Multiple of a Triangular Number is another Triangular Number I need to know if there are any necessary and/or sufficient condition(s) on the natural numbers $m$ and $n$ such that

$2T_n=T_{n+m}$, where $T_a$ is the $a$-th triangular number.


I have been able only to apply the formula $T_{a+b}=T_a+T_b+ab$ to the RHS to get $2T_n=T_n+T_m+mn \iff T_n=T_m+mn$. I've no more progress.

Any help will be appreciated.

This generalization is also much welcome:

$kT_n=T_{n+m}$, with $k\geq2$ is a natural number.

 A: Here are some ways of making progress
If we simply use $2T_n=T_m$ we find that $$n(n+1)=\frac {m(m+1)}2$$ and if we multiply by $8$ this becomes $$2(4n^2+4n+1)=4m^2+4m+1+1$$ or, with $N=2n+1, M=2m+1$ $$2N^2=M^2+1$$ which is a classic Pell's equation - $N=5, M=7$ will do, for example.
With $kT_n=T_m$ we get $$k(4n^2+4n+1)=(4m^2+4m+1)+(k-1)$$ or $$kN^2=M^2+(k-1)$$
Note that there will always be multiples of any triangle number, because $n=2T_m$ gives $T_n=T_m(2T_m+1)$ and $T_{n-1}=T_m(2T_m-1)$
A: We have, $$2T_n = T_{n+m} \Rightarrow 2\frac{n(n+1)}{2} = \frac{(n+m)(n+m+1)}{2}$$ $$\Rightarrow 2n(n+1) = (n+m)(n+m+1)$$ $$\Rightarrow 2n^2 + 2n = n^2 + 2nm + m^2 + m + n$$ $$\Rightarrow m^2 + m+2nm-n^2-n=0$$ That gives us, $$m=\frac{-(1+2n)\pm \sqrt{(1+2n)^2-4(1)(-n^2-n)}}{2} = \frac{-(1+2n)\pm \sqrt{4n^2+4n+1+4n^2+4n}}{2} =\frac{-(1+2n)\pm \sqrt{8n^2 +8n+1}}{2}$$ We have to find the condition for which $8n^2+8n+1$ is a perfect square. Those values of $n$ satisfying this condition are given in the sequence $A029549$ in the OEIS.   

Similarly, the sequence $A076140$ in the OEIS gives a list of those numbers which are $3$ times another triangular number. You can also see here for more information. Hope it helps.
