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
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 $a+b&gt;c$ (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:

$$\begin{align}P(a+b-c)&amp;=(a+b+c)(a+b-c)=(a+b)^2-c^2\\
&amp;=a^2+b^2+2ab-c^2=2ab=4A
\end{align}$$

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 $m&gt;n$.

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).
A=(1/2)(ab)
P=a+b+SQRT(a^2+b^2)
A=k*P implies ab=2ka+2kb+2k(SQRT(a^2+b^2))
(ab-2ka-2kb)^2=(ab-2k(a+b))^2=4k^2(a^2+b^2)
(ab)^2-4kab(a+b)+4k^2(a+b)^2=4k^2(a^2+b^2)
(ab)^2-4kab(a+b)+4k^2(a^2+b^2+2ab)= 4k^2(a^2+b^2)
(ab)(ab-4k(a+b)+8k^2)=0
(ab-4k(a+b)+8k^2)=0
(a-4k)(b-4k)-16k^2+8k^2=0
(a-4k)(b-4k)=8k^2
So we just need the divisors of 8k^2 to solve for a,b and hence c.