The only numbers that appears in both of these lists are $$36, 6^2 = 36, frac8*92 = 36, 1, 1^2=1, frac1*22=1$$
Are there any type of past this?
Well, $2x^2=y^2+y$. How carry out I go on from this? What execute I need to use?
What are the numbers wbelow the number you substitute in are equal, and their results 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, $x = 0.5pm 0.5$
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 at 15:52
VortexYTVortexYT
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From your equation you get$$8x^2=4y^2+4y$$or$$8x^2+1=(2y+1)^2.$$You have to solve$$z^2-8x^2=1$$in positive integers ($z$ will certainly automatically be odd so $y=frac12(z-1)$will certainly be an integer). This is a type of Pell"s equation. Its solutionsare $(x_n,z_n)$ where$$z_n+2sqrt 2 x_n=(3+2sqrt 2)^n.$$So $x_1=1$, $z_1=3$ providing $1$ as square-triangular.Then $x_2=6$, $z_2=17$ offering $36$ as square-triangular.Then $x_3=35$, $z_3=99$ giving $1225$ as square-triangular, and so on.
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edited May 6 "17 at 16:35
Will Jagy
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answered May 5 "17 at 16:00
Angina SengAngina Seng
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Following Shark, below is just how to settle the Pell equation $z^2 - 8 x^2 = 1$ by hand, although one deserve to easily guess the initially as $9-8=1:$
Method described by Prof. Lubin at 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 ongoing fractivity tableau: $$ eginarraycccccccccc & & 2 & & 1 & & 4 & \ \ frac 0 1 & frac 1 0 & & frac 2 1 & & frac 3 1 \ \ & 1 & & -4 & & 1 endarray $$
$$ eginarraycccc frac 1 0 & 1^2 - 8 cdot 0^2 = 1 & mboxdigit & 2 \ frac 2 1 & 2^2 - 8 cdot 1^2 = -4 & mboxdigit & 1 \ frac 3 1 & 3^2 - 8 cdot 1^2 = 1 & mboxdigit & 4 \ endarray $$
Anyway, provided a solution $(z,x)$ in positive integers to $z^2 - 8 x^2 = 1,$ we get 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 coordinates $z_n, x_n$ obey$$ z_n+2 = 6 z_n+1 - z_n, $$$$ x_n+2 = 6 x_n+1 - x_n. $$