Are square numbers also known as rectangular numbers, too? I've learned that a square number can be multiplied by itself twice and be used as a base raised to the exponent of $2$, such as $3^2, 9^2, 0^2,$ etc.  Square numbers can also fit into square shapes.  Can these also be rectangular numbers?  I've learned that rectangular numbers can have that amount of dots arranged in a perfect rectangle with at least two rows and at least two columns such as $6, 10, 24, 30,$ etc.  So, a rectangular number can be a composite number most of the time (it can't be one, since it's neither prime nor composite, and it doesn't fit into a rectangular shape).  Anyway, I've learned that a square is a kind of rectangle, so I'm guessing the answer to my question is yes.  Also, this means composite numbers can also be square numbers, excluding $1$, of course.  So, I guess it's possible that a square number can also be a rectangular number.  This is a good idea to exercise your minds, so am I on the right track somehow?
 A: All square numbers are rectangular numbers but not all rectangular numbers are square numbers. Kind of like how all terriers are dogs but not all dogs are terriers.
Just to be sure we're on the same page, these are the definitions I'm going by:


*

*Given non-negative integers $m$ and $n$, not necessarily distinct,

*Rectangular numbers are numbers of the form $m \times n$, more commonly written $mn$, and

*Square numbers are of the form $n \times n$, more commonly written $n^2$.


If $m = n$, then $mn = m^2 = n^2$, a square number. For example, $7^2 = 49$ is a square number. But if $m \neq n$, then $mn$ might be rectangular but not square. For example, $2 \times 18$ is a rectangular number that also happens to be a square number, though you'd have to rearrange it as $6 \times 6$ to fit in the square. A number like $40$, on the other hand, is rectangular but not square, as you can arrange it $2 \times 20$, $4 \times 10$ or $5 \times 8$, but $\sqrt{40} \approx 6.32$.
A: A much more common term
is triangular number,
which is a number 
of the form
$T_n = \dfrac{n(n+1)}{2}$.
These are the number of points 
in a triangular array
with rows of
$1, 2, ..., n$ points
(since
$1+2+...+(n-1)+n
=\dfrac{n(n+1)}{2}
$).
If you put two triangles
with $T_n$ and $T_{n-1}$
points together,
you get a square of side $n$
(since $\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}=n^2$).
