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Nice program and thanks!

n=1; cycle:a = (n+1)^2 / (n+23);c=if(ceil(a)==floor(a), goto3, goto4);printn,"=",a; n++; if(n<1000, goto cycle, discard=0;
The four numbers are:
n =21, 98, 219, 461

If

\(z^2 + z + 1 = 0\),

find \(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\).

\(\begin{array}{|rcll|} \hline \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z+2z^2 +2z^3 \\ \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z(z^2 + z + 1) \quad | \quad z^2 + z + 1 = 0 \\ \mathbf{0} &\mathbf{=} & \mathbf{1+z^2+z^4} \quad | \quad 0 = z^2 + z + 1 \\ z^2 + z + 1 & = & 1+z^2+z^4 \\ z & = & z^4 \quad | \quad : z \\ 1 & = & z^3 \\ \mathbf{z^3} &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline z^4 = z^3z= 1\cdot z &=& z \\ z^5 = z^3z^2= 1\cdot z^2 &=& z^2 \\ z^6 = \left(z^3\right)^2= 1^2 &=& 1 \\ \hline \end{array}\)

In general:

\(\begin{array}{|lcll|} \hline z^{0+3n} = 1 \\ z^{1+3n} = z \\ z^{2+3n} = z^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} = z^{1+3\cdot16} &=& z \\ z^{50} = z^{2+3\cdot16} &=& z^2 \\ z^{51} = z^{0+3\cdot17} &=& 1 \\ z^{52} = z^{1+3\cdot17} &=& z \\ z^{53} = z^{2+3\cdot17} &=& z^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} + z^{50} + z^{51} + z^{52} + z^{53} &=& z+z^2+1+z+z^2 \quad | \quad z^2 + z + 1 = 0 \\ z^{49} + z^{50} + z^{51} + z^{52} + z^{53} &=& z+z^2 \quad | \quad z^2 + z = -1 \\ \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)

Let the height of the Mesa =x

Let the distance to the base of the Mesa = y

x / y = tan(29),,,,,,,,,,,,,,,,,,,(1)

x / (y+100) = tan(25)........(2)

Now, can you solve the above 2 simultaneous equations? Try it.

pi cannot be written exactly, it is an irrational number. 

EP has given you the approximation.  

Hi guest, 

Should I interpret that as 'thanks'   ?

Thanks Chris :)

Your answers are the same as mine.

I think you are just having fun using your C++ programing    

Thanks Chris   

​complex roots

\(\begin{array}{|rcll|} \hline w &=& p+2i \\ w^{*} &=& p-2i \\\\ \left[ z-w \right] \left[z-w^{*} \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ z-(p+2i) \right] \left[z-(p-2i) \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ (z-p)-2i) \right] \left[(z-p)+2i) \right] &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2-(2i)^2 &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2+4 &=& z^2+(4+i+qi)z + 20 \\ z^2-2pz+(p^2+4) &=& z^2+\underbrace{(4+i+qi)}_{=-2p}z + \underbrace{20}_{=p^2+4} \\\\ p^2+4 &=& 20 \\ p^2 &=& 16 \\ \mathbf{p} & \mathbf{=} & \mathbf{\pm 4} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline p=4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& -8 \\ & qi &=& -12-i \\ & q &=& \dfrac{-12-i}{i} \\ & q &=& \dfrac{(-12-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(-12-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(-12-i)i}{-1} \\ & q &=& (12+i)i \\ & q &=& i^2+12i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1+12i} \\\\ & w &=& 4+2i \\ & w^* &=& 4-2i \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline p=-4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& 8 \\ & qi &=& 4-i \\ & q &=& \dfrac{4-i}{i} \\ & q &=& \dfrac{(4-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(4-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(4-i)i}{-1} \\ & q &=& (i-4)i \\ & q &=& i^2-4i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1-4i} \\\\ & w &=& -4+2i \\ & w^* &=& -4-2i \\ \hline \end{array}\)

The possible values of \(\mathbf{q}\) are \(\mathbf{-1+12i}\) and \(\mathbf{-1-4i}\)

and the quadratic equations are \(\mathbf{z^2-8z+20 = 0}\) and \(\mathbf{z^2+8z+20 = 0}\) .

\(|z^3|^2 = ||z|^2|^3 = 74^2 + c^2 = (x^2+y^2)^3\\ \text{Using software I get }x=1,~y=5,~c=-110\\ z = 1 +5i\)

what would be the 2nd part of the fraction of 

11.3 and 16.7

\(|z|=1 \Rightarrow z = \cos(x)+i\sin(x)\\ z^2 + z -1 = -\sin ^2(x)+\cos ^2(x)+\cos (x)-1 +i (\sin (x)+2 \sin (x) \cos (x))\\ |z^2 + z - 1|^2 =3-2 \cos (2 x) \text{ (after much simplification) }\\ \dfrac{d}{dx} |z^2 + z - 1|^2 = 4\sin(2x)=0 \Rightarrow x = \dfrac{(2k+1)\pi}{2},~k \in \mathbb{Z}\\ \text{Thus }z = \pm i \text{ and the max value of } |z^2 + z - 1|^2 = 5 \text{ so the max value of }\\ |z^2 + z - 1| = \sqrt{5}\)

a)

\(f(n) = 2^{4n+3}+3^{3n+1} \\ f(k+1)-16f(k) = \\ 2^{4(k+1)+3}+3^{3(k+1)+1)}-16\left( 2^{4k+3}+3^{3k+1}\right)\)

\(2^{4k+7}+3^{3k+4} - 16\cdot 2^{4k+3}-16\cdot 3^{3k+1}=\\ (16-16)2^{4k+3}+(27-16)3^{3k+1}=11 \cdot 3^{3k+1} =\\ 33 \cdot 3^{3k}\)

b)

\(P_1 = f(1) \pmod{11}=0\\ f(1) = 2^3 + 3^1 = 11\\ P_1=True\\ \text{Show }P_n \Rightarrow P_{n+1}\)

\(\text{Assume }P_n=True\\ f(n+1) = 2^{4(n+1)+3}+3^{3(n+1)+1} = \\ 16\cdot 2^{4n+3}+27\cdot 3^{3n+1}= \\ 16(2^{4n+3}+3^{3n+1}) + 11\cdot 3^{3n+1} = \\ 16f(n) + 11 \cdot 3^{3n+1} \)

\(f(n)\pmod{11} = 0 \text{ by assumption}\\ 11 \cdot 3^{3n+1} \pmod{11} =0 \text{ should be clear}\\ P_n \Rightarrow P_{n+1} \)

\(\text{I assume that }\overline{a}=Re(a) - i Im(a)\\ a+\overline{b}=2-7i\\ \overline{a+\overline{b}}=\overline{2-7i}\\ \overline{a}+b= 2+7i\)

\(f(x)=a x^3 + b x^2 + c x + d\\ f(0)=0 \Rightarrow d=0\)

\(\text{see if you can figure out how I use the rest of the conditions to derive the following}\\ \begin{pmatrix}1 &1 &1\\8 &4 &2 \\ 27 &9 &3\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}1\\2\\4\end{pmatrix}\)

Solving this system using Gaussian eliminations obtains

\((a,b,c) =\Large \left(\frac{1}{6},-\frac{1}{2},\frac{4}{3}\right)\\ f(5) = (a.b.c)\cdot (125,25,5) = 15\)

Thank you, CPhill !

pi equals the circumference of a circle divided by its diameter .......3.1415926535897932.........

Your last equation should be     y = 3/5 x - 2/5        Now find the y intercept     when x = 0    y =?

                                                                                    then find the x intercept    when y = 0   x = ?

When you find these two points ....draw a line through them.....

a_n  =  17-5n

a_n+1 = a_n - 5  ,   a₁ = 12

There are actually quite a few more. This is the procedure used to count them:
1- First, you have to count the number of primes between 2 and 10,000/2, or 5,000, of which there are 669 such prime numbers, since the largest of them,4,999 can be multiplied by the smallest of them, or 2, which will give,9,998 which is the largest such number under 10,000. Then 2 will be multiplied by every prime from 3 up to 4,999.

2- In addition to squaring all 25 primes under 100, then as indicated in (1) above, will continue thus: 2x3, 2x5, 2x7.......etc. till 2 x 4,987, 2 x 4,993, 2 x 4,999. Once we are done with 2, then will continue with 3: 3 x 5, 3 x 7, 3 x 11.....etc. till 3 x 3,323, 3 x 3,329, 3 x 3,331. Then will continue with 5: 5 x 7, 5 x 11, 5 x 13.....etc. till: 5 x 1,993, 5 x 1,997, 5 x 1999 and so on....

3-The above process will continue until we reach the two primes under sqrt(10,000), or thereabout. Actually 97 x 103 =9,991.

4 - The above process can be done accurately by a computer program written specifically for this purpose. Since I have a good knowledge of 2 programming languages, I wrote one just for this specific task.

5 - The final result looks like this:
(4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87......

........ 5006, 5007, 5017, 5027, 5029, 5033, 5041, 5042, 5045, 5053, 5057, 5062, 5063, 5065, 5069, 5071...........

........ 9974, 9977, 9979, 9983, 9985, 9986, 9987, 9989, 9991, 9993, 9995, 9997, 9998.

6 - So, the total number =2,625. Every one of these numbers has ONLY two prime factors by delibrate calculation!.

7 - And that is the END.

 We break this up into cases. To create 6, with three numbers, we can use: 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and. 2-2-2.

6-0-0: Only one way

5-1-0: Take the ball that is alone, so 6 ways.

4-2-0: 6C2=15 ways for the two balls in their own box.

4-1-1: Again 15 ways, pick two of the six balls, but they cannot go in the box of 4.

3-3-0: 5C2=10 ways, by selecting two of the remaining five balls.

3-2-1: 6*10=60 ways by picking any of the six balls to be by itself.

2-2-2: 15 ways? First, you think they are distinguishable, yet you overcount by a factor 3!=6.

Thus, adding them all up, we get \(1+6+15+15+10+60+15=\boxed{122}.\)

See this link: https://web2.0calc.com/questions/some-questions-please-elaborate-thanks#r1

Also isn't it r=-8 in part b?

Very nice "24".....!!!!