pythagorean triples

**frankdent1** · December 6th 2007, 10:33 PM

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

**PaulRS** · December 7th 2007, 01:41 AM

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

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

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

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

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

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

**badgerigar** · December 7th 2007, 02:58 AM

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

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

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

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