2011/035) Prove that if the sides of a triangle are prime numbers its surface can not be whole number.

proof:
Let the sides are a,b c, and a <=b <= c

now there are 2 cases

a = 2 or all are odd

now A^2 = ( a+b-c)(a+b+c)(a-b+c)(b+c-a)/ 4
if all are odd then all 4 terms on the RHS are odd then A^2 cannot be integer so A cannot be whole number

case 2:
for a= 2 and b = 2 or a= 2 and b != 2

a = 2 then b =2 => c = 2 or 3

a =2 b = 2 c = 2 => A^2 = 6*2^3/4 = 12 so A is not integer

a =2 , b= 2 c = 3 => A^2 = 7 * 1 * 3 * 3/4 so A is not integer


if b != 2 then b= c because if c >b then c>= b+2 or a+b= c

so we get A^2 = (2+2b)* (2b-2)* b^2 /4 = b^2(b^2-1)/4 cannot be a perfect square


so no solution