A number that is a triangle number and also a square number. This is a problem from standupmaths, from Youtube. 
Question:

Think of a number which is a triangle number and a square number.

This can be expressed using a single equation:
$$x^2=\frac{y(y+1)}{2}$$
If you use trial and improvement, you have the square numbers:
$[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289...]$
And triangle numbers:
$[1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210...]$
The only numbers that appears in both of these lists are $$36, 6^2 = 36, \frac{8*9}{2} = 36, 1, 1^2=1, \frac{1*2}{2}=1$$
Are there any beyond this?
Well, $2x^2=y^2+y$. How do I go on from this? What do I have to use?


What are the numbers where the number you substitute in are equal, and their results are equal?

Lets say: 

What if $x=y$?

$$x^2=\frac{x(x+1)}{2}$$
$2x^2=x(x+1)$
$2x^2=x^2+x$
$x^2-x=0$
$x(x-1)=0$
$x$ or $(x-1)$ has to be $0$.
Therefore, $x = 0.5\pm 0.5$ 
Second question: 

But is $0$ technically a square or triangle number?

 A: From your equation you get
$$8x^2=4y^2+4y$$
or
$$8x^2+1=(2y+1)^2.$$
You need to solve
$$z^2-8x^2=1$$
in positive integers ($z$ will automatically be odd so $y=\frac12(z-1)$
will be an integer). This is a form of Pell's equation. Its solutions
are $(x_n,z_n)$ where
$$z_n+2\sqrt 2 x_n=(3+2\sqrt 2)^n.$$
So $x_1=1$, $z_1=3$ giving $1$ as square-triangular.
Then $x_2=6$, $z_2=17$ giving $36$ as square-triangular.
Then $x_3=35$, $z_3=99$ giving $1225$ as square-triangular, etc.
A: Following Shark, here is how to solve the Pell equation $z^2 - 8 x^2 = 1$ by hand, although one can easily guess the first as $9-8=1:$
Method described  by Prof. Lubin  at Continued fraction of $\sqrt{67} - 4$ 
$$  \sqrt { 8} = 2 +     \frac{  \sqrt {8} - 2 }{ 1 }  $$ 
 $$    \frac{ 1 }{   \sqrt {8} - 2 }  =  \frac{    \sqrt {8} + 2 }{4 } = 1 +  \frac{    \sqrt {8} - 2 }{4 } $$ 
 $$    \frac{ 4 }{   \sqrt {8} - 2 }  =  \frac{    \sqrt {8} + 2 }{1 } = 4 +  \frac{    \sqrt {8} - 2 }{1 } $$ 
Simple continued fraction tableau:
 $$ 
 \begin{array}{cccccccccc}
 & & 2 & & 1 & & 4 & \\ 
 \\ 
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 2 }{ 1 }   & &   \frac{ 3 }{ 1 }    \\ 
  \\ 
 & 1 & & -4 & & 1
 \end{array}
 $$ 
$$ 
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 8 \cdot 0^2 = 1 &     \mbox{digit}  &  2  \\  
  \frac{ 2 }{ 1 }   & 2^2 - 8 \cdot 1^2 = -4 &     \mbox{digit}  &  1  \\  
  \frac{ 3 }{ 1 }   & 3^2 - 8 \cdot 1^2 = 1 &     \mbox{digit}  &  4  \\  
 \end{array}
 $$ 
Anyway, given a solution $(z,x)$ in positive integers to $z^2 - 8 x^2 = 1,$ we get the next in an infinite sequence by
$$  (z,x) \mapsto (3z + 8x, z + 3x),$$ so
$$ ( 1,0 ), $$
$$ (3,1),  $$
$$ ( 17,6),$$
$$ (99 ,35 ), $$
 $$ ( 577, 204 ), $$
$$ (3363 , 1189 ), $$
By Cayley -Hamilton, the coordinates $z_n, x_n$ obey
$$ z_{n+2} = 6 z_{n+1} - z_n, $$
$$ x_{n+2} = 6 x_{n+1} - x_n. $$
