# Thread: Triangular \ Square number

1. ## Triangular \ Square number

Let Tm, Tn be triangular numbers.

I need to show that (2m+1)^2*Tn + Tm is also triangular

Now,,,,I have tried time and time again to expand the number and then somehow factorise it to make another triangular number , but with no success.

(1) (2m+1)^2 = 4m^2 + 4m + 1
(2) Tn = n(n+1) / 2
(3) Tm = m(m+1) / 2

But trying to multiply (1)*(2) and then add the term for (3) just ends up with a very long result, which I can't factorise.

Any ideas please ??? Thanks a lot if anyone can explain it to me

ps : (2m+1)^2 is of course a square number, S(2m+1), but am not aware of any property for the product of a square number with a triangular number....unless I am missing something??

2. Originally Posted by yakirtal76

Let Tm, Tn be triangular numbers.

I need to show that (2m+1)^2*Tn + Tm is also triangular

Now,,,,I have tried time and time again to expand the number and then somehow factorise it to make another triangular number , but with no success.

(1) (2m+1)^2 = 4m^2 + 4m + 1
(2) Tn = n(n+1) / 2
(3) Tm = m(m+1) / 2

But trying to multiply (1)*(2) and then add the term for (3) just ends up with a very long result, which I can't factorise.

Any ideas please ??? Thanks a lot if anyone can explain it to me

ps : (2m+1)^2 is of course a square number, S(2m+1), but am not aware of any property for the product of a square number with a triangular number....unless I am missing something??
I haven't worked it out all the way, but I'm getting a bit of success using an induction type proof, starting with the m's. I didn't finish it by working on the n's, but it shouldn't be too hard to complete. Maybe not, but it's a thought.

-Dan

3. We have $(2m+1)^2 T_n + T_m = (2m+1)^2.n(n+1)/2 + m(m+1)/2$. If this is a triangular number $T_x$ then $x(x+1) = (2m+1)^2.n(n+1) + m(m+1)$. The equation $x(x+1) = F$ is quadratic in $x$ with solution $\frac{-1 \pm \sqrt{1+4F}}{2}$. In this case $F = (2m+1)^2.n(n+1) + m(m+1)$ and $4F+1 = (2m+1)^2(2n+1)^2$. We find that $x = \frac{(2m+1)(2n+1)-1}{2}$.