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Right angled Triangles will all side integers are called Pythagorean triangles… now using m,n rule
Let the two sides of triangle be m2-n2 and 2mn
Then hypotenuse= m2+n2, Where m , n are whole numbers
It is easy to see that m = n + 1.
The m,n formula in this case gives
a = m2 – n2 = (n+1)2 – n2 = 2 n + 1
b = 2 m n = 2 (n+1) n = 2 n2 + 2 n
h = m2 + n2 = (n+1)2 + n2 = 2 n2 + 2 n + 1
Area of triangle = ½ a x b
= ½ (2 n + 1) x (2 n2 + 2 n)
= ½ (2 n + 1) x 2n (n + 1)
= n (2 n2 + 3n + 1)
Which is always divisible by 6
As usual, assume for positive integers m,n the lengths of the side of the right angled triangles are (m^2-n^2), 2mn, (m^2+n^2) , with (m^2+n^2) being the hypotenuse. We have to prove that for all integers m,n (m>n)
area of the triangle S(m,n)=mn(m+n)(m-n) is divisible by 6.
First we prove that S(m,n) is divisible by 2
if either m or n is even then it holds naturally
else
if both m and n are odd then both (m+n) and (m-n) are even
Hence S(m,n) is divisible by 2. .... (1)
Next we show that S(m,n) is is divisible by 3
if either m or n is divisible by 3 then it holds naturally.
else if m= 3k + 2 and n= 3l + 1
or, m = 3k+ 1 and n = 3l +2 (for some integers k,l) then m+n is divisible by 3
on the other hand if m= 3k+1 and n =3l +1 or,
m= 3k+2 and n= 3l+2 (for some integers k,l) then m-n is divisible by 3
this exhausts all the possibilities
Hence S(m,n) is divisible by 3 also .... (2)
Combining (1) and (2) we conclude that the area of the triangle is divisible by 6. (Proved)
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