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shubhamsrg

  • 3 years ago

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

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  1. AJMole123
    • 3 years ago
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    not true..

  2. AJMole123
    • 3 years ago
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    in fact scratch that last read it wrong

  3. shubhamsrg
    • 3 years ago
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    any contradiction you can show me ?

  4. sauravshakya
    • 3 years ago
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    I guess its true

  5. shubhamsrg
    • 3 years ago
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    ofcorse it must be true!! its a CMI ques!! :P

  6. shubhamsrg
    • 3 years ago
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    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 ?

  7. A.Avinash_Goutham
    • 3 years ago
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    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 :-/

  8. sauravshakya
    • 3 years ago
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    thats no the case @A.Avinash_Goutham

  9. shubhamsrg
    • 3 years ago
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    and i didnt even get the case!! @A.Avinash_Goutham :P

  10. sauravshakya
    • 3 years ago
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    Because diagonal of all rectangles isnt a integer

  11. A.Avinash_Goutham
    • 3 years ago
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    yea i missed that

  12. AJMole123
    • 3 years ago
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    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.

  13. AJMole123
    • 3 years ago
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    you'll get (m^3)n -m(n^3) I think. Not that I checked how you got to mn(m+n)(m-n)

  14. experimentX
    • 3 years ago
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    that's divisible by 2 ... just have to prove it is also divisible by 3

  15. shubhamsrg
    • 3 years ago
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    yep..

  16. experimentX
    • 3 years ago
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    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

  17. experimentX
    • 3 years ago
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    http://openstudy.com/users/mukushla#/updates/501f6952e4b023c05561d562

  18. experimentX
    • 3 years ago
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    here it goes http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee

  19. experimentX
    • 3 years ago
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    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

  20. experimentX
    • 3 years ago
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    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

  21. experimentX
    • 3 years ago
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    for divisibility by 3, assume it to be of form 3a+1 or 3a+2 (for not divisible)

  22. sauravshakya
    • 3 years ago
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    I guess we have to prove that (2k+1)2k(k+1) is always divisible by 12

  23. sauravshakya
    • 3 years ago
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    right? @experimentX

  24. experimentX
    • 3 years ago
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    well ... i think so.

  25. shubhamsrg
    • 3 years ago
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    wait a min,,i think its done..induction will come to our aid here! :D

  26. mukushla
    • 3 years ago
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    it is not complete yet..

  27. phi
    • 3 years ago
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    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.

  28. sauravshakya
    • 3 years ago
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    Now, when k is multiple of 6 then, (2k+1)2k(k+1) is always divisible by 12

  29. sauravshakya
    • 3 years ago
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    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

  30. shubhamsrg
    • 3 years ago
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    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

  31. shubhamsrg
    • 3 years ago
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    see this please http://openstudy.com/study#/updates/502d0994e4b03d454ab5d0cc

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