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

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not true..
in fact scratch that last read it wrong
any contradiction you can show me ?

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I guess its true
ofcorse it must be true!! its a CMI ques!! :P
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 ?
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 :-/
thats no the case @A.Avinash_Goutham
and i didnt even get the case!! @A.Avinash_Goutham :P
Because diagonal of all rectangles isnt a integer
yea i missed that
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.
you'll get (m^3)n -m(n^3) I think. Not that I checked how you got to mn(m+n)(m-n)
that's divisible by 2 ... just have to prove it is also divisible by 3
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
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
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
for divisibility by 3, assume it to be of form 3a+1 or 3a+2 (for not divisible)
I guess we have to prove that (2k+1)2k(k+1) is always divisible by 12
well ... i think so.
wait a min,,i think its done..induction will come to our aid here! :D
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.
Now, when k is multiple of 6 then, (2k+1)2k(k+1) is always divisible by 12
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
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

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