<s>[INST] Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means. 
The problem stated is

Does a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ exist? 

(Here P is the perimeter of the triangle. Sum of the sides. And A is the area of the triangle. a*b/2 in a triangle with a b c and c is the hypotenuse)
Obviously for the simple cases such triangles exists. For an example 
when
$ A=2P \qquad $ the triangle $12,16,20$ works
$ A=P  \qquad \; \; $ the triangle $6,8,10$ works
$2A=P  \qquad $ the triangle $3,4,5$ works
I tried solving this by hand, first for the special case where $2P=A$. This ended up giving me 
$$ \frac{a \cdot b}{2} = A $$
and
$$ P = \frac{A}{2} = \frac{a \cdot b}{4} \qquad \text{also} \qquad P = a + b + c  $$ 
So 
$$ \frac{ab}{4} = a + b + c $$ 
By knowing that this is a right angle this leads to the equation (Using the Pythagorean theorem)
$$ a^2 + b^2 = c^2 $$
Now we have two equations and three unknowns, which also needs to be integers! Sadly I was not able to continue from here. I have only learned how to solve linear Diophantine equations. Not a system of nonlinear Diophantine equations. 
Just to restate my question below =) 

Is there a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ ? 

Regards, Werner
 [/INST] We have the following relations:
$$P=a+b+c,\quad 2A=ab,\quad a^2+b^2=c^2$$
In fact, there is one more that is implicit: the triangle inequality $a+b>c$ (which follows from $a^2+b^2=c^2$). We wish to relate the first and second relations, and so after staring at the problem for a while we realize the correct course of action is to multiply $P$ by $a+b-c$ (the latter guaranteed to be positive by the triangle inequality). This gives:
$$\begin{align}P(a+b-c)&=(a+b+c)(a+b-c)=(a+b)^2-c^2\\
&=a^2+b^2+2ab-c^2=2ab=4A
\end{align}$$
Hence in general for a right triangle we have $k=A/P=(a+b-c)/4$, so the question boils down to when the difference between the sum of the lengths of the legs and the length of the hypothenuse divides or is divisible by $4$, for a right triangle with integer side-lengths.
Here we actually need to generate pythagorean triples $(a,b,c)$ such that $a,b,c$ are integers and $a^2+b^2=c^2$; the wikipedia page here gives a good picture of Euclid's formula that I recommend thinking through. Euclid's formula is the following:
$$a=m^2-n^2,\quad b=2mn\quad c=m^2+n^2$$ for integers $m>n$.
Then the quantity $a+b-c$ is $m^2-n^2+2mn-m^2-n^2=2n(m-n)$, so $A/P=k=n(m-n)/2$. The options are now that either $2$ divides $n$ or $2$ divides $m-n$, so we either have $n=2d$ and $m=k/d+2d$ or $n=d$ and $m=2(k/d)+d$ for divisors $d$ of $k$.
(to get $P/A=k$ we need $2n(m-n)$ to divide $4$, so $n(m-n)$ must divide $2$, thus $n=1$, $m=2$, $n=1$, $m=3$, and $n=2$, $m=3$ are the only solutions, giving us $(3,4,5)$, $(8,6,10)$, $(5,12,13)$ with $k=2,1,2$ being the only possibilities).
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