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

- shubhamsrg

- katieb

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- anonymous

not true..

- anonymous

in fact scratch that last read it wrong

- shubhamsrg

any contradiction you can show me ?

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## More answers

- anonymous

I guess its true

- shubhamsrg

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

- shubhamsrg

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 ?

- anonymous

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 :-/

- anonymous

thats no the case @A.Avinash_Goutham

- shubhamsrg

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

- anonymous

Because diagonal of all rectangles isnt a integer

- anonymous

yea i missed that

- anonymous

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.

- anonymous

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

- experimentX

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

- shubhamsrg

yep..

- experimentX

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

- experimentX

- experimentX

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

- experimentX

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

- experimentX

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

- anonymous

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

- anonymous

right? @experimentX

- experimentX

well ... i think so.

- shubhamsrg

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

- anonymous

it is not complete yet..

- phi

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.

- anonymous

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

- anonymous

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

- shubhamsrg

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

- shubhamsrg

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

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