"Natural" Triangles

Question

I came across a question related to right triangles with side lengths that are natural numbers:

Q. A "natural" triangle is one where all the side lengths are natural numbers. Prove that a natural right triangle always has an area that is divisible by $6$.

My Attempt:

I labelled the sides as $a,b,$ and $c$ ($c$ is the hypotenuse). We have $a^2+b^2=c^2$. If we consider this equation modulo $4$ we see that atleast one of $a^2$ or $b^2$ must be divisible by $4$. This ensures that the area, which is given by $A=\frac{1}{2}ab$, is also a natural number. Now I considered the equation modulo $3$ and so, atleast one of $a^2$ or $b^2$ must be divisible by $3$. From this, we also get that atleast one of $a$ or $b$ must be divisible by 3. From here, I don't know what to do.

Note that I am aware of the solution that uses the general formula for all primitive pythagorean triplets, but that doesn't seem like something I could've come up with on my own during an exam, so please let me know if there are solutions that don't use that formula.

Answer 1

Addendum added to respond to the comment of Thomas Anton.

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You have solved the problem without realizing it.

Perhaps you were thrown by the significance of the $(1/2)$ factor in the formula $(1/2) \times a \times b$.

While it is necessary to show that one of the two numbers $a,b$ was a multiple of $4$, the problem is completed merely by showing that one of the two numbers $a,b$ is a multiple of $3$.

As you indicated, a simple (mod $3$) argument does the trick, and so you are done.

Therefore, there is no need to reference the general formula for generating pythagorean triples.

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Addendum

Responding to the comment of Thomas Anton.

When in the OP was it shown that one of the two numbers a, b is a multiple of 4? It was shown that one was a multiple of 2.

Nice catch.

Unclear if the following $\pmod{8}$ argument was intended by the OP (i.e. original poster).

Assume, without loss of generality, that
$a = (4r + 2), b = (2s + 1).$

This implies that

$$a^2 + b^2 = 16r^2 + 16r + 4 + 4s^2 + 4s + 1. \tag1 $$

Note that $~4s^2 + 4s = 4s(s+1),~$ which must be a multiple of $8$. Therefore, the expression in (1) above must be congruent to $5 \pmod{8}$.

This implies that $~c^2 \equiv 5\pmod{8},~$ which is impossible.

Therefore, under the assumption that $a$ is even, either $a$ is a multiple of $4$ or $a$ and $b$ are both even.

Answer 2

HINT.-Without Pythagorean triples we have $x^2+y^2=z^2$ with $x,y,z$ coprimes implies $x$ or $y$ is even because if not we get easily the absurd odd equal even.

So let $x=2a,y=2b+1$ and $z=2c+1$ which implies $a^2+b^2+b=c^2+c\Rightarrow a^2=(c-b)(c+b+1)$. It follows $c-b=\alpha^2$ and $c+b+1=\beta^2$ so $2b+1=\beta^2-\alpha^2$ where $a=\alpha\beta$. On the other hand the area is $$A=\dfrac{xy}{2}=a(2b+1)=\alpha\beta(\beta^2-\alpha^2)=\alpha\beta(\beta+\alpha)(\beta-\alpha)$$ which is always divisible by $6$.

Answer 3

Okay, evidently this problem is harder than I thought based on the amount it's been talked about, so I'm just clarifying my hint in the comments, which is pretty much how Peter expanded on it, as well as how I came up with it. So, the OP's strategy was to use modular arithmetic to find the divisibility of $a^2$ and $b^2$. They used this to get divisibility by $3$ and $4$. Problem is, we want divisibility of $a$ and $b$. Divisibility of $a^2$ by $\prod_{i=1}^np_i^{c_i}$ where each $p_i$ is prime and $c_i$ is greater than $0$ is only equivalent to divisibility of $a$ by $\prod_{i=1}^np_i^{\left\lceil\frac{c_i}{2}\right\rceil}$. So, divisibility by $3$, a prime power of index $1$, of $a^2$ is good enough for $a$, but since $4$ factors $2^2$ it only gives us divisibility by $2$. However, going up to $8=2^3$ is the first number that could be enough to guarantee divisibility by $4=2^2$, so I thought checking mod $8$ would be a good idea.

It's a little more complicated, because you can't show one of the numbers must be divisible by $4$, at least with the same methods the OP used. But the quadratic residues are $0$, $1$ and $4$. We have $0+0=0$, $0+1=1$, $0+4=4$, $1+1=2$, $1+4=5$ and $4+4=0$. So if $c^2$ is equal to $1$ or $4$ mod $8$ we're done. The problem is if $c^2$ is equal to $0$ mod $8$, we have the sum $4+4$. There are two easy ways to deal with this, the first is to go, okay this only gives $a$ is divisible by $2$, but the same for $b$, and we only care about divisibility of $ab$. The second is to go, hey, this makes the whole equation divisible by $2$, so if we factor out by the highest power, we get either $a$ or $b$ divided by some power of $2$ is divisible by $4$, so we get divisibility of one by $4$ no matter what.