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Show that the area of a rightangled triangle with all side lengths integers is an
integer divisible by 6.
 one year ago
 one year ago
Show that the area of a rightangled triangle with all side lengths integers is an integer divisible by 6.
 one year ago
 one year ago

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AJMole123Best ResponseYou've already chosen the best response.0
in fact scratch that last read it wrong
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
any contradiction you can show me ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
ofcorse it must be true!! its a CMI ques!! :P
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
here;s what i tied to do.. we know that (m+n)^2 = (mn)^2 +( 2sqrt(mn) )^2 thus lets consider 3 sides of the right triangle as (m+n) , ( mn) , 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)(mn) how do i prove that mn(m+n)(mn) is always divisible by 6 ?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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 :/
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
thats no the case @A.Avinash_Goutham
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
and i didnt even get the case!! @A.Avinash_Goutham :P
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
Because diagonal of all rectangles isnt a integer
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
yea i missed that
 one year ago

AJMole123Best ResponseYou've already chosen the best response.0
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.
 one year ago

AJMole123Best ResponseYou've already chosen the best response.0
you'll get (m^3)n m(n^3) I think. Not that I checked how you got to mn(m+n)(mn)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
that's divisible by 2 ... just have to prove it is also divisible by 3
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
http://openstudy.com/users/mukushla#/updates/501f6952e4b023c05561d562
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
here it goes http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
for divisibility by 3, assume it to be of form 3a+1 or 3a+2 (for not divisible)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
I guess we have to prove that (2k+1)2k(k+1) is always divisible by 12
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
right? @experimentX
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
well ... i think so.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
wait a min,,i think its done..induction will come to our aid here! :D
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
it is not complete yet..
 one year ago

phiBest ResponseYou've already chosen the best response.1
mn(m+n)(mn) 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 mn = 3p3q 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 mn yes that is all the cases.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
Now, when k is multiple of 6 then, (2k+1)2k(k+1) is always divisible by 12
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
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
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
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
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
see this please http://openstudy.com/study#/updates/502d0994e4b03d454ab5d0cc
 one year ago
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