The just numbers that shows up in both of this lists are $$36, 6^2 = 36, \frac8*92 = 36, 1, 1^2=1, \frac1*22=1$$

Are there any type of beyond this?

Well, $2x^2=y^2+y$. Just how do i go on native this? What carry out I need to use?

What space the numbers whereby the number you instead of in room equal, and also their outcomes are equal?

Lets say:

What if $x=y$?

$$x^2=\fracx(x+1)2$$

$2x^2=x(x+1)$

$2x^2=x^2+x$

$x^2-x=0$

$x(x-1)=0$

$x$ 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?




You are watching: Read the numbers and decide what the next number should be. 256 196 144 100 64 36 16

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asked may 5 "17 in ~ 15:52
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VortexYTVortexYT
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From her equation girlfriend get$$8x^2=4y^2+4y$$or$$8x^2+1=(2y+1)^2.$$You need to solve$$z^2-8x^2=1$$in optimistic integers ($z$ will instantly be odd therefore $y=\frac12(z-1)$will it is in an integer). This is a kind of Pell"s equation. That solutionsare $(x_n,z_n)$ where$$z_n+2\sqrt 2 x_n=(3+2\sqrt 2)^n.$$So $x_1=1$, $z_1=3$ providing $1$ as square-triangular.Then $x_2=6$, $z_2=17$ offering $36$ together square-triangular.Then $x_3=35$, $z_3=99$ providing $1225$ as square-triangular, etc.


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will certainly Jagy
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answered may 5 "17 at 16:00
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Following Shark, below is just how to resolve the Pell equation $z^2 - 8 x^2 = 1$ by hand, back one can quickly guess the first as $9-8=1:$

Method described by Prof. Lubin in ~ Continued fraction of $\sqrt67 - 4$

$$ \sqrt 8 = 2 + \frac \sqrt 8 - 2 1 $$ $$ \frac 1 \sqrt 8 - 2 = \frac \sqrt 8 + 2 4 = 1 + \frac \sqrt 8 - 2 4 $$ $$ \frac 4 \sqrt 8 - 2 = \frac \sqrt 8 + 2 1 = 4 + \frac \sqrt 8 - 2 1 $$

Simple continued fraction tableau: $$ \beginarraycccccccccc & & 2 & & 1 & & 4 & \\ \\ \frac 0 1 & \frac 1 0 & & \frac 2 1 & & \frac 3 1 \\ \\ & 1 & & -4 & & 1 \endarray $$

$$ \beginarraycccc \frac 1 0 & 1^2 - 8 \cdot 0^2 = 1 & \mboxdigit & 2 \\ \frac 2 1 & 2^2 - 8 \cdot 1^2 = -4 & \mboxdigit & 1 \\ \frac 3 1 & 3^2 - 8 \cdot 1^2 = 1 & \mboxdigit & 4 \\ \endarray $$

Anyway, given a solution $(z,x)$ in optimistic integers come $z^2 - 8 x^2 = 1,$ we acquire the next in an boundless 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 works with $z_n, x_n$ obey$$ z_n+2 = 6 z_n+1 - z_n, $$$$ x_n+2 = 6 x_n+1 - x_n. $$