How to deduce an equation from this 3 Diophantine equations with 5 variables? I have three equations: 
${m \choose 2} + nk = {x \choose 2}$
${n \choose 2} + mk = {y \choose 2}$
$x + y = m + n + k$
$m, n, k, x, y$ are natural numbers. I want to deduce from this 3 equations either $x = y$ or $m = n$. From where I got these equations makes me sure that this is only possible if $x = y$ and $m = n$. Just deducing either $x=y$ or $m=n$ is enough. 
I can show that if I show that $x + y$ is not divisible by 3. So it will be enough if we can show that $x + y$ is not divisible by 3.
 A: The first two equalities imply $x>m$ and $y>n$ so one can substitute $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$:
${X \choose 2}=nX+nY-mX\tag{1}$
${Y \choose 2}=mX+mY-nY\tag{2}$
From (1) follows: $\quad m=n+n\frac{Y}{X}-\frac{1}{X}{X \choose 2}$,
then eliminate $m$ from (2): $\quad {Y \choose 2}+{X \choose 2}+\frac{Y}{X}{X \choose 2}=nX+nY+n\frac{Y^2}{X}$,
and finally: $\quad 2n=X-\frac{X^2+2XY}{X^2+XY+Y^2}$.
Clearly $X-2n=\frac{X^2+2XY+0Y^2}{X^2+XY+Y^2}\in(0,2)$, but since it is an integer it can only be $1$.
This proves that $X=2n+1$ and by symmetry $Y=2m+1$.
Substituting for $X$ and $Y$ in (1) or (2) finally yields $m=n$, so $X=Y$ and $x=y$.
A: Rewrite the system of equations.
$$\left\{\begin{aligned}&m(m-t)+2n(x+y-m-n)=x(x-t)\\&n(n-t)+2m(x+y-m-n)=y(y-t)\end{aligned}\right.$$
The solution of this system can be written as.
$$n=(-8a^3+12ba^2-6ab^2+2b^3)p+(4a^2-5ab+b^2)s$$
$$m=(8a^3+2ba^2-ab^2)p+(4a^2-5ab+b^2)s$$
$$t=(-8a^3+12ba^2-6ab^2+2b^3)p+(12a^2-6ab+3b^2)s$$
$$x=(10ba^2-5ab^2+4b^2)p-3(a-b)bs$$
$$y=(8a^3+8ba^2+2ab^2)p+3a(4a-b)s$$
And you can see that the linear Diophantine equation  $t=1$ Has no solutions in integers.
