Ramsey2879

Edited to add:

 

The first question is a bit ambiguous. Is the problem to find the number of bit strings of length n that contain exactly 4 consecutive zeros, or to find the number of bit strings of length n that contain at least 4 consecutive zeros? The answers are obviously quite different.

I think it would be OK to work this out since this post has aged a bit and doing a someone's "homework" doesn't seem to apply anymore. Lets assume the string consists of only 0's and 1's ; that there must be at least a leading 1, and must contain at least one string of at least 4 consecutive zeros. Question: is this what is wanted-The number of strings of length N of this type?

My proof extends to still a more general case that includes Pythagorean Triples

Theorem

For [math]A,B,x,y,m \in Z | (A,B = Constants)[/math]

Let [math] N = F_{x,y} = x^{2} + Bxy -y^{2}[/math] where gcd(x,y) = 1

Then there exists a coprime pair [math](x_{1},y_{1})[/math] such that

[math]F_{x_{1},y_{1}} = N^{2^{m}}[/math]

I have a recursive formula that gives [math](x_{1},y_{1})[/math]

It even gives values if [math] N = F_{x,y} = Ax^{2} + Bxy -y^{2}[/math] e.g. [math](x_{1},y_{1})[/math] such that [math]F = N^{2}[/math]

If A is a perfect square and B = 0 then my formula relates to Pythagorean triples

e.g.

[math]2^{2} - 1^{2} = 3[/math]

[math]5^{2} - 4^{2} = 3^{2}[/math]

[math]41^{2}-40^{2}=3^{4}[/math]

[math]3281^{2}-3280^{2}=3^{8}[/math]

[math]3^{2}-2^{2}=5[/math]

[math]13^{2}-12^{2}=5^{2}[/math]

[math]313^{2}-312^{2}=5^{4}[/math]

[math]195313^{2}-195312^{2}=5^{8}[/math]

You want to treat part 3d. as if there were absolute value bars outside of both sums being multiplied. So you would have | (x-1)*(x+2) | = 3. Now you can just multiply the two terms and get |x^2 +2x -x -2 |=3, now just add like-terms and solve.

You can also solve if you substitute a+b for x and set a-1 = -(a+2). This gives a = -1/2. So (-3/2 + b)*(3/2+b) = +/- 3 gives b = +/-sqrt{21/4} since x is real. x = a+b = (-1+/-sqrt{21})/2

i have an assignment question where i'm supposed to prove that the square of an odd number is an odd number using both a direct proof and an indirect proof.

 

the direct proof was easy enough, but i'm kinda stuck on the indirect part.

 

so basically i'm supposed to prove that "if n^2 is even then n is even."

 

so i started by stating that .

 

n^2 = 2k

n * n = 2k

n = 2 (k/n) and i'm saying that (k/n) is always an integer

 

but something doesnt seem right about this, especially since i'm not really sure how to justify that (k/n) is an integer.

 

can someone put me on the right path?

 

thank you

I think you need to use the fundamental theorem of arithmetic. i.e. that there is one and only one way to express an integer greater than 1 as a product of primes. (a difference in the order in which the primes are multiplied does not count as a different way). Thus 30 = 2*3*5 = 3*2*5= 5*2*3 etc. There is one 2, one 3, and one 5 in each expression of 30 as a product of primes so that counts as only one way. What does this say about n*n?

Actually my problem is to find my proof that there are an infinite number of solutions to the Diophantine equation 5a^2 + 5ab + b^2 = p^n where a,b are coprime and p is a prime ending in 1 or 9. There is a relationship between any three consecutive term of a Fibonacci type series and the form 5a^2 + 5ab + b^2 that is invariant with the index number of the first term. That is a key to my proof. I leave the proof for you to figure out, but will make suggestions if you reach a dead end and have no idea where to turn.

First off then, how does the form 5a^2 + 5ab + b^2 relate to three terms of a Fibonacci type series in an invariant manner? (By Fibonacci type, I mean F(n) = F(n-1) + F(n-2) )