Area of a right angled triangle is an even integer Actual Question is :
Let $ABC$ be a triangle in the plane such that $BC$ is perpendicular to $AC$. Let $a,b,c$ be the lengths of $BC$, $AC$, and $AB$ respectively. Suppose that $a,b,c$ are integers and have no common divisor other than $1$. Which of the following statements are necessarily true? 
a. Either $a$ or $b$ is an even integer.
b. The area of the triangle $ABC$ is an even integer.
c. Either $a$ or $b$ is divisible by $3$. 
With this information, I could see that right angle is at $C$ and $AB=c$, $BC=a$, $AB=c$.
First of all I should check that either $a$ or $b$ is even.
Suppose not, then we would have $(2n+1)^2+(2m+1)^2=c^2$
i.e., $4n^2+4m^2+4n+4m+2=c^2 \Rightarrow 2(2n^2+2m^2+2n+2m+1)=c^2$
i.e., $2$ divides $c^2$ i.e., $2$ divides $c$ i.e., $2^2$ divide $c^2$
i.e., $4$ divides $c^2$ but then, $4n^2+4m^2+4n+4m+2=c^2$ so, going modulo $4$ we see that $\bar{2}\equiv\bar{0} \text{ mod } 4$ which is false.
So, at least one of $a,b$ has to be even integer.
This does not immediately imply that area is even integer (sq.units) because 
$\text {Area} =\frac{1}{2} \cdot a \cdot b$ Suppose $a$ or $b$ has only $1$ as power of $2$ in its prime decomposition, then $2$ in numerator and denominator gets cancelled out which may not give an even integer.
I would be thankful if some one can help me to clear this.
Thank you

For the even area problem, you had proved that one leg is even. But we can do better. We have $a^2+b^2=c^2$. Since $c$ is odd, $c^2\equiv 1\pmod{8}$. 
Let $b$ be odd. Then $b^2\equiv 1\pmod{8}$. So we cannot have $a\equiv 2\pmod{4}$. For if $a\equiv 2\pmod{4}$ then $a^2\equiv 4\pmod{8}$, and therefore $a^2+b^2\equiv 5\pmod{8}$.
Thus $a\equiv 0\pmod{4}$, and therefore the area is even. 
For the divisibility by $3$, show that if neither $a$ nor $b$ is divisible by $3$, then $a^2\equiv 1\pmod{3}$ and $b^2\equiv 1\pmod{3}$. Argue that this is not possible.  
