## shubhamsrg 3 years ago Show that the area of a right-angled triangle with all side lengths integers is an integer divisible by 6.

1. AJMole123

not true..

2. AJMole123

in fact scratch that last read it wrong

3. shubhamsrg

any contradiction you can show me ?

4. sauravshakya

I guess its true

5. shubhamsrg

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

6. 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 ?

7. A.Avinash_Goutham

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

thats no the case @A.Avinash_Goutham

9. shubhamsrg

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

10. sauravshakya

Because diagonal of all rectangles isnt a integer

11. A.Avinash_Goutham

yea i missed that

12. AJMole123

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

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

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

15. shubhamsrg

yep..

16. 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

17. experimentX
18. experimentX
19. 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

20. 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

21. experimentX

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

22. sauravshakya

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

23. sauravshakya

right? @experimentX

24. experimentX

well ... i think so.

25. shubhamsrg

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

26. mukushla

it is not complete yet..

27. 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.

28. sauravshakya

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

29. sauravshakya

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

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