I didn't get the question.. does the area divide 12? in that case if the lengths are say (6,8,10) we have area 24.. if you mean 12 divides the area take (3,4,5) then we have area 6..

There is a result which says that 6 always divides the area.. Consider a^2 + b^2 =c^2 if (a,b) are both even 2 divides the area.. if not let wlog a be odd.. So a is 1 or 9 mod 16.. If b is not 0 mod 16 then c^2 becomes 2 or 10 or 5 or 14 or 10 or 13 mod 16.. which is impossible as the quadratic residues of 16 are 0,1,4,9.. Hence b is 0 mod 16.. Hence the area is divisible by 2. Quadratic residues of 3 are 0,1..hence a and b cant both be non-divisible by 3. Hence area is divisible by 6.. Moreover it can be shown that 5 divides a*b*c.. To show this follow the same technique of inspecting the quadratic residues and showing a contradiction if none is divisible by 5..

if a^2 + b^2 = c^2 then, a = p (m^2 - n^2) b = 2p mn c = p (m^2 + n^2) p = (a,b) 2|mn now we can found a*b = 2*p^2*m*n*(m-n)*(m+n). 3 | n(m-n)m(m+n) 2 | mn so 12|ab

If both of a,b odd then c^2=2 mod 4 not possible If both of a,b not divisible by 3 then c^2=2 mod 3 not possible .so 6 divides ab.

$if(a,b,c)=(3,4,5)$ Then $ar(ABC)=3*4/2=6$ $2*6=12|12$ example

Solution: By Euclid's formula of the Pythagorean triplet, we have that, $$p=m^2-n^2,b=2mn, h=m^2+n^2$$We see that, $$2|b, \textbf{ If } gcd(m,n) \not = 3k, \texttt{ then } 3|m^2-n^2, \textbf{ otherwise if } gcd(m,n)=3k \implies 3|p,b$$Hence, the area $A=\frac{1}{2}p.b$ when doubled is always multiple of $12$.

mod3,mod4

ck204 wrote: I didn't get the question.. does the area divide 12? in that case if the lengths are say (6,8,10) we have area 24.. if you mean 12 divides the area take (3,4,5) then we have area 6..