I have to solve the equation $$\frac{n_1(3n_1-1)}{2}=\frac{n_2(3n_2-1)}{2}$$ and find out the couple $(x;y)\in\Bbb N^2$ for wich $|x-y|$ is minimised with $n_1≠n_2$. What I've thought so far is that when we write $ab = cd$, at least, $a$ or $b$ should divide $c$ or $d$, or, $c$ or $c$ or $d$ should divide $a$ or $b$. I know that's a bit confusing. By giving an example, we can write $$12=4.3=2.6$$
Here, $2|3$ and $3|6$. But I don't know if this property is always thrue or useful in this situation with $n_1(3n_1-1)=n_2(3n_2-1)$. How can I solve this equation?
Use $x$ and $y$ instead of subscripts. If the equality holds then $3x^2-3y^2-(x-y)=0$, or equivalently $$(x-y)\left(3(x+y)-1\right)=0.$$
There are no integers $x$ and $y$ such that $3(x+y)-1=0$. For if $3(x-y)-1=0$, then $3(x+y)=1$. But $3$ does not divide $1$.
Thus if $(x,y)$ is a solution of the equation in integers, then $x=y$.