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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

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@j.w.Perry : Thank you for the edit. :) – Praphulla Koushik Nov 26 '13 at 7:02
Praphulla Koushik: Thanks for the excellent well researched post. :)) – J. W. Perry Nov 26 '13 at 7:03
@labbhattacharjee : That was helpful.. Thank you :) – Praphulla Koushik Nov 26 '13 at 11:55
up vote 2 down vote accepted

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.

share|cite|improve this answer
I am not sure if $c$ is odd :( I have to check it... :( – Praphulla Koushik Nov 26 '13 at 6:41
It is. For if $c$ is even, then $c^2\equiv 0\pmod{4}$. But that would force $a$ and $b$ to be both even, impossible since the gcd of $a,b,c$ is $1$. – André Nicolas Nov 26 '13 at 6:43
yes yes :) I was just about to write the same.. thank you :) – Praphulla Koushik Nov 26 '13 at 6:44
Thank you... I got it now! :) – Praphulla Koushik Nov 26 '13 at 6:45
You are welcome. You can also use the fact that all primitive Pythagorean triples are given by $(s^2-t^2, 2st, s^2+t^2)$ where $s$ and $t$ are relatively prime, of opposite parity, with $s\gt t$. That's a good thing to know in general, but your problem can be solved with less machinery, as you have seen. – André Nicolas Nov 26 '13 at 6:55

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