Question

Show that the area of a right-angled triangle with all side lengths integers is an
integer divisible by 6.

Answer 1

not true..

Answer 2

in fact scratch that last read it wrong

Answer 3

any contradiction you can show me ?

Answer 4

I guess its true

Answer 5

ofcorse it must be true!! its a CMI ques!! :P

Answer 6

here;s what i tied to do..

we know that (m+n)^2 = (m-n)^2 +( 2sqrt(mn) )^2
thus lets consider 3 sides of the right triangle as
(m+n) , ( m-n) , 2 sqrt(mn)

i found this absurd a bit so i took m^2 and n^2 in place of m and n i.e.
(m^2 + n^2)^2 = (m^2 - n^2)^2 + (2mn)^2
thus the sides are
m^2 + n^2 , m^2 -n^2 and 2mn
for integral values of m and n,, we get integral sides..
so area of the triangle = 1/2(m^2 - n^2)(2mn) = mn(m+n)(m-n)

how do i prove that mn(m+n)(m-n) is always divisible by 6 ?

Answer 7

i wonder if it's true
if u have a r8 triangle say u shud be able to form a rectangle using 2 r8 triangles
so if area is divisible by 6 (of triangle) area of every rectangle is divsible by 12 :-/

Answer 8

thats no the case @A.Avinash_Goutham

Answer 9

and i didnt even get the case!! @A.Avinash_Goutham :P

Answer 10

Because diagonal of all rectangles isnt a integer

Answer 11

yea i missed that

Answer 12

well provided n != m and n >0 and m > 0. which have to hold
m or n >= 2. Try from there I don't have much time. I came to find an expert on ordinary differential eqns.

Answer 13

you'll get (m^3)n -m(n^3) I think. Not that I checked how you got to mn(m+n)(m-n)

Answer 14

that's divisible by 2 ... just have to prove it is also divisible by 3

Answer 15

yep..

Answer 16

this guy had posted the general solution of a^2+b^2 =c^2
i can't remember ... and can't broswse this Q's either

Answer 17

http://openstudy.com/users/mukushla#/updates/501f6952e4b023c05561d562

Answer 18

here it goes
http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee

Answer 19

the integer solution solution of this
\[ \color{red}{ x^2 + y^2 = z^2} \]
is \[\color{red}{(x, y, z)=(2k+1 \ , \ 2k^2+2k\ , \ 2k^2+2k+1) \ \ k=1,2,3,… } \]
x = 2k + 1
y = 2k(k + 2)

show that (2k+1)k(k+1) is always divisible by 2 and 3

Answer 20

assume k to be odd ... k+1 is divisible by 2
if it is not divisible by 3, 2k+1 is divisible by 3

assume k to be even,
either k+1 or 2k+1 is always divisible by 3

Answer 21

for divisibility by 3, assume it to be of form 3a+1 or 3a+2 (for not divisible)

Answer 22

I guess we have to prove that (2k+1)2k(k+1) is always divisible by 12

Answer 23

right? @experimentX

Answer 24

well ... i think so.

Answer 25

wait a min,,i think its done..induction will come to our aid here! :D

Answer 26

it is not complete yet..

Answer 27

mn(m+n)(m-n) is always divisible by 6 ?

first show divisible by 2:
if m,n even then yes
if either m,n even then yes
if both odd, then (m+n) is even,so yes

now show divisible by 3:

if m= 3p or n= 3q then yes
if m= 3p+1 and n= 3q+1 then m-n = 3p-3q yes
if m= 3p+1 and n= 3q+2 then m+n= 3p+3q+3 yes

if m= 3p+2 and n= 3q+1 then m+n yes
if m= 3p+2 and n= 3q+2 then m-n yes

that is all the cases.

Answer 28

Now, when k is multiple of 6 then,
(2k+1)2k(k+1) is always divisible by 12

Answer 29

when k is even, except the numbers that are divisible by 6 then,
k is divisible by 2
and either (k+1) or (2k+1) will surely be divisible by 3
Thus, (2k+1)2k(k+1) is always divisible by 12

Answer 30

and by induction ,
after proving for k=1
k(k+1)(2k+1) = 6q -->let it be true
for k+1,
(k+1) ( k+2) ( 2k +1 +2)
=(k+1)(k+2)(2k+1) + 2 (k+1)(k+2)
=(k+1)(k)(2k+1) + 2(k+1)(2k+1) + 2(k+1)(k+2)
= 6q + 2(k+1)(2k+1 + k+2) = 6q + 2(k+1)(3)(k+1) = 6 * something
hence proved..
thanks guys..
especially @mukushla

Answer 31

see this please
http://openstudy.com/study#/updates/502d0994e4b03d454ab5d0cc