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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.

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  • $\begingroup$ The solutions seem to be of the form $[m,n,k,x,y] = [a,a,4a+2,3a+1,3a+1]$ for $a$ a natural number and $a \geq 2$. $\endgroup$ – Dirk Aug 2 '17 at 8:57
  • $\begingroup$ It seems that you are right, I just want to show that either $m = n$, or $x = y$ $\endgroup$ – DavitS Aug 2 '17 at 9:07
  • $\begingroup$ You might use this to transform your system into another system with variables, say, $a,b,c,d,e$, such that $a=b=c=d=e \geq 2$ are the only solutions of your new system. Maybe then showing that all unknowns in the new system have to be the same is easier than the problem you face now (just a thought, might or might not work). $\endgroup$ – Dirk Aug 2 '17 at 9:12
  • $\begingroup$ I also can deduce that if I show that $x + y$ is not divisible by $3$, i.e. $x + y != 0 mod 3$ $\endgroup$ – DavitS Aug 2 '17 at 9:18
  • $\begingroup$ If k=0 is allowed it can not be shown that x=y or m=n has to hold. consider for example k=0 then m=x and n=y and all natural numbers are allowed for either of these pairs $\endgroup$ – zen Aug 2 '17 at 9:25
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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 $x=y$.

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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.

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  • $\begingroup$ It isn't clear to me just what you claim to have proved. The original equations do have solutions over the integers. $\endgroup$ – Brendan McKay Aug 9 '17 at 12:50
  • $\begingroup$ @BrendanMcKay Has no solutions for $t=1$. I changed the task. Instead of solving the system. The task is reduced to finding solutions of linear equations. $\endgroup$ – individ Aug 9 '17 at 13:03
  • $\begingroup$ $k=3,m=0,n=1,x=3,y=1$ and infinitely many other solutions. You need to impose the positivity condition to eliminate the solutions. $\endgroup$ – Brendan McKay Aug 16 '17 at 10:56
  • $\begingroup$ @BrendanMcKay Well, Yes. Of course. I didn't notice that... $$t=(4a^2-2ab+b^2)(3s+(b-a)p)$$ Then other numbers to divide by a common factor. $\endgroup$ – individ Aug 16 '17 at 14:11

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