. 3)$ , $\displaystyle x^2\equiv{0;1}(\bmod. 4)$ and $\displaystyle x^2\equiv{0;1;-1}(\bmod. 5)$ (these are the options)

Suppose none of them is divisible by $\displaystyle 3$ then$\displaystyle x^2\equiv{1}(\bmod{3})$, $\displaystyle y^2\equiv{1}(\bmod{3})$ and $\displaystyle z^2\equiv{1}(\bmod{3})$ . But this is absurd since $\displaystyle z^2=x^2+y^2\equiv{2}(\bmod{3})$

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

3. ## ugerly proof

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 $\displaystyle a^2+b^2=c^2$ =>

c=$\displaystyle r^2+s^2$
b=2rs
a=$\displaystyle r^2-s^2$=(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 $\displaystyle r \equiv s$ mod 3 then 3|(r-s) => 3|a
This leaves the possibilities $\displaystyle r \equiv 2$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
$\displaystyle r\equiv4$mod 5, $\displaystyle s\equiv 3$ mod 5: r = 5x+4, s = 5y+3. $\displaystyle r^2+s^2$
=$\displaystyle 25x^2+40x+16+25y^2+30y+9$
=$\displaystyle 25x^2+40x+25y^2+30y+25\equiv0$ mod 5, so 5|c
The previous approach can be used on the following possible combinations:
$\displaystyle r\equiv3 mod 5, s\equiv 4$ mod 5
$\displaystyle r\equiv4 mod 5, s\equiv 2$ mod 5
$\displaystyle r\equiv2 mod 5, s\equiv 4$ mod 5
$\displaystyle r\equiv3 mod 5, s\equiv 1$ mod 5
$\displaystyle r\equiv1 mod 5, s\equiv 3$ mod 5

If $\displaystyle r\equiv4 mod 5 s\equiv 1$ mod 5 then $\displaystyle r+s\equiv4+1\equiv0$ mod 5, so 5|a
The previous approach can also be used on the following
$\displaystyle r\equiv1 mod 5, s\equiv$ 4 mod 5
$\displaystyle r\equiv3 mod 5, s\equiv 2$ mod 5
$\displaystyle r\equiv2 mod 5, s\equiv 3$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