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

square-numbers
Share
Cite
Follow
asked May 5 "17 at 15:52 VortexYTVortexYT
\$endgroup\$
1

4
\$egingroup\$
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.

Share
Cite
Follow
edited May 6 "17 at 16:35 Will Jagy
answered May 5 "17 at 16:00 Angina SengAngina Seng
\$endgroup\$
1
1

See more: How Many Glasses Of Champagne Per Bottle, Champagne Toast Calculator And Event Drinks Guide

\$egingroup\$
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. \$\$