<s>[INST] 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
 [/INST] 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.  
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